Question 13 Marks
- A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.
- Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
Answer
- $\mu=\frac{\sin\big(\frac{A+\delta_m}{2}\big)}{\sin\big(\frac{A}{2}\big)}$
$=\frac{\sin\big(\frac{60+30}{2}\big)}{\sin\big(\frac{60^\circ}{2}\big)}=\sqrt{2}$
Also $\mu=\frac{c}{v}\Rightarrow v=\frac{3\times10^8}{\sqrt{2}}\text{m/s}$
$=2.122\times10^8\text{m/s}$
-
At face AC, let the angle of incidence be $r_2$. For grazing ray, $e = 90^\circ$
$\Longrightarrow\mu=\frac{1}{sin r_2}\Longrightarrow r_2=\sin^{-1}\big(\frac{1}{\sqrt{2}}\big)=45^\circ$
Let angle of refraction at face AB be $r_1$. Now $r_1+r_2=A$
$\therefore r_1=A-r_2=60^\circ-45^\circ=15^\circ$
Let angle of incidence at this face be $i$
$\mu=\frac{\sin i}{\sin r_1}$
$\Longrightarrow\sqrt{2}=\frac{\sin i}{\sin 15^\circ}$
$\therefore i=\sin^{-1}(\sqrt{2}.\sin 15^\circ)$ View full question & answer→Question 23 Marks
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
AnswerFocal length of the objective lens, $f_0 = 140\ cm$
Focal length of the eyepiece, $f_e = 5\ cm$
Height of the tower,$ h_1= 100\ m$
Distance of the tower (object) from the telescope, u= 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
$\theta=\frac{\text{h}_1}{\text{u}}$
$=\frac{100}{3000}=\frac{1}{30} \ \text{rad}$
The angle subtended by the image produced by the objective lens is given as:
$\theta=\frac{\text{h}_2}{\text{f}_\text{o}}=\frac{\text{h}_2}{140}$ rad
Where,
$h_2=$ Height of the image of the tower formed by the objective lens
$\frac{1}{30}=\frac{\text{h}_2}{140}$
$\therefore \ \text{h}_2=\frac{140}{30}=4.7 \ \text{cm}$
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
View full question & answer→Question 33 Marks
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25\ mm^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
AnswerArea of the virtual image of each square, $A= 6.25\ mm^2$
Area of each square, $A_0= 1\ mm^2$
Hence, the linear magnification of the object can be calculated as:
$\text{m}=\sqrt{\frac{\text{A}}{\text{A}_0}}$
$\text{m}=\sqrt{\frac{6.25}{1}}=2.5$
But, $\text{m}=\frac{\text{imagr dis tan ce(v)}}{\text{Object dis tan ce(u)}}$
$\therefore \ \text{v}=\text{mu}$
$= 2.5\ u ...(1)$
Focal length of the magnifying glass, f= 10 cm
According to the lens formula, we have the relation:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{10}=\frac{1}{2.5\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{u}}\Big(\frac{1}{2.5}-\frac{1}{1}\Big)=\frac{1}{\text{u}}\Big(\frac{1-2.5}{2.5}\Big)$
$\therefore \ \text{u}=\frac{1.5\times10}{2.5}=-6 \ \text{cm}$
and v = 2.5 u
= 2.5 × 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
View full question & answer→Question 43 Marks
- At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
- What is the magnification in this case?What is the magnification in this case?
- Is the magnification equal to the magnifying power in this case? Explain.
AnswerMaximum magnifying power is obtained when the image is at the near point (25 cm).
Thus, Image distance, v = -25 cm
Focal length, f = +10 cm
Object distance, u = ?
Using the formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\therefore \ \frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$=\frac{1}{-25}-\frac{1}{10}$
$=\frac{-2-5}{50}$
$=\frac{-7}{50}$
i.e., $\text{u}=-\frac{50}{7}=-7.14 \ \text{cm}$.
So the lens should be held at a distance of 7.14 cm away so as to have a maximum possible magnifying power.
View full question & answer→Question 53 Marks
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when,
- the telescope is in normal adjustment (i.e., when the final image is at infinity)?
- the final image is formed at the least distance of distinct vision (25 cm)?
AnswerFocal length of the objective lens, $f_0 = 140\ cm$
Focal length of the eyepiece, $f_e = 5\ cm$
Least distance of distinct vision, $d = 25\ cm$
- When the telescope is in normal adjustment, its magnifying power is given as:
When the final image is formed at d, the
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{140}{5}=28$
- magnifying power of the telescope is given as:
$\frac{\text{f}_0}{\text{f}_\text{e}}\Big[1+\frac{\text{f}_\text{e}}{\text{d}}\Big]$
$=\frac{140}{5}\Big[1+\frac{5}{25}\Big]$
$= 28[1 + 0.2]$
$= 28 × 1.2 = 33.6$ View full question & answer→Question 63 Marks
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
AnswerFor a normal eye, near point of distinct vision = 25 cm.
i.e., u = -25 cm
Converging power = +40 D
To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres. This gives the rough idea of the distance between the retina and cornea eye-lens.
$\therefore$ to focus an object at near point,
Focal length of eye - lenc $=\frac{100}{\text{P}}=\frac{100}{60}=\frac{10}{6}=\frac{5}{3} \ \text{cm}$
$\text{v}=-\frac{5}{3} \ \text{cm}$
Now, using the formula,
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$=\frac{3}{5}+\frac{1}{25}=\frac{16}{25}$
i.e., $\text{f}=\frac{25}{16}$, corresponding to a converging power given by,
$\text{P}=\frac{\frac{100}{25}}{16}=64$ dipotre.
Hence, we can say that the range of accommodation of the eye-lens is roughly 20 to 24 dioptre.
View full question & answer→Question 73 Marks
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
AnswerSize of the object, $h_1 = 3\ cm$
Object distance, u = -14 cm
Focal length of the concave lens, f = -21 cm
Image distance = v
According to the lens formula, we have the relation:
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=-\frac{1}{21}-\frac{1}{14}=\frac{-2-3}{42}=\frac{-5}{42}$
$\therefore \ \text{v}=-\frac{42}{5}=-8.4 \ \text{cm}$
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
$\text{m}=\frac{\text{Im age height(h_2)}}{\text{Object height(h_1)}}=\frac{\text{v}}{\text{u}}$
$\therefore \ \text{h}_2=\frac{-8.4}{-14}\times3=0.6\times3=1.8 \ \text{cm}$
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.
View full question & answer→Question 83 Marks
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
- deviate a pencil of white light without much dispersion.
- disperse (and displace) a pencil of white light without much deviation.
Answer
- Angular dispersion produced by two prisms should be zero so that, no dispersion take place.
The sum of angular dispersion by crown glass prism and angular dispersion by flint glass prism is equal to zero.
i.e., $(\mu_\text{b}-\mu_\text{r})\text{A}+(\mu_\text{b}'-\mu\text{r}')\text{A}'=0.$
Since $(\mu_\text{b}'-\mu_\text{r}')$ for flint glass is more than that for crown glass, therefore,
A' < A
Hence, the combination of prism is such that the, flint glass prism of smaller angle has to be suitably combined with crown glass prism of larger angle.
- For almost no deviation,
$(\mu_\text{y}-1)\text{A}'+(\mu_\text{y}'-1)\text{A}'=0.$
To disperse without deviation, we need flint glass prism of greater so that the deviation due to two prisms are equal and opposite. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism because the refractive index is more for flint glass as compared to the refractive index of crown glass. View full question & answer→Question 93 Marks
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
AnswerSize of the candle, h= 2.5 cm
Image size = h'
Object distance, u= - 27 cm
Radius of curvature of the concave mirror, R= - 36 cm
$\therefore \ \text{h}'=-\frac{\text{v}}{\text{u}}\times\text{h}$
Focal length of the concave mirror, $=-\Big(\frac{-54}{27}\Big)\times2.5=-5 \ \text{cm}$
Image distance = v
The image distance can be obtained using the mirror formula:
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$-\frac{1}{-18}-\frac{1}{-27}=\frac{-3+2}{54}=-\frac{1}{54}$
$\therefore \ \text{v}=-54 \ \text{cm}$
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image. The magnification of the image is given as:
$\text{m}=\frac{\text{h}'}{\text{h}}=-\frac{\text{v}}{\text{u}}$
$\therefore \ \text{h}'=-\frac{\text{v}}{\text{u}}\times\text{h}$
$=-\Big(\frac{-54}{27}\Big)\times2.5=-5 \ \text{cm}$
The height of the candle's image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
View full question & answer→Question 103 Marks
- A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
- If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^6m,$ and the radius of lunar orbit is $3.8 \times 10^8m.$
AnswerFocal length of the objective lens,$ f_0 = 15 m = 15 \times 10^2cm$
Focal length of the eyepiece, $f_e = 1.0 cm$
- The angular magnification of a telescope is given as:
$\alpha=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{15\times10^2}{1.0}=1500$
Hence, the angular magnification of the given refracting telescope is 1500.
- Diameter of the moon, $d = 3.48 \times 10^6m$
Radius of the lunar orbit, $r_0 = 3.8 \times 10^8 m$
Let d' be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
$\frac{\text{d}}{\text{r}_0}=\frac{\text{d}'}{\text{d}_0}$
$\frac{3.48\times10^6}{3.8\times10^8}=\frac{\text{d}'}{15}$
$\therefore \ \text{d}'=\frac{3.48}{3.8}\times10^{-2}\times15$
$= 13.74 \times 10^{-2} m = 13.74 cm$
Hence, the diameter of the moon's image formed by the objective lens is 13.74 cm. View full question & answer→Question 113 Marks
- A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
- If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^6 \mathrm{~m}$, and the radius of lunar orbit is $3.8 \times 10^8 \mathrm{~m}$.
AnswerFocal length of the objective lens, $\mathrm{f}_0=15 \mathrm{~m}=15 \times 10^2 \mathrm{~cm}$
Focal length of the eyepiece, $\mathrm{f}_{\mathrm{e}}=1.0 \mathrm{~cm}$
- The angular magnification of a telescope is given as:
$\alpha=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{15\times10^2}{1.0}=1500$
Hence, the angular magnification of the given refracting telescope is 1500.
- Diameter of the moon, $d = 3.48 \times 10^6m$
Radius of the lunar orbit, $r_0 = 3.8 \times 10^8 m$
Let d' be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
$\frac{\text{d}}{\text{r}_0}=\frac{\text{d}'}{\text{d}_0}$
$\frac{3.48\times10^6}{3.8\times10^8}=\frac{\text{d}'}{15}$
$\therefore \ \text{d}'=\frac{3.48}{3.8}\times10^{-2}\times15$
$= 13.74 \times 10^{-2} m = 13.74 cm$
Hence, the diameter of the moon's image formed by the objective lens is 13.74 cm. View full question & answer→Question 123 Marks
A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
AnswerPower of spectacles, P = -1 D
$\therefore$ Focal length, f = -100 cm
That is, the far point of the person is at 100 cm.
Near point of the eye might have been normal (i.e., 25 cm).
The objects at infinity produce virtual images at 100 cm (using spectacles).
To see objects between 25 cm to 100 cm, the person uses the ability of accommodation of his eye-lens. This ability is partially lost in old age.
The near point of the eye may recede to 50 cm. He has, therefore, to use glasses of suitable power for reading.
Here,
Object distance, u = -25 cm
Image distance, v = -50 cm
Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}=-\frac{1}{50}+\frac{1}{25}$
i.e., focal length, f = 50 cm
power $\text{P}=\frac{100}{\text{f}}=\frac{100}{50}=+2$ dipotre.
View full question & answer→Question 133 Marks
Answer the following question:
Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
AnswerThe angular magnification produced by the eyepiece of a compound microscope is
$\Bigg[\Big(\frac{25}{\text{f}_0}\Big)+1\Bigg]$
Where, $f_e =$ Focal length of the eyepiece
It can be inferred that if $f_e$ is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as
$\frac{1}{(|\text{u}_0|\text{f}_0)}$
Where, $u_0 =$ Object distance for the objective lens
$f_0 =$ Focal length of the objective
The magnification is large when $u_0 > f_0$. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since $u_0$ is small, $f_0$ will be even smaller. Therefore, $f_0$ and $f_0$ are both small in the given condition.
View full question & answer→Question 143 Marks
Double - convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?
AnswerRefractive index of glass, µ = 1.55
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens $= R_1$
Radius of curvature of the other face of the lens $= R_2$
Radius of curvature of the double-convex lens = R
$\therefore \ \text{R}_1=\text{R} \ \text{and} \ \text{R}_2=-\text{R}$
The value of R can be calculated as:
$\frac{1}{\text{f}}=(\mu-1)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{20}=(1.55-1)\Big[\frac{1}{\text{R}}+\frac{1}{\text{R}}\Big]$
$\frac{1}{20}=0.55\times\frac{2}{\text{R}}$
$\therefore \ \text{R}=0.55\times2\times20=22 \ \text{cm}$
Hence, the radius of curvature of the double-convex lens is 22 cm.
View full question & answer→Question 153 Marks
Double - convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?
AnswerRefractive index of glass, $µ = 1.55$
Focal length of the double-convex lens, $f = 20\ cm$
Radius of curvature of one face of the lens $= R_1$
Radius of curvature of the other face of the lens $= R_2$
Radius of curvature of the double-convex lens = R
$\therefore \ \text{R}_1=\text{R} \ \text{and} \ \text{R}_2=-\text{R}$
The value of R can be calculated as:
$\frac{1}{\text{f}}=(\mu-1)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{20}=(1.55-1)\Big[\frac{1}{\text{R}}+\frac{1}{\text{R}}\Big]$
$\frac{1}{20}=0.55\times\frac{2}{\text{R}}$
$\therefore \ \text{R}=0.55\times2\times20=22 \ \text{cm}$
Hence, the radius of curvature of the double-convex lens is 22 cm.
View full question & answer→Question 163 Marks
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
AnswerActual depth of the needle in water, $h_1 = 12.5\ cm$
Apparent depth of the needle in water, $h_2 = 9.4\ cm$
Refractive index of water = µ
The value of µ can be obtained as follows:
$\mu=\frac{\text{h}_1}{\text{h}_2}$
$=\frac{12.5}{9.4}=1.33$
Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index, µ' = 1.63
The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation:
$\mu'=\frac{\text{h}_1}{\text{y}}$
$\therefore \ \text{y}'=\frac{\text{h}_1}{\mu'}$
$=\frac{12.5}{1.63}=7.67 \ \text{cm}$
Hence, the new apparent depth of the needle is 7.67 cm. It is less than $h_2$. Therefore, to focus the needle again, the microscope should be moved up.
$\therefore$ Distance by which the microscope should be moved up = 9.4 - 7.67
= 1.73 cm
View full question & answer→Question 173 Marks
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
AnswerFocal length of the objective lens, $f_0 = 144\ cm$
Focal length of the eyepiece, $f_e = 6.0\ cm$
The magnifying power of the telescope is given as:
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{144}{6}=24$
The separation between the objective lens and the eyepiece is calculated as:
$f_0 + f_e$
$= 144 + 6 = 150\ cm$
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
View full question & answer→Question 183 Marks
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
AnswerFocal length of the objective lens, $f_0 = 144\ cm$
Focal length of the eyepiece, $f_e = 6.0\ cm$
The magnifying power of the telescope is given as:
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{144}{6}=24$
The separation between the objective lens and the eyepiece is calculated as:
$f_0 + f_e$
$= 144 + 6 = 150\ cm$
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
View full question & answer→Question 193 Marks
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
AnswerIn the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
- Focal length of the convex lens, f = 20 cm
Image distance = v
According to the lens formula, we have the relation:
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{12}=\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}$
$\therefore \ \text{v}=\frac{60}{8}=7.5 \ \text{cm}$
Hence, the image is formed 7.5 cm away from the lens, toward its right.
- Focal length of the concave lens, f = –16 cm
Image distance = v
According to the lens formula, we have the relation:
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}$
$\therefore \ \text{v}=48\ \text{cm}$
Hence, the image is formed 48 cm away from the lens, toward its right. View full question & answer→Question 203 Marks
- Draw a ray diagram for the formation of image by a compound microscope.
- You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?
| Lenses |
Power (D) |
Aperture (cm) |
| $L_1$ |
3 |
8 |
| $L_2$ |
6 |
1 |
| $L_3$ |
10 |
1 |
- Define resolving power of a microscope and write one factor on which it depends.
Answer
- Ray Diagram for compound microscope:
-
- Objective: Lens $L_3$
- Eye Piece: Lens $L_2$
- $\text{R}_\text{p}=\frac{2{\mu}\sin \beta}{1.22\lambda}$
View full question & answer→Question 213 Marks
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm apart. A point object is placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw the ray diagram showing the image formation.
AnswerFor the lens:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
U= - 40 cm, f= +20 cm This gives v= + 40 cm.
This image acts as a (virtual) object for the convex mirror
$\therefore u= (+40-15)\text{cm}=25\text{ cm}$
$ \text{Also f} =+\frac{20}{2}\text{cm}=+10\text{ cm}$
From
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
We get
$v=\frac{50}{3}\text{cm}-\cong16.67\text{cm}$
The final image is, therefore formed at a distance of 16.67 cm $ (\frac{50}{3}\text{cm})$ to the right of the convex mirror.
(at a distance of 31.67 cm $(\frac{95}{3}\text{cm})$ to the right of the convex lens.)
View full question & answer→Question 223 Marks
- Draw a ray diagram showing the formation of image by a reflecting telescope.
- Write two advantages of a reflecting telescope over a refracting telescope.
Answer
-
Ray Diagram.
Arrow marking.
Labelling.
- Advantages:
- Spherical aberration is absent.
- Chromatic aberration is absent.
- Mounting is easier.
- Polishing is done on only one side.
- Light gathering power is more.
View full question & answer→Question 233 Marks
A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
AnswerFor the lens $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ As u is infinity, v = 20cm
For the concave mirror, the image, formed by the lens, acts as the object.
Hence, u = - (50 – 20)cm = -30cm
Also, f = -10cm $\frac{1}{\text{f}} = \frac{1}{\text{v}} + \frac{1}{\text{u}}$ $-\frac{1}{\text{10}} = \frac{1}{\text{v}} - \frac{1}{\text{30}}$$\therefore$ v = -15cm
The final image is, therefore, at a distance of 15cm to the left of the concave mirror.
View full question & answer→Question 243 Marks
- Draw a ray diagram depicting the formation of the image by an astronomical telescope in normal adjustment.
- You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? Give reason.
| Lenses |
Power (D) |
Aperture (cm) |
| $L_1$ |
3 |
8 |
| $L_2$ |
6 |
1 |
| $L_3$ |
10 |
1 |
Answer
- Ray diagram of astronomical telescope.
- Objective Lens: Lens $L_{1}$.
Eyepiece Lens: Lens $L_2.$
Reason: The objective should have large aperture and large focal length while the eyepiece should have small aperture and small focal length.
View full question & answer→Question 253 Marks
- Monochromatic light of wavelength 589 nm is incident from air on a water surface. If $\mu$ for water is 1.33, find the wavelength, frequency and speed of the refracted light.
- A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm.
Answer
- $\lambda=\frac{589\text{ }\text{nm}}{1.33}=442.8\text{ }\text{nm}$
Frequency $\nu=\frac{3\times10^8\text{ }\text{ms}^{-1}}{589\text{ }\text{nm}}=5.09\times10^{12}\text{Hz}$
Speed $\nu=\frac{3\times10^8}{1.33}\text{m/s}=2.25\times10^{8}\text{m/s}$
- $\frac{1}{f}=\bigg[\frac{\mu_2}{\mu_1}-1\bigg]\bigg[\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\bigg]$
$\therefore\text{ }\frac{1}{20}=\bigg[\frac{1.55}{1}-1\bigg]\frac{2}{\text{R}}$
$\therefore\text{ }\text{R}=(20\times1.10)\text{cm}=22\text{ }\text{cm}$ View full question & answer→Question 263 Marks
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
Answer
For the convex lens
u = - 60 cm, f = + 20 cm
$\frac{1}{\text{v}} - \frac{1}{\text{u}} = \frac{1}{\text{f}} \text{ gives } \text{v} = + 30 \text{ cm }$
For the convex mirror
u= + (30 – 15) cm = 15 cm,
$\text{F} = + \frac{20}{2}\text{cm} = 10 \text{ cm}$
$\frac{1}{\text{v}} + \frac{1}{\text{u}} = \frac{1}{\text{f}} \text{ gives }\text{v} = + 30 \text{ cm}$
Final image is formed at the distance of 30 cm from the convex mirror (or 45 cm from the convex lens ) to the right of the convex mirror. The final image formed is a virtual image. View full question & answer→Question 273 Marks
Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it.
Write two important limitations of a refracting telescope over a reflecting type telescope.
Answer
It is defined as the ratio of the angle ($\beta$) subtended by the final image on the eye to the angle($\alpha$) subtended by the object on eye.
$\text{M} = \frac{\tan\beta}{\tan\alpha} = \bigg(\frac{\beta}{\alpha}\bigg)$
Magnifying power $\text{M} = \frac{-\text{f}_{0}}{\text{f}_{e}}$ (for comfortable view)
$ = \frac{-\text{f}_{0}}{\text{f}_{e}}\bigg(1+\frac{\text{f}_{e}}{\text{D}}\bigg)$ (for strained eye)
Limitations:
- Image is not free from chromatic aberration and spherical aberration.
- Aperture of the objective lens should be large for high resolving power.
View full question & answer→Question 283 Marks
You are given three lenses $L_1, L_2$ and $L_3$ each of focal length 20 cm . An object is kept at 40 cm in front of $L_1$, as shown. The final real image is formed at the focus I of $L_3$. Find the separations between $L_1, L_2$ and $L_3$.
AnswerFor lens $L_1$
$\frac{1}{\text{f}_{1}} = \frac{1}{\text{v}_{1}} - \frac{1}{\text{u}_{1}}$
$\frac{1}{20} = \frac{1}{\text{v}_{1}} -\frac{1}{-40}=>\text{v}_{1} = 40 \text{ cm}$
For $L_3$
$\frac{1}{\text{f}_{3}} = \frac{1}{\text{v}_{3}} - \frac{1}{\text{u}_{3}}$
$u_3 = ?, f_3 = + 20 cm, v_3 = 20 cm$
$\frac{1}{20} = \frac{1}{20} + \frac{1}{\text{u}_{3}}$
$\text{u}_{3} = \infty$
It shows that $L_2$ must render the rays parallel to the common axis. It means that the image $\left(I_1\right)$, formed by $L_1$, must be at a distance of 20 cm from $L_2$ (at the focus of $L_2$ )
Therefore, distance between $L_1$ and $L_2(=40+20)=60 \mathrm{~cm}$ and distance between $L_2$ and $L_3$ can have any value.
View full question & answer→Question 293 Marks
Use the mirror equation to show that.
- An object placed between f and 2f of a concave mirror produces a real image beyond 2f.
- A convex mirror always produces a virtual image independent of the location of the object.
- An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Answer$\frac{1}{\text{v}} + \frac{1}{\text{u}} = \frac{1}{\text{f}}$
for concave mirror, f < 0 or f = - ve
for convexmirror, f > 0 or f = + ve
Concave Mirror
Let f = - c
Also, Let u = nf = - nc
$\frac{1}{\text{v}} = \frac{1}{\text{f}} - \frac{1}{\text{u}} = -\frac{1}{\text{c}} + \frac{1}{\text{nc}} = \frac{-\text{n} + 1 }{\text{nc}}$
$\therefore\text{v} = \frac{\text{nc}}{(1 - \text{n})}$
- When object is between f and 2f, we have 1 < n < 2
$\therefore\text{v} \text{ is -ve real image }$
(For n =1 and n =2), magnitude of v becomes $ \infty$ and 2c, respectively
$\therefore$ Real image is formed beyond 2F.
- Obiect between pole and F we have 0 < 0 < 1
v is + ve $\Rightarrow$(virtual image) and | v | > c
$\therefore$ We get a virtual, enlarged image
- Convex Mirror
f = + d Let u = - pd (p can have any value)
1/v = 1/d + 1/pd = (1+p)/pd
$\text{v} = \frac{\text{pd}}{(\text{p} + 1) }$
$\therefore$ v is always +ve and always less than d
$\therefore$ Convex mirror always produces a virtua1 image between pole and focus. View full question & answer→Question 303 Marks
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
Answer$\frac{1}{\text{v}_{0}} - \frac{1}{\text{u}_{0}} = \frac{1}{\text{f}_{0}}$
$\text{f}_{0} = 4 \text{cm};\text{u}_{e} = - 6 \text{cm}$
$\frac{1}{\text{v}_{0}} = \frac{1}{4}- \frac{1}{6} = \frac{1}{12} \therefore\text{v}_{0} = 12 \text{ cm}$
magnification by objective, $\text{m}_{0} = \frac{\text{v}_{0}}{|\text{u}_{0}|}$
$ = \frac{12}{6} = 2 $
Magnification by eyepiece
$\text{m}_{e} = \bigg(1 + \frac{\text{D}}{\text{f}_{e}}\bigg) \text{ or }\frac{\text{D}}{\text{f}_{e}}$
$ = \bigg(1 + \frac{25}{10}\bigg)\text{ or } \frac{25}{10}$
$ = 3.5 \text{ or } 2.5$
magnification power of the microscope
$m = m_0 \times m_e$
$= 2 × 3.5$ or $2 × 2.5$
$= 7$ or $5$
Length of the microscope,
$L = |V_0| + |u_0| or L = |V_0| + f_e$
$u_e = ? v_e = D = – 25 cm, f_e = 10 cm$
$\frac{1}{\text{u}_{e}} = \frac{1}{\text{v}_{e}} - \frac{1}{\text{f}_{e}} = - \frac{1}{25} - \frac{1}{10} = - \frac{7}{50}$
$\text{u}_{e} = - \frac{50}{7}\text{ cm}$
$\therefore\text{L} = 12 \text{ cm} + \frac{50}{7}\text{ cm}$
$ = 19.1 \text{ cm}$
Alternate Answer
$L = |V_0| + |f_0|$
$= 12 + 10 = 22\ cm.$
View full question & answer→Question 313 Marks
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33.
- Will it behave as a converging or a diverging lens in the two cases?
- How will its focal length change in the two media?
Answer$\frac{1}{\text{f}} = (\mu - 1)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
-
- Diverging lens or concave lens.
- Converging lens or convex lens.
-
- Focal length will become negative and its magnitude would increase.
- Focal length increases.
View full question & answer→Question 323 Marks
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.42 \times 10^6\ m$ and the radius of the lunar orbit is $3.8 \times 10^6\ m.$
AnswerAngular magnification, $\text{m} = \frac{\text{f}_{0}}{\text{f}_{e}}$
$ = \frac{1500\text{ cm}}{1\text{ cm}}$
= 1500
$\frac{\text{Diameter of image of moon (d)}}{\text{Focal length of objective}\text{(f}_{0})} = \frac{\text{Diameter of the moon}}{\text{Radius of Lunar orbit}}$
$\text{d} = 1500\times\frac{3.42\times10^{6}}{3.80\times10^{6}}$
= 13.5 cm.
View full question & answer→Question 333 Marks
An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.
Answerfor real image m $ = - 2 = \frac{\text{v}}{\text{u}}\Rightarrow\text{v} = - 2\text{u}$
given | u | + | v | = 90 cm
$\Rightarrow3 |\text{u}| = 90 \text{cm}$
$\Rightarrow|\text{u}| = 30 \text{cm}$
We have for a lens
$\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$
$\therefore\frac{1}{\text{f}} = \frac{1}{60} - \frac{1}{-30}$
$\frac{1}{\text{f}} = \frac{1}{20}$
$\Rightarrow\text{f} = 20 \text{ cm }$
Nature of lens: Convex/Converging.
View full question & answer→Question 343 Marks
- Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working.
- An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment?
Answer
Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image at infinity.
Calculation of magnifying power:
Given: Power of eyepiece = 10 D
Power of objective = 1 D
Magnifying power in normal adjustment: $\text{m} = \frac{\text{f}_{o}}{\text{f}_{e}} = \frac{\text{p}_{e}}{\text{p}_{o}} = \frac{10}{1} = 10 $ View full question & answer→Question 353 Marks
- Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working.
- Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
Answer
The objective forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual.
To achieve a large magnification of a small object; both the objective and eyepiece should have small focal lengths.
View full question & answer→Question 363 Marks
A double convex lens of glass of refractive index 1.6 has its both surfaces of equal radii of curvature of 30 cm each. An object of height 5cm is placed at a distance of 12.5 cm from the lens. Calculate the size of the image formed.
Answer$\frac{1}{\text{f}} = (\mu -1)\bigg(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\bigg) $
Alternate Answer
$\text{f} = \frac{\text{R}}{2(\mu -1)}$
f = 25 cm
$\frac{1}{\text{f}} =\frac{1}{\text{v}} -\frac{1}{\text{u}}$
v = - 25 cm
$\frac{\text{I}}{0} =\frac{\text{v}}{\text{u}} = 2 $
Alternate Answer
$\frac{\text{I}}{0} = \frac{\text{f}}{\text{f+ u}}$
$\therefore \text{I} = + 10 \text{cm}$.
View full question & answer→Question 373 Marks
A beam of light converges to a point P. A lens is placed in the path of the convergent beam12 cm from P. At what point does the beam converge if the lens is.
- a convex lens of focal length 20 cm,
- a concave lens of focal length 16 cm?
Do the required calculations.
Answer$\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$
For convex lens u = 12cm.
Calculation $\text{f} = + 20\text{cm}$ $\text{v} =7.5 \text{cm}$
For the concave lens u = +12cm
f = -16cm then v = 48cm.
View full question & answer→Question 383 Marks
Three rays (1, 2, 3) of different colours fall normally on one of the sides of an isosceles right angled prism as shown. The refractive index of prism for these rays is 1.39, 1.47 and 1.52 respectively. Find which of these rays get internally reflected and which get only refracted from AC. Trace the paths of rays. Justify your answer with the help of necessary calculations.
Answer$\text{i}=45^\circ$ (on face AC)
For TIR
$i>i_c$
$\Rightarrow\sin i>\sin i_c$
$\Rightarrow\frac{1}{\sin i}<\frac{1}{\sin i_c}$
$\Rightarrow\mu>\frac{1}{\sin i}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\because\mu=\frac{1}{\sin i_c}$
$\mu>\sqrt{2}=1.414$ for TIR
$\therefore$ Ray (1) is refracted from AC And rays (2) and (3) are internally reflected.
View full question & answer→Question 393 Marks
In the following diagram, an object ‘O’ is placed 15 cm in front of a convex lens $L_1$ of focal length 20 cm and the final image is formed at ‘I’ at a distance of 80 cm from the second lens $L_2.$ Find the focal length of the lens $L_2.$
AnswerFor $L_1$
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1 }$
$\Rightarrow \frac{1}{v_1}=\frac{1}{20}-\frac{1}{15}=-\frac{1}{60}$
$\Rightarrow v_1=-60\text{ }cm$
For lens $L_2$
$u = (-20 – 60) cm = -80 cm$
$v = 80 cm$
$\therefore\text{ }|u|=|v|=2\text{f}_2$
$\Rightarrow f_2=\frac{80}{2}=40\text{ }cm$
View full question & answer→Question 403 Marks
Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eye-piece. Write its two important advantages over a refracting telescope.
Answer
- Large gathering power.
- Large magnifying power.
- No chromatic aberration.
- Spherical aberration is also removed.
- Easy mechanical support.
- Large resolving power.
View full question & answer→Question 413 Marks
- Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also.
- Using mirror formula, explain why does a convex mirror always produce a virtual image.
Answer
- Given R = -20 cm, and magnification m= -2
Focal length of the mirror $\text{f} = \frac{\text{R}}{2} = - 10\text{cm}$
Magnification $(\text{m}) = -\frac{\text{v}}{\text{u}}$
$ - 2 = - \frac{\text{v}}{\text{u}}$
$ = > {v} = 2\text{u}$
Using mirror formula
$\frac{1}{\text{f}} = \frac{1}{\text{v}} + \frac{1}{\text{u}}$
$\Rightarrow - \frac{1}{10} = \frac{1}{2\text{u}} + \frac{1}{\text{u}}$
$\Rightarrow\text{u} = - 15 \text{cm}$
$\therefore\text{v} = 2 \times -15 \text{cm} = - 30 \text{cm}$
- $\frac{1}{\text{f}} =\frac{1}{\text{v}} + \frac{1}{\text{u}}$
Using sign convention, for convex mirror, we have
$\text{f} > 0 , \text{u} < 0$
From the formula
$\frac{1}{\text{v}} =\frac{1}{\text{f}} - \frac{1}{\text{u}}$
$\because \text{f}$ is positive and $\text{u}$ is negative,
$\Rightarrow\text{v}$ is always positive, hence image is always virtual. View full question & answer→Question 423 Marks
- A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
- If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 x 108 m.
AnswerAngular Magnification $\text{m} =\frac{\text{f}_{\circ}}{\text{f}_{e}}$ $ = \frac{15}{10^{-2}} = 1500$
Angular size of the moon $ =\bigg(\frac{3.48\times10^{6}}{3.8\times10^{8}}\bigg) = \frac{3.48}{3.8}\times10^{-2} \text{radian}$ $\therefore$ Angular size of the image $ =\bigg(\frac{3.48}{3.8}\times10^{-2}\times1500\bigg) =\text{radian}$ Diameter of the image $ =\frac{3.48}{3.8}\times15\times$ focal length of eye piece $ =\frac{3.48}{3.8}\times15\times1 \text{cm}$ = 13.7 cm. View full question & answer→Question 433 Marks
- Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
- The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a-certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.
Answer
For eyepiece $\text{m}_{e} = \frac{\text{v}_{e}}{\text{u}_{e}}$
$\text{u}_{e} = \frac{\text{v}_{e}}{\text{m}_{e}} = \frac{-20}{5}\text{cm} = - 4 \text{cm}$
Also, $\frac{1}{\text{f}_{e}}= \frac{1}{\text{v}_{e}} - \frac{1}{\text{u}_{e}}$
$\frac{1}{\text{f}_{e}} = \frac{-1}{20} + \frac{1}{4}$
$\text{f}_{e} = 5 \text{cm}$
$\text{m} = \text{m}_{e}\times \text{m}_{o}$
$-20 = 5 \times \text{m}_{o} = > \text{m}_{0} = -4$
Also, $|\text{v}_{o}| + |\text{u}_{e}| = 14$
$= > \text{v}_{o}= (14 – 4) \text{ cm} = 10\text{ cm}$
$\text{m}_{0} = 1-\frac{\text{v}_{o}}{\text{f}_{0}} = > -4= 1-\frac{10}{\text{f}_{o}}$
$= > \text{f}_{o} = 2\text{ cm}$
where subscripts e and o are used for eyepiece and objective respectively. View full question & answer→Question 443 Marks
Three light rays red (R), green (G) and blue (B) are incident on a right-angled prism ‘abc’ at face ‘ab’. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Out of the three which colour ray will emerge out of face ‘ac’? Justify your answer. Trace the path of these rays after passing through face ‘ab’.
AnswerAngle of incidence at face ac for all three colours = 45°A ray will get transmitted if the angle of incidence, for it, is less than thecritical angle for it.
$\therefore 45^{\circ}$ is the critical angle for $\mu = 1.414.$
Hence only the red colour $($with angle more than $45^\circ)$ ray will get transmitted $\big(\mu = 1.39\big)$. The green $\big(\mu = 1.44\big)$ and blue colour $\big(\mu = 1.47\big)$ rays will undergo total internal reflection.
View full question & answer→Question 453 Marks
Define the term 'resolving power' of an astronomical telescope. How does it get affected on?
- Increasing the aperture of the objective lens?
- Increasing the wavelength of the light used?
Justify your answer in each case.
AnswerResolving power (R.P) is the reciprocal of limit of angular resolution OR any other suitable definition.
- Increases:
Resolving power $ = \frac{\text{d}}{1.22\lambda}$
$\therefore$ R.P is directly proportional to ‘d’ (For a given $\lambda$)
- Decreases:
$\therefore$ R.P is inversely proportional to $'\lambda'$ (For a given d). View full question & answer→Question 463 Marks
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in:
- Medium A of refractive index 1.65
- Medium B of refractive index 1.33
Explain, giving reasons, whether it will behave as a converging lens or a diverging lens in each of these two media.
Answer$\frac{1}{\text{f}}_{\text{m}}=\bigg(\frac{_{a}\mu_{g}}{_{a}\mu_{m}}-1\bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg )$
- $\text{For } _{a}{\mu}_{g} = 1.5 \text { and } _{a}{\mu}_{m}= 1.6\text{f}_{m} \text{will be negative}$
hence it will be a converging lens.
- $\text{For} _{a}\mu_{g}= 1.5 \text {and}_{a}\mu_{m}=1.33\text{f}_{m}\text{will be positive}$
hence it will be a converging lens. View full question & answer→Question 473 Marks
- Show using a proper diagram how unpolarised light can be linearly polarised by reflection from a transparent glass surface.
- The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index $\frac{3}{2},$ placed in water of refractive index $\frac{4}{3}.$ Will this ray suffer total internal reflection on striking the face AC? Justify your answer.
Answer
-
When unpolarised light ray is incident at an angle such that the angle between reflected of refracted rays is 90°, then reflected ray is linearly polarised. In that case incident angle is called polarising angle or Brewster angle$(i_P$ or $i_B).$
- For Total internal reflection $\Big(\frac{1}{\sin\text{i}_{\text{c}}}=\mu_{\text{DR}}\Big)$
$\sin\text{i}_{\text{c}}=\Big(\frac{1}{\sin\text{i}_{\text{c}}}=\mu_{\text{DR}}\Big)$
$\sin\text{i}_{\text{c}}=\mu_{\text{wg}}$
$\sin\text{i}_{\text{c}}=\frac{4}{3}\div\frac{3}{2}$
$=\frac{4}{3}\times\frac{2}{3}$
$\sin\text{i}_{\text{c}}=\frac{8}{9}=0.88$
Now, in this case
$\sin\text{i}=\sin60^\circ=\frac{\sqrt{3}}{2}=0.867$
$\because\ \sin\text{i}<\sin\text{i}_{\text{c}}$
$\text{So,}\ \text{i}<\text{i}_{\text{c}}$
So, ray will not suffer Total internal reflection. View full question & answer→Question 483 Marks
A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1·5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.
AnswerLet $f_1 =$ focal length of biconvex lens.
$f_2 =$ focal length of plano concave liquid.
$f =$ focal length of combination of above two lenses.
$f = x f_1 = y .........(i)$ From combination of thin lenses,
$\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}=\frac{1}{\text{f}}$ $\frac{1}{\text{f}_2}=\frac{1}{\text{f}}-\frac{1}{\text{f}_1}$ $=\frac{1}{\text{x}}-\frac{1}{\text{y}}$ $\frac{1}{\text{f}_2}=\frac{\text{y}-\text{x}}{\text{xy}}\ ....(\text{ii})$
For bicovex lens, $\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}+\frac{1}{\text{R}}\Big)$$\frac{1}{\text{f}_1}$
$=(1.5-1)\Big(\frac{2}{\text{R}}\Big)$
$f_1 = R = y .......(iii)$
For planoconcave liquid, $\frac{1}{\text{f}_2}=(\mu_1-1)\Big(-\frac{1}{\text{R}}-\frac{1}{\infty}\Big)$
From equation (ii) & (iii), $\Rightarrow\ \frac{\text{y}-\text{x}}{\text{xy}}=(\mu_1-1)\Big(-\frac{1}{\text{y}}\Big)$
$\Rightarrow\ \frac{\text{y}-\text{x}}{\text{x}}=1-\mu_1$
$\mu_1=1-\frac{\text{y}-\text{x}}{\text{x}}$
$\mu_1=1-\frac{\text{x}-\text{y}+\text{x}}{\text{x}}$
$\mu_1=\frac{2\text{x}-\text{y}}{\text{x}}$
View full question & answer→Question 493 Marks
A light ray incident at grazing angle on the face AB of a prism ABC, follows the path shown in the figure. Obtain the relation between the angle of prism A and the refractive index $\mu$ of its material.
AnswerFor grazing incidence at the first face.
$\mu=\frac{\frac{\sin}{\pi^2}}{\text{r}_1}=\frac{1}{\sin\text{r}_1}$
$\Rightarrow\text{r}_1=\sin^{-1}\Big(\frac{1}{\mu}\Big)\dots(1)$
At the second face, $\frac{1}{\mu}=\frac{\sin\text{r}_2}{\frac{\sin}{\pi^2}}=\frac{\sin\text{r}_2}{1}$
$\Rightarrow\text{r}_2=\sin^{-1}\Big(\frac{1}{\mu}\Big)\dots(2)$
Also $\text{A}=\text{r}_1+\text{r}_2=2\sin^{-1}\Big(\frac{1}{\mu}\Big)$
View full question & answer→Question 503 Marks
Why are the magnification properties of microscopes and telescopes defined in terms of the ratio of angles and not in terms of the ratio of sizes of objects and images?
AnswerInstruments like telescopes and microscopes deal with objects placed at different distances. Due to some physical factors, there is a relative change in heights not in the angle which the light emerging from them subtends on the lens. So, the magnification properties of instruments are defined in terms of the ratio of angles.
View full question & answer→Question 513 Marks
The magnifying power of a simple microscope is given by $1+\frac{\text{D}}{\text{f}},$ where D is the least distance for clear vision. For farsighted persons, D is greater than the usual. Does it mean that the magnifying power of a simple microscope is greater for a farsighted person as compared to a normal person? Does it mean that a farsighted person can see an insect more clearly under a microscope than a normal person?
AnswerThe magnifying power of a simple microscope depends on the ratio $\frac{\text{D}}{\text{f}}$ for a farsighted person. Here, D for a farsighted person is greater than that for a normal person, but the value of f remains the same. Therefore, the magnifying power of a simple microscope is greater for a farsighted person compared to that for a person with normal vision. Also, a farsighted person can see the insect more clearly under the microscope than a person with normal vision.
View full question & answer→Question 523 Marks
Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?
AnswerRefractive index $(\mu)$ of the material from which prism is made = 1.732
We know refractive index is given by:
$\mu=\frac{\Big[\frac{\delta_{\text{min}}+\text{A}}{2}\Big]}{\sin\Big[\frac{\text{A}}{2}\Big]}$
Where $\delta_{\text{min}}$ is the angleof minimum deviation and A is the angle of prism = 60º
$\Rightarrow 1.732\times \sin(30^\circ)=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$
$\Rightarrow \frac{1.732}{2}=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$
$\Rightarrow \Big(\frac{\delta_{\text{min}}+60^\circ}{20}\Big)=60^\circ$
$\delta_{\text{min}}=60^\circ$
$\delta_{\text{min}}=2\text{i}-\text{A}$
$2\text{i}=120^\circ$
$\text{i}=60^\circ$
Hence, the required angle of deviation is 60°. View full question & answer→Question 533 Marks
A thin prism of crown glass $(\mu_\text{r}=1.515,\mu_\text{v}=1.525)$ and a thin prism of flint glass $(\mu_\text{r}=1.612,\mu_\text{v}=1.632)$ are placed in contact with each other. Their refracting angles are 5.0° each and are similarly directed. Calculate the angular dispersion produced by the combination.
AnswerGiven that,
$\mu_\text{cr}=1.515,\mu_\text{cv}=1.525$ and $\mu_\text{fr}=1.612,\mu_\text{fv}=1.632$ and A = 5° Since, they are similarly directed, the total deviation produced is given by, $\delta=\delta_\text{c}+\delta_\text{r}=(\mu_\text{c}-1)\text{A}+(\mu_\text{r}-1)\text{A}$ $=(\mu_\text{c}+\mu_\text{r}-2)\text{A}$ So, angular dispersion of the combination is given by, $\delta_\text{v}-\delta_\text{y}=(\mu_\text{cv}+\mu_\text{fv}-2)\text{A}-(\mu_\text{cr}+\mu_\text{fr}-2)\text{A}$ $=(\mu_\text{cv}+\mu_\text{fv}-\mu_\text{cr}-\mu_\text{fr})\text{A}$ $=(1.525+1.632-1.515-1.612)5=0.15^\circ$ View full question & answer→Question 543 Marks
Locate the image of the point P as seen by the eye in the figure.
Answer
The presence of air medium in between the sheets does not affect the shift. The shift will be due to 3 sheets of different refractive index other than air. $\Delta \text{t}=\Big[1-\frac{1}{\mu_1}\Big]\text{t}_1+\Big[1-\frac{1}{\mu_2}\Big]\text{t}_2+\Big[1-\frac{1}{\mu_3}\Big]\text{t}_3$ $=\Big(1-\frac{1}{1.2}\Big)(0.2)+\Big(1-\frac{1}{13}\Big)(0.3)+\Big(1-\frac{1}{14}\Big)(0.4)$ $=0.2\text{cm}$ above point P. View full question & answer→Question 553 Marks
A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.
Answer
Given, P = 5 diopter (convex lens)
$\Rightarrow\text{f}=\frac{1}{5}\text{m}=20\text{cm}$
Since, a virtual image is formed, u and v both are negative.Given, $\frac{\text{v}}{\text{u}}=4$
$\Rightarrow\text{v}=4\text{u} \ ...(1)$
From lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{4\text{u}}-\frac{1}{\text{u}}\Rightarrow\frac{1}{20}=\frac{1-4}{4\text{u}}=-\frac{3}{4\text{u}}$
$\Rightarrow\text{u}=-15\text{cm}$
$\therefore$ Object is placed 15cm away from the lens. View full question & answer→Question 563 Marks
A small piece of wood is floating on the surface of a 2.5m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water $=\frac{4}{3}.$
Answer
Height of the lake = 2.5m
When the sun is just setting, $\theta$ is approximately = 90°
$\therefore \ \frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}\Rightarrow\frac{1}{\sin\text{r}}=\frac{\frac{4}{3}}{1}\Rightarrow\sin\text{r}=\frac{3}{4}\Rightarrow\text{r}=49^{\circ}$
As shown in the figure, $\frac{\text{x}}{2.5}=\tan\text{r}=1.15$
$\Rightarrow\text{x}=2.5\times1.15=2.8\text{m}.$ View full question & answer→Question 573 Marks
A light ray is incident at an angle of 45° with the normal to a $\sqrt{2}\text{cm}$ thick plate $(\mu = 2.0)$. Find the shift in the path of the light as it emerges out from the plate.
Answer
Applying snell's law,
$\frac{\sin \text{i}}{\sin \text{r}}=\frac{1}{\mu}$ $\Rightarrow \frac{\sin \text{i}}{\sin \text{r}}=\frac{2}{1}$ As shown in the figure, $\frac{\sin45^{\circ}}{\sin\text{r}}=\frac{2}{1}\Rightarrow\sin\text{r}=\frac{\sin45^{\circ}}{2}=\frac{1}{2\sqrt{2}}\Rightarrow\text{r}=21^{\circ}$ Here, BD = shift in path = AB sin 24° $=0.406\times\text{AB}=\frac{\text{AE}}{\cos21^{\circ}}\times0.406=0.62\text{cm}.$ View full question & answer→Question 583 Marks
A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of their apparent sizes.
| Tree |
Height(m) |
Distance form the eye(m) |
| A |
2.8 |
50 |
| B |
2.5 |
80 |
| C |
1.8 |
70 |
| D |
2.8 |
100 |
AnswerThe visual angles made by the tree with the eyes can be calculated be below.
$\theta=\frac{\text{Height of the tree}}{\text{Distance from the eye}}=\frac{\text{AB}}{\text{OB}}$
$\Rightarrow\theta_\text{A}=\frac{2}{50}=0.04$
Similarly, $\theta_\text{B}=\frac{2.5}{80}=0.03125$
$\theta_\text{c}=\frac{2.8} {70}=0.02571$
$\theta_\text{D}=\frac{2.8}{100}=0.028$
Since, $\theta_\text{A}>\theta_\text{B}>\theta_\text{D}>\theta_\text{C}$ the arrangement in decreasing order is given by A, B, D and C.
View full question & answer→Question 593 Marks
The focal lengths of a convex lens for red, yellow and violet rays are 100cm, 98cm and 96cm respectively. Find the dispersive power of the material of the lens.
AnswerThe focal length of a lens is given by
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow(\mu-1)=\frac{1}{\text{f}}\times\frac{1}{\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)}=\frac{\text{k}}{\text{f}}\ ...(1)$
So, $\mu_\text{r}-1=\frac{\text{K}}{100}\ ...(2)$
$\mu_\text{y}-1=\frac{\text{K}}{98}\ ...(3)$
And $\mu_\text{v}-1=\frac{\text{K}}{96}\ ...(4)$
So, Dispersive power $=\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}=\frac{\frac{\text{K}}{96}-\frac{\text{K}}{100}}{\frac{\text{K}}{98}}$ $=\frac{98\times4}{9600}=0.0408$
View full question & answer→Question 603 Marks
For a given position of an object, an equiconvex lens forms its real image at a distance of 60cm from the lens. A convex mirror is now kept in between this image position and the convex lens. The final image formed by this combination coincides with the object when the distance between the convex lens and the convex mirror equals 25cm. Calculate the radius of curvature of the convex mirror.
AnswerGiven that the image formed by the combination of convex lens and the convex mirror coincides with the object.
This means that the position of the image formed by the convex lens alone corresponds to the position of centre of curvature of the convex mirror.
Hence, the distance between the convex mirror and this image (formed by the lens) which is equal to 60-25 represents radius of curvature of the mirror.
Thus, radius of curvature of the mirror = 35cm.
View full question & answer→Question 613 Marks
The following data was recorded for values of object distance and the corresponding values of image distance in the experiment on study of real image formation by a convex lens of power +5D. One of these observations is incorrect. Identify this observation and give reason for your choice:
|
S .No
|
Object distance (cm)
|
Image distance (cm)
|
|
1
|
25
|
97
|
|
2
|
30
|
61
|
|
3
|
35
|
37
|
|
4
|
45
|
35
|
|
5
|
50
|
32
|
|
6
|
55
|
30
|
AnswerPower of lens = +5D
Focal length of lens,
The observations at serial number (3) i.e., (object distance 35cm and image distance 37cm is incorrect), because if the object is placed at a distance between f and 2f its image will be formed beyond 2f, while in this observation the object and image distances, both are between f and 2f.
View full question & answer→Question 623 Marks
A man uses a concave mirror for shaving. He keeps his face at a distance of 25cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.
Answer$\text{u}=-25\text{cm}$
$\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\frac{\text{v}}{\text{u}}\Rightarrow1.4=-\Big(\frac{\text{v}}{-25}\Big)\Rightarrow\frac{14}{10}=\frac{\text{v}}{25}$
$\Rightarrow\text{v}=\frac{25\times14}{10}=35\text{cm}$
Now, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{35}-\Big(\frac{1}{-25}\Big)$
$\Rightarrow\frac{5-7}{175}=\frac{-2}{175}$
$\Rightarrow\text{f}=-87.5\text{cm}$
So, focal length of the concave mirror is 87.5cm.
View full question & answer→Question 633 Marks
One end of a cylindrical glass rod $(\mu=1.5)$ of radius 1.0cm is rounded in the shape of a hemisphere. The rod is immersed in water $\Big(\mu=\frac{4}{3}\Big)$ and an object is placed in the water along the axis of the rod at a distance of 8.0cm from the rounded edge. Locate the image of the object.
Answer
Radius of the cylindrical glass tube = 1cm
We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
Here, $\text{u}=-8\text{cm}, \ \mu_2=\frac{3}{2},\mu_1=\frac{4}{3},\text{R}=+1\text{cm}$
So, $\frac{3}{2\text{v}}+\frac{4}{3\times8}\Rightarrow\frac{3}{2\text{v}}+\frac{1}{6}=\frac{1}{6}$
$ \ \text{v}=\infty$
$\therefore$ The image will be formed at infinity. View full question & answer→Question 643 Marks
A convex lens produces a double size real image when an object is placed at a distance of 18cm from it. Where should the object be placed to produce a triple size real image?
AnswerA real image is formed. So, magnification m = -2(inverted image)
$\therefore\frac{\text{v}}{\text{u}}=-2\Rightarrow\text{v}=-2\text{u}=(-2)(-18)=36$
From lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{36}-\frac{1}{-18}=\frac{1}{\text{f}}$
$\Rightarrow\text{f}=12\text{cm}$
Now, for triple sized image $\text{m}=-3=\Big(\frac{\text{v}}{\text{u}}\Big)$
$\therefore \ \frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{-3\text{u}}-\frac{1}{\text{u}}=\frac{1}{12}$
$\Rightarrow3\text{u}=-48\Rightarrow\text{u}=-16\text{cm}$
So, object should be placed 16cm from lens.
View full question & answer→Question 653 Marks
A concave mirror has a focal length of 20cm. Find the position or positions of an object for which the imagesize is double of the object-size.
Answer
For the concave mirror,
$\text{f}=-20\text{cm}, \ \text{M}=-\frac{\text{v}}{\text{u}}=2$
$\Rightarrow\text{v}=-2\text{u}$
| $1^{st} case:$ |
$2^{nd} case:$ |
| $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$ |
$\frac{-1}{2\text{u}}-\frac{1}{\text{u}}=-\frac{1}{\text{f}}$ |
| $\Rightarrow\frac{1}{2\text{u}}-\frac{1}{\text{u}}=-\frac{1}{\text{f}}$ |
$\Rightarrow\frac{3}{2\text{u}}=\frac{1}{\text{f}}$ |
| $\Rightarrow\text{u}=\frac{\text{f}}{2}=10\text{cm}$ |
$\Rightarrow\text{u}=\frac{3\text{f}}{2}=30\text{cm}$ |
$\therefore$ The positions are 10cm or 30cm from the concave mirror. View full question & answer→Question 663 Marks
The magnifying power of a converging lens used as a simple microscope is $\Big(1+\frac{\text{D}}{\text{f}}\Big).$ A compound microscope is a combination of two such converging lenses. Why don't we have magnifying power $\Big(1+\frac{\text{D}}{\text{f}_\text{o}}\Big) \Big(1+\frac{\text{D}}{\text{f}_\text{o}}\Big)?$ In other words, why can the objective not be treated as a simple microscope but the eyepiece can?
AnswerIn a simple microscope, the converging lens is used to magnify the object. It is done by the eyepiece in a compound microscope. But the purpose of the objective lens is the same, i.e., to form an enlarged, real and inverted image of the object at a distance less than the focal length of the eyepiece. So, its magnification power cannot be expressed in a way it is expressed for a simple microscope.
View full question & answer→Question 673 Marks
A nearsighted person cannot see beyond 25cm. Assuming that the separation of them glass from the eye is 1cm, find the power of lens needed to see distant objects.
AnswerFor the near sighted person,
v = distance of image from glass
= distance of image from eye - separation between glass and eye
= 25cm - 1cm = 24cm = 0.24m
So, for the glass, $\text{u}=\infty$ and v = -24cm = -0.24m
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-0.24}-\frac{1}{\infty}=-4.2\text{D}$
View full question & answer→Question 683 Marks
A person has near point at 100cm. What power of lens is needed to read at 20cm if he/ she uses.
- Contact lens,
- Spectacles having glasses 2.0cm separated from the eyes?
AnswerThe person has near point 100cm. It is needed to read at a distance of 20cm.
- When contact lens is used,
u = -20cm = -0.2m,
v = -100cm = -1m
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-1}-\frac{1}{-0.2}=-1+5=+4\text{D}$
- When spectacles are used,
u = -(20 - 2) = -18cm = -0.18m
v = -100cm = -1m
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-1}-\frac{1}{-0.18}=-1+5.55=+4.5\text{D}$ View full question & answer→Question 693 Marks
A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1m is refracted into it at an angle of 30°. Calculate the time taken by the light rays, to cross the slab. Speed of light in vacuum $= 3 \times 10^8m/s$
Answer
We know, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{3\times10^8}{\text{v}}=\frac{\sin45^{\circ}}{\sin30^{\circ}}=\sqrt{2}$
$\Rightarrow\text{v}=\frac{3\times10^8}{\sqrt{2}}\text{m}/\text{sec}$
Distance travelled by light in the slab is,
$\text{x}=\frac{1\text{m}}{\cos30^{\circ}}=\frac{2}{\sqrt{3}}\text{m}$
So, time taken $=\frac{2\times\sqrt{2}}{\sqrt{3}\times3\times10^8}=0.54\times10^{-8}=5.4\times10^{-9}\text{sec}.$ View full question & answer→Question 703 Marks
A magnifying glass is a converging lens placed close to the eye. A farsighted person uses spectacles having converging lenses. Compare the functions of a converging lens used as a magnifying glass and as spectacles.
AnswerA converging lens in a magnifying glass is of small focal length which is used to magnify an object which is placed close to the lens. On the other hand, converging lens used as spectacles is of varying focal length which depends upon the actual near point of the long-sighted person. It forms image at the near point of the defected eye which is further focussed by the eye lens.
View full question & answer→Question 713 Marks
If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?
AnswerNo, Yes, Yes.
Real Image: Rays are really getting focussed at point P so intensity is high.
Virtual Image: Ray appears to come from point P and no real concentration of ray at P.
View full question & answer→Question 723 Marks
The diameter of the sun is $1.4 \times 10^9m$ and its distance from the earth is $1.5 \times 10^{11}m$. Find the radius of the image of the sun formed by a lens of focal length 20cm.
Answer$u = -1.5 \times 10^{11} m; v = +20 \times 10^{-2} m$
Since, f is very small compared to u, distance is taken as $\infty.$
So, image will be formed at focus.
$\Rightarrow v = +20 \times 10^{-2}m$
$\therefore$ We know, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}$
$\Rightarrow\frac{20\times10^{-2}}{1.5\times10^{11}}=\frac{\text{D}_{\text{image}}}{1.4\times10^9}$
$\Rightarrow\text{D}_{\text{image}}=1.86\text{mm}$
So, radius $=\frac{\text{D}_{\text{image}}}{2}=0.93\text{mm}.$
View full question & answer→Question 733 Marks
A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?
AnswerGiven $\text{A}=60^{\circ}$ and $\mu=30^{\circ}$
We know that,
$\mu=\frac{\sin\Big(\frac{\text{A}+\delta_{\text{m}}}{2}\Big)}{\frac{\sin\text{A}}{2}}=\frac{\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}}{\sin30^{\circ}}=2\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}$
Since, one ray has been found out which has deviated by 30°, the angle of minimum deviation should be either equal or less than 30°. (It can not be more than 30°).
So, $\mu\leq2\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}$ $\big($because $\mu$ will be more if $\delta_{\mu}$ will be more$\big)$
or, $\mu\leq2\times\frac{1}{\sqrt{2}}$
or, $\mu\leq\sqrt{2}.$
View full question & answer→Question 743 Marks
Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.
Answer
Thickness of glass $=3\text{cm}, \ \mu_{\text{g}}=1.5$
Image shif $=3\Big(1-\frac{1}{1.5}\Big)$
[Treating it as a simple refraction problem because the upper surface is flat and the spherical surface is in contact with the object]
$=3\times\frac{0.5}{1.5}=1\text{cm}.$
The image will appear 1 cm above the point P. View full question & answer→Question 753 Marks
A double convex lens, made from a material of refractive index $\mu_1,$ is immersed in a liquid of refractive index $\mu_2$ where $\mu_1>\mu_2.$ What change, if any, would occur in the nature of the lens?
AnswerFocal length of lens (refractive index $\mu_1$) in a liquid of refractive index $\mu_2$ is
$\text{f}_1=\frac{\mu_1-1}{\frac{\mu_1}{\mu_2}-1}\times\text{f}_{\text{a}}$
Given $\mu_2>\mu_1, \text{ i.e., }\frac{\mu_1}{\mu_2}<1$
So $\text{f}_{\text{l}}=\frac{\mu_1-1}{1-\frac{\mu_1}{\mu_2}}\text{f}_{\text{a}}$
So the focal length of lens in liquid will be of opposite sign of the focal length of lens in air; i.e., nature of lens will change. Hence, lens would now behave like a diverging (concave) lens.
View full question & answer→Question 763 Marks
A convex lens, of focal length 20cm, has a point object placed on its principle axis at distance of 40cm from it. A plane mirror is placed 30cm behind the convex lens. Locate the position of image formed by this combination.
AnswerWe first consider the effect of the lens. For the lens, we have u = -40cm and f = +20cm.
$\frac{1}{\text{v}}-\frac{1}{(-140)}=\frac{1}{20}$
$\Rightarrow \mathrm{v}+40 \mathrm{~cm}$ Had there been the lens only the image would have been formed at $\mathrm{Q}_1$. The plane mirror M is at a distance of 30 cm from the lens $L$. We can, therefore, think of a $Q_1$ as a virtual object, located at a distance of 10 cm , behind the plane mirror $M$. The plane mirror therefore forms a real image (of this virtual object $Q_1$ ) at $Q,10 \mathrm{~cm}$ in front of it. This is show in the figure.
View full question & answer→Question 773 Marks
Draw a plot showing the variation of power of a lens with the wavelength of the incident light.
AnswerRefractive index $=\text{A}+\frac{\text{B}}{\lambda^2},$ where $\lambda$ is the wavelength. Power of a lens $\text{P}=\frac{1}{\text{f}}=(\text{n}_{\text{g}-1})\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$ Clearly, power of a lens $\propto(\text{n}_{\text{g}}-1).$ This implies that the power of a lens. decreases with increase of wavelength $\big(\text{P}\propto\frac{1}{\lambda^2}\text{nearly}\Big).$ The plot is shown in fig. alongside.
View full question & answer→Question 783 Marks
A 3cm tall object is placed at a distance of 7.5cm from a convex mirror of focal length 6cm. Find the location, size and nature of the image.
Answer
Given AB = 3cm, u = -7.5cm, f = 6cm.
Using $\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
Putting values according to sign conventions,
$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{-7.5}=\frac{3}{10}$
$\Rightarrow\text{v}=\frac{10}{3}\text{cm}$
$\therefore$ magnification $=\text{m}=-\frac{\text{v}}{\text{u}}=\frac{10}{7.5\times3}$
$\Rightarrow\frac{\text{A}'\text{B}'}{\text{A}\text{B}}=\frac{10}{7.5\times3}\Rightarrow\text{A}'\text{B}'=\frac{100}{72}=\frac{4}{3}=1.33\text{cm}$
$\therefore$ Image will form at a distance of $\frac{10}{3}\text{cm}$ From the pole and image is 1.33cm (virtual and erect). View full question & answer→Question 793 Marks
An object is to be seen through a simple microscope of focal length 12cm. Where should the object be placed so as to produce maximum angular magnification? The least distance for clear vision is 25cm.
AnswerFor the given simple microscope, f = 12cm and D = 25cm For maximum angular magnification, the image should be produced at least distance of clear vision. So, v = -D = -25cm Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\Rightarrow \frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}=\frac{1}{-25}-\frac{1}{12}=-\frac{37}{300}$ $\Rightarrow \text{u}=-8.1\text{cm}$So, the object should be placed 8.1cm away from the lens.
View full question & answer→Question 803 Marks
The critical angle for a given piece of glass is 45°. Calculate the polarising angle for it. Also calculate the angle of refraction when light is incident on this glass at an angle of incidence equal to $i_p.$
AnswerWe know $\sin\text{i}_{\text{c}}=\frac{1}{\mu}$
$\mu=\frac{1}{\sin{\text{i}_{\text{c}}}}=\frac{1}{\sin45^\circ}=\sqrt{2}$
According to Brewster' s law
$\tan\text{i}_{\text{p}}=\mu=\sqrt{2}$
$\Rightarrow\text{i}_{\text{p}}=\tan^{-1}\sqrt{2}=(1.414)\cong51^\circ40^\circ$
When light is incident at an angle $i_p$ the corresponding angle of refraction 'r' is givan by
$\text{i}_{\text{p}}+\text{r}=90^\circ$
$\therefore\text{r}=90^\circ-(51^\circ40^\circ)=(38^\circ20^\circ)$
View full question & answer→Question 813 Marks
An equiconvex lens of focal length ‘f’ is cut into two identical plane convex lenses. How will the power of each part be related to the focal length of the original lens?A double convex lens of +5D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.
AnswerPower of a lens, $\text{P}=\frac{1}{\text{f}(\text{in metre})}$
After cutting, the power of each part will be half of the power of original lens. Therefore, focal length = 2f
$\therefore$ power of each part, $\text{P}'=\frac{1}{2\text{f}}$
$=\text{P}=\frac{1}{\text{f}}$
$\Rightarrow5=\frac{1}{\text{f}}$
$\text{f}=\frac{1}{5}\text{m}=20\text{m}$
Now, $\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
Since $R_1= + R$ and $R_2 = -R$
$\therefore\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\frac{1}{20}=(1.5-1)\times\frac{2}{\text{R}}$
$\text{R}=20\text{m}.$
View full question & answer→Question 823 Marks
A concave mirror having a radius of curvature 40cm is placed in front of an illuminated point source at a distance of 30cm from it. Find the location of the image.
Answer
u = -30cm, R = -40cm
From the mirror equation,
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{2}{\text{R}}$
$\Rightarrow \frac{1}{\text{v}}=\frac{2}{\text{R}}-\frac{1}{\text{u}}=\frac{2}{-30}-\frac{1}{-30}=-\frac{1}{60}$
or, v = -60cm
So, the image will be formed at a distance of 60cm in front of the mirror. View full question & answer→Question 833 Marks
Find the angle of deviation suffered by the light ray shown in figure. The refractive index $\mu=1.5$ for the prism material.
Answer
Given, $\mu=1.5$
And angle of prism $=4^{\circ}$
$\therefore \ \mu=\frac{\sin\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\frac{\sin\text{A}}{2}}=\frac{\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\big(\frac{\text{A}}{2}\big)}$ $\big($for small angle $\sin\theta=\theta\big)$
$\Rightarrow\mu=\frac{\text{A}+\delta_{\text{m}}}{2}\Rightarrow1.5=\frac{4^{\circ}+\delta_{\text{m}}}{4^{\circ}}$
$\Rightarrow\delta_{\text{m}}=4^{\circ}\times(1.5)-4^{\circ}=2^{\circ}$ View full question & answer→Question 843 Marks
A vessel contains water upto a height of 20cm and above it an oil upto another 20cm. The refractive indices of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.
Answer
Shift due to water $\Delta\text{t}_\text{w}=\Big(1-\frac{1}{\mu}\Big)\text{d}\Big(1-\frac{1}{1.33}\Big)20=5\text{cm}$
Shift due to oil, $\Delta\text{t}_0=\Big(1-\frac{1}{1.33}\Big)20=4.6\text{cm}$
Total shift $\Delta\text{t}=5+4.6=9.6\text{cm}$
Apparent depth = 40 - (9.6) = 30.4cm below the surface. View full question & answer→Question 853 Marks
A lens forms a real image of an object. The distance from the object to the lens is u cm and the distance of the image from the lens is v cm. The given graphs shows variation of v with u.
- What is the nature of the lens?
- Using the graph, find the focal length of this lens.
Answer
The lens is convex because image formed by the lens is real.
Focal length: From graph when u = -40cm, v = 15cm.
$\therefore \frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$ (for a lens)
$=\frac{1}{15}+\frac{1}{40}=\frac{40+15}{40\times15}$
$\text{f}=\frac{40\times15}{55}=\frac{120}{11}\approx11\text{cm.}$ View full question & answer→Question 863 Marks
Does focal length of a lens depend on the colour of the light used? Does focal length of a mirror depend on the colour?
AnswerYes, the focal length of a lens depends on the colour of light.
According to lens-maker's formula,
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
Here, f is the focal length, $\mu$ is the refractive index, R is the radius of curvature of lens.
The refractive index $(\mu)$ depends on the inverse of square of wavelength.
The focal length of a mirror is independent of the colour of light.
View full question & answer→Question 873 Marks
A concave mirror of radius R is kept on a horizontal. Water $\big($refractive index $=\mu\big)$ is poured into it upto a height h. Where should an object be placed so that its image is formed on itself?
Answer
Let the object be placed at a height x above the surface of the water.
The apparent position of the object with respect to mirror should be at the centre of curvature so that the image is formed at the same position.
Since, $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{1}{\mu}$ (with respect to mirror)
Now, $\frac{\text{x}}{\text{R}-\text{h}}=\frac{1}{\mu}$
$\Rightarrow\text{x}=\frac{\text{R}-\text{h}}{\mu}$ View full question & answer→Question 883 Marks
A particle goes in a circle of radius 2.0cm. A concave mirror of focal length 20cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30cm. Calculate the radius of the circle formed by the image.
Answer
$\text{u}=-30\text{cm}, \ \text{f}=-20\text{cm}$
We know, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{30}\Big)=\Big(-\frac{1}{20}\Big)\Rightarrow\text{v}=-60\text{cm}$
Image of the circle is formed at a distance 60cm in front of the mirror.
$\therefore \ \text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{R}_{\text{image}}}{\text{R}_{\text{object}}}\Rightarrow\frac{-60}{-30}=\frac{\text{R}_{\text{image}}}{2}$
$\Rightarrow\text{R}_{\text{image}}=4\text{cm}$
Radius of image of the circle is 4cm. View full question & answer→Question 893 Marks
A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25cm) of a normal eye.
- What is its focal length?
- What will be its magnifying power if the image is formed at infinity?
AnswerThe simple microscope has, m = 3, when image is formed at D = 25cm
- $\text{m}=1+\frac{\text{D}}{\text{f}}\Rightarrow3=1+\frac{25}{\text{f}}$
$\Rightarrow\text{f}=\frac{25}{2}=12.5\text{cm}$
- When the image is formed at infinity (normal adjustment)
Magnifying power $=\frac{\text{D}}{\text{f}}=\frac{25}{12.5}= 2.0$ View full question & answer→Question 903 Marks
A pole of length 1.00m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.
Answer
Shadow length $= \text{BA}' = \text{BD} + \text{A}'\text{D} = 0.5 + 0.5 \tan \text{r}$
Now, $1.33=\frac{\sin45^{\circ}}{\sin\text{r}}\Rightarrow\sin\text{r}=0.53$
$\Rightarrow\cos\text{r}=\sqrt{1-\sin^2\text{r}}=\sqrt{1-(0.53)^2}=0.85$
So, $\tan \text{r} = 0.6235$
So, shadow length = (0.5)(1 + 0.6235) = 81.2cm View full question & answer→Question 913 Marks
An object P is focused by a microscope M. A glass slab of thickness 2.1cm is introduced between P and M. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?
AnswerThe thickness of the glass is $\text{d} = 2.1\text{cm}$ and $\mu=1.5$ Shift due to the glass slab
$\Delta\text{T}=\Big(1-\frac{1}{\mu}\Big)\text{d}$
$=\frac{1}{3}(2.1)=0.7\text{cm}$
So, the microscope should be shifted 0.70cm to focus the object again.
View full question & answer→Question 923 Marks
Two prisms of identical geometrical shape are combined with their refracting angles oppositely directed. The materials of the prisms have refractive indices 1.52 and 1.62 for violet light. A violet ray is deviated by 1.0° when passes symmetrically through this combination. What is the angle of the prisms?
AnswerTwo prisms of identical geometrical shape are combined.
Let A = Angle of the prisms
$\mu'_\text{v}=1.52$ and $\mu_\text{v}=1.62,\delta_\text{v}=1^\circ$
$\delta_\text{v}=(\mu_\text{v}-1)\text{A}-(\mu'_\text{v}-1)\text{A}$ [Since A = A']
$\Rightarrow\delta_\text{v}=(\mu_\text{v}-\mu'_\text{v})\text{A}$
$\Rightarrow\text{A}=\frac{\delta_\text{v}}{\mu_\text{v}-\mu'_\text{v}}=\frac{1}{1.62-1.52}=10^\circ$ View full question & answer→Question 933 Marks
A point object is placed on the principal axis of a convex lens (f = 15cm) at a distance of 30cm from it. A glass plate $(\mu=1.50)$ of thickness 1cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.
Answer
For the lens, f = 15cm, u = 30cm
From lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}\Rightarrow\text{v}=30\text{cm}$
The image is formed at 30cm of right side due to lens only.
Again, shift due to glass slab is,
$=\Delta\text{t}=\Big(1-\frac{1}{15}\Big)1 \ \big[\text{Since,} \ \mu_{\text{g}=1.5 \ \text{and} \ \text{t}=1\text{cm}}\big]$
$=1-\Big(\frac{2}{3}\Big)=0.33\text{cm}$
$\therefore$ The image will be formed at 30 + 0.33 = 30.33cm from the lens on right side. View full question & answer→Question 943 Marks
A narrow beam of light passes through a slab obliquely and is then received by an eye. The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.
AnswerThe image position shifts because of variation in refractive index. Thus it appears to be twinkling to the eyes. When $\mu$ changes $\alpha$ changes and also x changes so ray appear to come from different position.
View full question & answer→Question 953 Marks
Locate the image formed by refraction in the situation shown in figure.
Answer
$\mu_1=1, \ \mu_2=1.5,$
R = 20cm (Radius of curvature), u = -25cm
$\therefore \ \frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.5}{\text{v}}=\frac{0.5}{20}-\frac{1}{25}=\frac{1}{40}-\frac{1}{25}=\frac{-3}{200}$
$\Rightarrow \ \text{v}=200\times0.5=-100\text{cm}.$
So, the image is 100cm from (P) the surface on the side of S. View full question & answer→Question 963 Marks
The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?
AnswerWe know that, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{D}}{\text{f}}$
$\Rightarrow\ \text{f}=\frac{\text{D}}{\text{m}}$
We are given that, v = D = 25cm and u = f
Substituting D = 25cm and m = 10, we get
$\Rightarrow\ \text{f}=\frac{25}{10}=2.0.225\text{m}$
Now, $\text{P}=\frac{1}{0.025}=40\text{D}$
View full question & answer→Question 973 Marks
A converging lens and a diverging mirror are placed at a separation of 15cm. The focal length of the lens is 25cm and that of the mirror is 40cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?
Answer
If the image in the mirror will form at the focus of the converging lens, then after transmission through the lens the rays of light will go parallel.
Let the object is at a distance x cm from the mirror
$\therefore$ u = -x cm; v = 25 - 15 = 10cm (because focal length of lens = 25cm)
f = 40cm
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{x}}=\frac{1}{10}-\frac{1}{40}$
$\Rightarrow\text{x}=\frac{400}{30}=\frac{40}{3}$
$\therefore$ The object is at distance $\Big(15-\frac{40}{3}\Big)=\frac{5}{3}=1.67\text{cm}$ from the lens. View full question & answer→Question 983 Marks
A thin prism is made of a material having refractive indices 1.61 and 1.65 for red and violet light. The dispersive power of the material is 0.07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4.0° in favourable conditions. Calculate the angle of the prism.
AnswerGiven that, $\mu_\text{r}=1.61,\mu_\text{v}=1.65,\omega=0.07\ \text{and}\ \delta_\text{y}=4^\circ$
Now, $\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}$
$\Rightarrow0.07=\frac{1.65-1.61}{\mu_\text{y}-1}$
$\Rightarrow\mu_\text{y}-1=\frac{0.04}{0.07}=\frac{4}{7}$
Again,$\delta=(\mu-1)\text{A}$
$\Rightarrow\text{A}=\frac{\delta_\text{y}}{\mu_\text{y}-1}=\frac{4}{\Big(\frac{4}{7}\Big)}=7^\circ$
View full question & answer→Question 993 Marks
A candle flame 1.6cm high is imaged in a ball bearing of diameter 0.4cm. If the ball bearing is 20cm away from the flame, find the location and the height of the image.
Answer
Height of the object AB = 1.6cm
Diameter of the ball bearing = d = 0.4cm
⇒ R = 0.2cm
Given, u = 20cm
We know, $\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{2}{\text{R}}$
Putting the values according to sign conventions $\frac{1}{-20}+\frac{1}{\text{v}}=\frac{2}{0.2}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{20}+10=\frac{201}{20}\Rightarrow\text{v}=0.1\text{cm}=1\text{mm}$ inside the ball bearing.
Magnification $=\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\frac{\text{v}}{\text{u}}=-\frac{0.1}{-20}=\frac{1}{200}$
$\Rightarrow\text{A}'\text{B}'=\frac{\text{A}\text{B}}{200}=\frac{16}{200}=+0.008\text{cm}=+0.8\text{mm}.$ View full question & answer→Question 1003 Marks
A certain material has refractive indices 1.56, 1.60 and 1.68 for red, yellow and violet light respectively.
- Calculate the dispersive power.
- Find the angular dispersion produced by a thin prism of angle 6° made of this material.
AnswerGiven that,
$\mu_\text{r}=1.56,\mu_\text{y}=1.60,\ \text{and}\ \mu_\text{v}=1.68$
- Dispersive power $=\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{1.68-1.56}{1.60-1}$
$=\frac{0.12}{0.60}=0.2$
- Angular dispersion $=(\mu_\text{v}-\mu_\text{r})\text{A}=0.12\times6^\circ=7.2^\circ$
View full question & answer→Question 1013 Marks
The near point and the far point of a child are at 10cm and 100cm. If the retina is 2.0cm behind the eye-lens, what is the range of the power of the eye-lens?
AnswerThe child has near point and far point 10cm and 100cm respectively.Since, the retina is 2cm behind the eye-lens, v = 2cm
For near point u = -10cm = -0.1m, v = 2cm = 0.02m
So, $\frac{1}{\text{f}_\text{near}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{-1}=50+10=60\text{D}$
For far point, u = -100cm = = -1m,
v = 2cm = 0.02m
So, $\frac{1}{\text{f}_\text{far}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-1}=50+1=51\text{D}$
So, the rage of power of the eye-lens is +60D to +51D.
View full question & answer→Question 1023 Marks
Two lenses of power 10D and -5D are placed in contact.
- Calculate the power of lens combination.
- Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?
Answer
- Given $P_1 = 10\ D,$
$P_2 = -5D$
Power of Combination, $P = p_1 + P_2 = 10D - 5D = 5D$
- Focal length (Convergent lens) $\text{f}=\frac{1}{\text{P}}=\frac{1}{5}\text{m}$
$=0.20\text{m}=20\text{cm}$
(Convergent lens)
Magnification $\text{m}=\frac{\text{v}}{\text{u}}=+2$
$\Rightarrow\text{v}=2\text{u}$
From lens formula (u is negative) $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{20}=\frac{1}{2\text{u}}-\frac{1}{\text{u}}$
$\Rightarrow-\frac{1}{2\text{u}}=\frac{1}{20}$
$\Rightarrow\text{u}=-10\text{cm}.$ View full question & answer→Question 1033 Marks
A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the mean ray is zero. The refractive index of flint and crown glasses for the mean ray are 1.620 and 1.518 respectively. If the refracting angle of the flint prism is 6.0°, what would be the refracting angle of the crown prism?
AnswerGiven that,
Refractive index of flint glass $=\mu_\text{f}=1.620$
Refractive index of crown glass $=\mu_\text{c}=1.518$
Refracting angle of flint prism $=\text{A}_\text{f}=6.0^\circ$
For zero net deviation of mean ray
$(\mu_\text{f}-1)\text{A}_\text{f}=(\mu_\text{c}-1)\text{A}_\text{c}$
$\Rightarrow\text{A}_\text{c}=\frac{\mu_\text{f}-1}{\mu_\text{c}-1}\text{A}_\text{f}=\frac{1.620-1}{1.518-1}(6.0)^\circ=7.2^\circ$
View full question & answer→Question 1043 Marks
You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
AnswerYes, plane and convex mirrors can form real images if the object is virtual i.e., rays incident on the mirror is convergent as shown in figs. (i) and (ii).
View full question & answer→Question 1053 Marks
Find the position of the image formed by the lens combination given in the Fig. $9.20$.
AnswerImage formed by the first lens
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
$\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}$
or $ v_1=15 \ cm$
The image formed by the first lens serves as the object for the second.
This is at a distance of $(15-5) \ cm =10 \ cm$ to the right of the second lens.
Though the image is real, it serves as a virtual object for the second lens,
Which means that the rays appear to come from it for the second lens.
$\frac{1}{v_2}-\frac{1}{10}=\frac{1}{-10}$
or $ v_2=\infty$
The virtual image is formed at an infinite distance to the left of the second lens.
This acts as an object for the third lens.
$\frac{1}{v_3}-\frac{1}{u_3}=\frac{1}{f_3}$
or $ \frac{1}{v_3}=\frac{1}{\infty}+\frac{1}{30}$
or $ v_3=30 \ cm$
The final image is formed $30 \ cm$ to the right of the third lens.
View full question & answer→Question 1063 Marks
(i) If $f=0.5 m$ for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$. Its focal length is $12 cm$. What is the refractive index of glass? (iii) A convex lens has $20 cm$ focal length in air. What is focal length in water? (Refractive index of air-water $=1.33$, refractive index for air-glass $=1.5$.)
Answer(i) Power $=+2$ dioptre.
(ii) Here, we have $f=+12 cm , R_1=+10 cm , R_2=-15 cm$.
Refractive index of air is taken as unity.
We use the lens formula of Eq. (9.22). The sign convention has to be applied for $f, R_1$ and $R_2$.
Substituting the values, we have
$
\frac{1}{12}=(n-1)\left(\frac{1}{10}-\frac{1}{-15}\right)
$
This gives $n=1.5$.
(iii) For a glass lens in air, $n_2=1.5, n_1=1, f=+20 cm$. Hence, the lens formula gives
$
\frac{1}{20}=0.5\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$
For the same glass lens in water, $n_2=1.5, n_1=1.33$. Therefore,
$
\frac{1.33}{f}=(1.5-1.33)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$
Combining these two equations, we find $f=+78.2 cm$.
View full question & answer→Question 1073 Marks
Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R=2 m$. If the jogger is running at a speed of $5 m s ^{-1}$, how fast the image of the jogger appear to move when the jogger is (a) $39 m$, (b) $29 m$, (c) $19 m$, and (d) $9 m$ away.
AnswerFrom the mirror equation, Eq. (9.7), we get
$
v=\frac{f u}{u-f}
$
For convex mirror, since $R=2 m , f=1 m$. Then
for $u=-39 m , v=\frac{(-39) \times 1}{-39-1}=\frac{39}{40} m$
Since the jogger moves at a constant speed of $5 m s ^{-1}$, after $1 s$ the position of the image $v$ (for $u=-39+5=-34)$ is $(34 / 35) m$.
The shift in the position of image in $1 s$ is
$
\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}=\frac{1}{280} m
$
Therefore, the average speed of the image when the jogger is between $39 m$ and $34 m$ from the mirror, is $(1 / 280) m s ^{-1}$
Similarly, it can be seen that for $u=-29 m ,-19 m$ and $-9 m$, the speed with which the image appears to move is
$
\frac{1}{150} m s ^{-1}, \frac{1}{60} m s ^{-1} \text { and } \frac{1}{10} m s ^{-1} \text {, respectively. }
$
Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.
View full question & answer→Question 1083 Marks
An object is placed at $(i) 10 \ cm, (ii) 5 \ cm$ in front of a concave mirror of radius of curvature $15 \ cm$. Find the position, nature, and magnification of the image in each case.
AnswerThe focal length $f=-15 / 2 \ cm =-7.5 \ cm$
$(i)$ The object distance $u=-10 \ cm$.
Then Eq. $(9.7)$ gives
$\frac{1}{v}+\frac{1}{-10}=\frac{1}{-7.5}$
$\text { or } v=\frac{10 \times 7.5}{-2.5}=-30 \ cm$
The image is $30 \ cm$ from the mirror on the same side as the object.
Also, magnification $m=-\frac{v}{u}$
$=-\frac{(-30)}{(-10)}=-3$
The image is magnified, real and inverted.
$(ii)$ The object distance $u=-5 \ cm$.
Then from Eq. $(9.7),$
$\frac{1}{v}+\frac{1}{-5}=\frac{1}{-7.5}$
or $ v=\frac{5 \times 7.5}{(7.5-5)}=15 \ cm$
This image is formed at $15 \ cm$ behind the mirror. It is a virtual image.
Magnification $m=-\frac{v}{u}$
$=-\frac{15}{(-5)}=3$
The image is magnified, virtual and erect.
View full question & answer→Question 1093 Marks
- For a glass prism $(\mu =\sqrt{3} )$ the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism.
- Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index $\mu = \sqrt{3} . $
Answer- $\mu = \frac{\sin(\frac{\text{A} + \text{D}}{2})}{\sin\frac{\text{A}}{2}}$
$ = \frac{\sin(\frac{2\text{A}}{2})}{\sin\frac{\text{A}}{2}} =2 \cos\text{A}\sqrt{2} = \sqrt{3}$$\therefore\text{A} = 60 ^{o}$- $\mu = \sqrt{3} = \frac{1}{\sin\text{i}_{c}}$
$\therefore\sin\text{i}_{c} = \frac{1}{\sqrt{3}}\cong0.58$
Lies between 30oand 45o
Hence, TIR takes place.
Alternate Answer
$\sin\text{C} = \frac{1}{\sqrt{3}}$which is less than$\frac{1}{\sqrt{2}}$
$\therefore $Angle of incidence > ic
$\therefore\text{TIR}$ View full question & answer→