Question 512 Marks
Can a plane mirror ever form a real image?
Answer
View full question & answer→It object is virtual image is real.
Questions · Page 2 of 2
2 Marks Questions
Question 522 Marks
A 1cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5cm. Find its distance from the mirror if the image formed is 0.6cm in size.
Answer
View full question & answer→$\text{m}=-\frac{\text{v}}{\text{u}}=0.6$ and $\text{f}=7.5\text{cm}=\frac{15}{2}\text{cm}$
From mirror equation,
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{0.6\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{5}{3\text{u}}-\frac{1}{\text{u}}=\frac{5}{15}$
$\Rightarrow\text{u}=5\text{cm}$
From mirror equation,
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{0.6\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{5}{3\text{u}}-\frac{1}{\text{u}}=\frac{5}{15}$
$\Rightarrow\text{u}=5\text{cm}$
Question 532 Marks
A concave lens is placed in water. Will there be any change in focal length? Give reason.
Answer
View full question & answer→Focal length of lens in water $\text{f}_{\text{w}}=\frac{\text{n}_{\text{g}}-1}{\frac{\text{n}_{\text{g}}-1}{\text{nw}}-1}$
$\text{As n}_{\text{g}}>\text{n}_{\text{w}},\frac{\text{n}_{\text{g}}}{\text{n}_{\text{w}}}>1,\text{ so }\text{f}_{\text{w}}>\text{f}_{\text{a}}$
$\text{As n}_{\text{g}}>\text{n}_{\text{w}},\frac{\text{n}_{\text{g}}}{\text{n}_{\text{w}}}>1,\text{ so }\text{f}_{\text{w}}>\text{f}_{\text{a}}$
Question 542 Marks
Can virtual image be formed on the retina in a seeing process?
Answer
View full question & answer→The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.
Question 552 Marks
Can a virtual image be photographed by a camera?
Answer
Yes, when you stand in front of plane mirror image is virtual and can be photographed.
View full question & answer→Yes, when you stand in front of plane mirror image is virtual and can be photographed.Question 562 Marks
An optical fibre $(\mu=1.72)$ is surrounded by a glass coating $(\mu=1.50).$ Find the critical angle for total internal/ reflection at the fibre-glass interface.
Answer
View full question & answer→For calculation of critical angle,
$\frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^\circ}=\frac{15}{1.72}$
$=\frac{75}{86}$
$\Rightarrow\theta_\text{C}=\sin^{-1}\Big(\frac{75}{86}\Big)$
$\frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^\circ}=\frac{15}{1.72}$
$=\frac{75}{86}$
$\Rightarrow\theta_\text{C}=\sin^{-1}\Big(\frac{75}{86}\Big)$
Question 572 Marks
A nearsighted person cannot clearly see beyond 200cm. Find the power of the lens needed to see objects at large distances.
Answer
View full question & answer→For the near sighted person,$\text{u}= \infty$ and $\text{v}=-200\text{cm}=-2\text{m}$
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$
$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$
So, power of the lens is -0.5D
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$
$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$
So, power of the lens is -0.5D
Question 582 Marks
A convex lens made of a material of refractive index $n_1$ is kept in a medium of refractive index $n_2$. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if: (i) $\mathrm{n}_1>\mathrm{n}_2$ (ii) $\mathrm{n}_1=\mathrm{n}_2$ (iii) $\mathrm{n}_1<\mathrm{n}_2$.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}\Big)$
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}\Big)$
Answer
View full question & answer→- In case (i) $\mathrm{n}_1>\mathrm{n}_2$, the lens behaves as convergent lens.
- In case (ii) $n_1=n_2$, the lens behaves as a plane plate.
- In case (iii) $\mathrm{n}_1<\mathrm{n}_2$, the lens behaves as a divergent lens.
Question 592 Marks
A point source is placed at a depth h below the surface of water (refractive index = μ).
- Show that light escapes through a circular area on the water surface with its centre directly above the point source.
- Find the angle subtended by a radius of the area on the source.
Answer
$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{\sin\theta_\text{C}}{\sqrt{1-\sin^2\theta_\text{C}}}=\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}} \ \Big(\because \ \sin\theta_\text{C}=\frac{1}{\mu}\Big)$
$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{\mu^2-1}}$ or $\text{x}=\frac{\text{h}}{\sqrt{\mu^2-1}}$
So, light escapes through a circular area on the water surface directly above the point source.
View full question & answer→
- Let, x = radius of the circular area
$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{\sin\theta_\text{C}}{\sqrt{1-\sin^2\theta_\text{C}}}=\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}} \ \Big(\because \ \sin\theta_\text{C}=\frac{1}{\mu}\Big)$
$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{\mu^2-1}}$ or $\text{x}=\frac{\text{h}}{\sqrt{\mu^2-1}}$
So, light escapes through a circular area on the water surface directly above the point source.
- Angle subtained by a radius of the area on the source, $\theta_\text{C}=\sin^{-1}\Big(\frac{1}{\mu}\Big)$
Question 602 Marks
Light falls from glass $(\mu=1.5)$ to air. Find the angle of incidence for which the angle of deviation is 90°.
Answer
View full question & answer→Since, $\mu=1.5,$ Critial angle $=\sin^{-1}\Big(\frac{1}{\mu}\Big)=\sin^{-1}\Big(\frac{1}{1.5}\Big)=41.8^{\circ}$
We know, the maximum attainable deviation in refraction is (90° - 41.8°) = 47.2°
So, in this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° - 2i = 90°
⇒ 2i = 90° ⇒ i = 45°.
We know, the maximum attainable deviation in refraction is (90° - 41.8°) = 47.2°
So, in this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° - 2i = 90°
⇒ 2i = 90° ⇒ i = 45°.
Question 612 Marks
A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?
Answer
View full question & answer→The simple microscope has magnification of 5 for normal relaxed eye (D = 25cm).Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.
Question 622 Marks
A concave mirror and a converging lens have the same focal length in air. Which one of the two will have greater focal length when both are immersed in water?
Answer
View full question & answer→Converging lens; the focal length of a spherical mirror remains unaffected.
For converging lens $\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
When it is immersed in water $\mathrm{n}_2$ (in water) $>\mathrm{n}_2$ (air)
$\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)$ decreases hence focal length of converging lens increases in water.
For converging lens $\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
When it is immersed in water $\mathrm{n}_2$ (in water) $>\mathrm{n}_2$ (air)
$\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)$ decreases hence focal length of converging lens increases in water.
Question 632 Marks
A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
Answer
View full question & answer→No,
Question 642 Marks
A farsighted person cannot see objects placed closer to 50cm. Find the power of the lens needed to see the objects at 20cm.
Answer
View full question & answer→For the far sighted person,u = -20cm, v = -50cm
from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$
So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$
from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$
So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$
Question 652 Marks
A convex lens (n = 1.5) of focal length fa is immersed.
- In water n = 1.33 and.
- In carbon disulphide n = 1.6, how does the lens behave in the two cases?
Answer
View full question & answer→- When lens is immersed in water, it behaves as a convex lens but its focal length will increase.
- When convex lens is immersed in carbon-disulphide, it will behave as a concave lens.
Question 662 Marks
Can you ever have a situation in which a light ray goes undeviated through a prism?
Answer
View full question & answer→Yes.
Question 672 Marks
The equation $\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu-1}$ asderived for a prism having small refracting angle. Is it also valid for a prism of large refracting angle? Is it also valid for a glass slab or a glass sphere?
Answer
View full question & answer→Dispersive power depends on angular deviation, and angular deviation is valid only for a small refracting angle and a small angle of incidence. Therefore, dispersive power is not valid for a prism of large refracting angle. It is also not valid for a glass slab or a glass sphere, as it has a large refracting angle.
Question 682 Marks
A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?
- Looking at small germs.
- Looking at circular spots.
- Looking at a vertical tube containing some water.
Answer
If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.
View full question & answer→- Looking at a vertical tube containing some water
If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.
Question 692 Marks
A spherical surface of radius 30cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?
Answer
Since, paraxial rays become parallel after refraction i.e. image is formed at $\infty.$
$\text{v}=\infty, \ \mu_1=1.33, \ \text{u}=?, \ \mu_2=1.48, \ \text{R}=30\text{cm}$
$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.48}{\infty}-\frac{1.33}{\text{u}}=\frac{1.48-1.33}{30}\Rightarrow-\frac{1.33}{\text{u}}-\frac{0.15}{30}$
$\Rightarrow\text{u}=-266.0\text{cm}$
$\therefore$ Object should be placed at a distance of 266cm from surface (convex) on side A.
View full question & answer→Since, paraxial rays become parallel after refraction i.e. image is formed at $\infty.$
$\text{v}=\infty, \ \mu_1=1.33, \ \text{u}=?, \ \mu_2=1.48, \ \text{R}=30\text{cm}$
$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.48}{\infty}-\frac{1.33}{\text{u}}=\frac{1.48-1.33}{30}\Rightarrow-\frac{1.33}{\text{u}}-\frac{0.15}{30}$
$\Rightarrow\text{u}=-266.0\text{cm}$
$\therefore$ Object should be placed at a distance of 266cm from surface (convex) on side A.
Question 702 Marks
The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32cm below the top surface of the slab. Calculate the dispersive power of the material.
Answer
Given that, $\mu_\text{v}-\mu_\text{r}=0.014$
$\text{Again,}\ \mu_\text{y}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{2.00}{1.30}=1.515$
So, dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{0.014}{1.515-1}=0.027$
View full question & answer→Given that, $\mu_\text{v}-\mu_\text{r}=0.014$$\text{Again,}\ \mu_\text{y}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{2.00}{1.30}=1.515$
So, dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{0.014}{1.515-1}=0.027$
Question 712 Marks
Can the dispersive power $\omega=\frac{\mu_\text{v}-\mu\text{r}}{\mu-1}$ be negative? What is the sign of co if a hollow prism is immersed into water?
Answer
View full question & answer→No, it cannot be negative, as the refractive index for violet light is always greater than that for red light. Also, refractive index is inversely proportional to $\lambda^2$. The sign of wwill be positive, as $\mu$ is still greater than 1 and es $\mu_\text{v}>\mu_\text{r}.$
Question 722 Marks
The minimum deviations suffered by red, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.
Answer
View full question & answer→Given that, $\delta_\text{r}=38.4^\circ,\delta_\text{y}=38.7^\circ\ \text{and}\ \delta_\text{v}=39.2^\circ$
Dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}$ $=\frac{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)-\Big(\frac{\delta_\text{r}}{\text{A}}\Big)}{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)}\ [\because\delta=(\mu-1)\text{A}]$
$=\frac{\delta_\text{v}-\delta_\text{r}}{\delta_\text{y}}=\frac{39.2-3.84^\circ}{38.7}=0.0204$
Dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}$ $=\frac{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)-\Big(\frac{\delta_\text{r}}{\text{A}}\Big)}{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)}\ [\because\delta=(\mu-1)\text{A}]$
$=\frac{\delta_\text{v}-\delta_\text{r}}{\delta_\text{y}}=\frac{39.2-3.84^\circ}{38.7}=0.0204$
Question 732 Marks
A simple microscope using a single lens often shows coloured image of a white source. Why?
Answer
View full question & answer→A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.
Question 742 Marks
The refractive index of a material of a concave lens is $n_1$. It is immersed in a medium of refractive index $n_2 A$ parallel beam of light is incident on the lens. Trace the path of emergent rays when (i) $n_2=n_1$ (ii) $n_2>n_1$ (iii) $n_2<n_1$.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
Answer
View full question & answer→$\text{For} \text{ n}_1 = \text{n}_2\text{ f} = \infty$
$\text{For} \text{ n}_1 < \text{n}_2 \text{ f} > 0$
$\text{For} \text{ n}_1 > \text{n}_2 \text{ f} < 0$
The path of rays in three cases is shown in fig.
$\text{For} \text{ n}_1 < \text{n}_2 \text{ f} > 0$
$\text{For} \text{ n}_1 > \text{n}_2 \text{ f} < 0$
The path of rays in three cases is shown in fig.
Question 752 Marks
Suggest a method to produce a rainbow in your house.
Answer
View full question & answer→A rainbow can be produced using a prism. Another way of producing a rainbow is to dip a mirror inside water, keeping it inclined along the wall of a tumbler. The light coming from water after reflecting from the mirror will give a rainbow.
Question 762 Marks
The angular magnification of a system is less than one. Does it mean that the image formed is inverted?
Answer
View full question & answer→No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.
Question 772 Marks
Find the maximum angle of refraction when a light ray is refracted from glass $(\mu=1.50)$ to air.
Answer
View full question & answer→From the definition of critical angle, if refracted angle is more than 90°, then reflection occurs, which is known as total internal reflection.
So, maximum angle of refraction is 90°.
So, maximum angle of refraction is 90°.
Question 782 Marks
If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?
Answer
View full question & answer→Slower $\text{V}_{\text{img}}=\text{m}^2\text{v}_0$ [m < 1 as far away]
It objects moves from infinity to 2f as distance moved by object is more in same time hence velocity of object is more.
It objects moves from infinity to 2f as distance moved by object is more in same time hence velocity of object is more.
Question 792 Marks
A Galilean telescope is 27cm long when focused to form an image at infinity. If the objective has a focal length of 30cm, what is the focal length of the eyepiece?
Answer
View full question & answer→For the given Galilean telescope, (When the image is formed at infinity) $\mathrm{f}_0=30 \mathrm{~cm}, \mathrm{~L}=27 \mathrm{~cm}$ Since $L=f_0-\left|f_e\right|$
[Since, concave eyepiece lens is used in Galilean Telescope]
$\Rightarrow f_e=f_0-L=30-27=3 \mathrm{~cm}$
[Since, concave eyepiece lens is used in Galilean Telescope]
$\Rightarrow f_e=f_0-L=30-27=3 \mathrm{~cm}$
Question 802 Marks
A light ray is incident normally on the face AB of a right-angled prism ABC $(\mu=1.50)$ as shown in figure. What is the largest angle $\phi$ for which the light ray is totally reflected at the surface AC? 
Answer
Let $\theta_\text{c}$ be the critical angle for the glass
$\frac{\sin\theta_{\text{c}}}{\sin90^{\circ}}=\frac{1}{\text{x}}\Rightarrow\sin\theta_{\text{c}}=\frac{1}{1.5}=\frac{2}{3}\Rightarrow\theta_{\text{c}}=\sin^{-1}\Big(\frac{2}{3}\Big)$
From figure, for total internal reflection, $90^{\circ}-\phi>\theta_{\text{c}}$
$\Rightarrow\phi<90^{\circ}-\theta_{\text{c}}\Rightarrow\phi<\cos^{-1}\Big(\frac{2}{3}\Big)$
So, the largest angle for which light is totally reflected at the surface is $\cos^{-1}\Big(\frac{2}{3}\Big).$
View full question & answer→Let $\theta_\text{c}$ be the critical angle for the glass
$\frac{\sin\theta_{\text{c}}}{\sin90^{\circ}}=\frac{1}{\text{x}}\Rightarrow\sin\theta_{\text{c}}=\frac{1}{1.5}=\frac{2}{3}\Rightarrow\theta_{\text{c}}=\sin^{-1}\Big(\frac{2}{3}\Big)$
From figure, for total internal reflection, $90^{\circ}-\phi>\theta_{\text{c}}$
$\Rightarrow\phi<90^{\circ}-\theta_{\text{c}}\Rightarrow\phi<\cos^{-1}\Big(\frac{2}{3}\Big)$
So, the largest angle for which light is totally reflected at the surface is $\cos^{-1}\Big(\frac{2}{3}\Big).$
Question 812 Marks
An extended object is placed at a distance of 5.0cm from a convex lens of focal length 8.0cm.
- Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens.
- Find the position of the image from the lens formula and see how close the drawing is to the correct result.
Answer
$\Rightarrow\frac{1}{8}-\frac{1}{5}=\frac{-3}{40}$
$\Rightarrow\text{v}=-13.3\text{cm}$ (virtual image).
View full question & answer→
- Given, u = -5cm, f = 8cm
$\Rightarrow\frac{1}{8}-\frac{1}{5}=\frac{-3}{40}$
$\Rightarrow\text{v}=-13.3\text{cm}$ (virtual image).
Question 822 Marks
A telescope has been adjusted for relaxed eye. You are asked to adjust it for least distance of distinct vision, then how will you change the distance between two lenses?
Answer
View full question & answer→For relaxed eye, $\text{L} = \text{f}_0 +\text{f}_{\text{e}}$
For least distance of distinct vision $\text{L}' = \text{f}_0 + \text{u}_{\text{e}}, \text{u}_{\text{e}} < \text{f}_{\text{e}}$
Therefore, L' < L, that is, the distance will be decreased.
For least distance of distinct vision $\text{L}' = \text{f}_0 + \text{u}_{\text{e}}, \text{u}_{\text{e}} < \text{f}_{\text{e}}$
Therefore, L' < L, that is, the distance will be decreased.
Question 832 Marks
Why is there no dispersion in the light refracted through a rectangular glass slab?
Answer
View full question & answer→It can be considered to be equivalent to two prisms in reverse way. In the case of a rectangular glass slab the rays of all colours combine together in the reverse prism. Hence, there is no dispersion.
Question 842 Marks
An object is placed at the principal focus of a concave lens of focal length f. Where will its image be formed?
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
Answer
View full question & answer→Here u = -f and for a concave lens f = -f
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
That image will be formed between optical centre and focus of lens; towards the side of the object.
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
That image will be formed between optical centre and focus of lens; towards the side of the object.
Question 852 Marks
A person wears glasses of power -2.5D. Is the person farsighted or nearsighted? What is the far point of the person without the glasses?
Answer
View full question & answer→The person wears glasses of power -2.5DSo, the person must be near sighted
$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$
$\Rightarrow \text{v}=-40\text{cm}$
So, the far point of the person is 40cm.
$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$
$\Rightarrow \text{v}=-40\text{cm}$
So, the far point of the person is 40cm.
Question 862 Marks
The focal length of a convex lens made of glass is 20cm. What will be its new focal length when placed in a medium of refractive index 1.25?
Answer
View full question & answer→$\frac{1}{\text{f}}=(\mu-1)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\frac{1}{20}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{\text{f}}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\text{f}'=40\text{cm}$
$\Rightarrow\frac{1}{20}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{\text{f}}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\text{f}'=40\text{cm}$
Question 872 Marks
The equation of refraction at a spherical surface is $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\text{R}}.$ Taking $\text{R}=\infty,$ show that this equation leads to the equation $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{\mu_2}{\mu_1}$ for refraction at a plane surface.
Answer
View full question & answer→$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\infty}$
$\frac{\mu_2}{\text{v}}=\frac{\mu_1}{\text{u}}$
$\Rightarrow\frac{\mu_2}{\mu_1}=\frac{\text{v}}{\text{u}}.$
$\frac{\mu_2}{\text{v}}=\frac{\mu_1}{\text{u}}$
$\Rightarrow\frac{\mu_2}{\mu_1}=\frac{\text{v}}{\text{u}}.$
Question 882 Marks
Light is incident from glass $(\mu=1.5)$ to air. Sketch the variation of the angle of deviation $\delta$ with the angle of incident i for 0 < i < 90°.
Answer
Refractive index of glass $\mu_{\text{g}}=1.5$
Given, 0° < i < 90°
Let, $\theta_\text{C}$ → Critical angle.
$\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\mu_{\text{a}}}{\mu_{\text{g}}}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^{\circ}}=\frac{1}{1.5}=0.66$
$\Rightarrow\text{C}=40^{\circ}48''$
The angle of deviation due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48''. The angle of deviation due to total internal reflection further increases for 40°48'' to 45° and then it decreases.
View full question & answer→Refractive index of glass $\mu_{\text{g}}=1.5$
Given, 0° < i < 90°
Let, $\theta_\text{C}$ → Critical angle.
$\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\mu_{\text{a}}}{\mu_{\text{g}}}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^{\circ}}=\frac{1}{1.5}=0.66$
$\Rightarrow\text{C}=40^{\circ}48''$
The angle of deviation due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48''. The angle of deviation due to total internal reflection further increases for 40°48'' to 45° and then it decreases.
Question 892 Marks
A right-angled crown glass prism with critical angle 41° is placed before an object, PQ in two positions as shown in the figures (i) and (ii). Trace the paths of the rays from P and Q passing through the prisms in the two cases.
Answer
View full question & answer→The formation of images is shown in figures (i) and (ii).
Question 902 Marks
The eyepiece of an astronomical telescope has a focal length of 10cm. The telescope is focused for normal vision of distant objects when the tube length is 1.0m. Find the focal length of the objective and the magnifying power of the telescope.
Answer
View full question & answer→For the given astronomical telescope in normal adjustment, $\mathrm{F}_{\mathrm{e}}=10 \mathrm{~cm}, \mathrm{~L}=1 \mathrm{~m}=100 \mathrm{~cm}$
So $, \mathrm{f}_0=\mathrm{L}-\mathrm{f}_{\mathrm{e}}=100-10=90 \mathrm{~cm}$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$
So $, \mathrm{f}_0=\mathrm{L}-\mathrm{f}_{\mathrm{e}}=100-10=90 \mathrm{~cm}$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$
Question 912 Marks
If three identical prisms are combined, is it possible to pass a beam that emerges undeviated? Undispersed?
Answer
View full question & answer→No, it is not possible even when prisms are be combined with their refractive angle reversed with respect to each other. There will be at least a net deviation and dispersion equal to the dispersion and deviation produced by a single prism.
Question 922 Marks
"Monochromatic light should be used to produce pure spectrum". Comment on this statement.
Answer
View full question & answer→No, monochromatic light cannot be used to produce a pure spectrum. A spectrum is produced when a light of different wavelengths is deviated through different angles and gets separated. Monochromatic light, on the other hand, has a single wavelength.
Question 932 Marks
How does the power of a convex lens vary, if the incident red light is replaced by violet light?
Answer
View full question & answer→Power of a lens increases if red light is replaced by violet light because.
$\text{P}=\frac{1}{\text{f}}=(_{\text{a}\text{n}_{\text{g}}-1})\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_1}\Big)$ and refractive index is maximum for violet light in visible region of spectrum.
$\text{P}=\frac{1}{\text{f}}=(_{\text{a}\text{n}_{\text{g}}-1})\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_1}\Big)$ and refractive index is maximum for violet light in visible region of spectrum.