4 Marks Questions

Question 14 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?

Answer

Given that :Size of the object $h=3.0 cm$
Distance of object from concave lens $=u=-14 cm$
Focal length of concave lens $=f=-21 cm$
Distance of image $v=$ ?
Size of the image $h^{\prime}=$ ?
(i) Using relation $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
∴ $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ or $\frac{1}{v}=\frac{1}{-21}+\frac{1}{-14}$
$=\frac{-1}{21}-\frac{1}{14}=\frac{-2-3}{42}$
or $\frac{1}{v}=\frac{-5}{42}$
or $v=-\frac{42}{5}=-8.4 cm$
Image will be imaginary will be at distance 8.4 cm from lens towards object.
We know that:
Magnification $m=\frac{h^{\prime}}{h}=\frac{\text { Size of image }}{\text { Size of object }}$
But $\frac{h^{\prime}}{h}=\frac{v}{u}$
∴ $h^{\prime}=\frac{v}{u} \times h=\frac{-8.4}{-14} \times 3=1.8 cm$
Therefore size of image will be small, virtual and erect
(ii) When $u \rightarrow \infty$ then $v=f$ but cannot go beyond $f$, while
$m=\frac{v}{u}=0$
when object go to away from lens.
Let $u=f=-21 cm$
then $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ or $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=-\frac{1}{21}-\frac{1}{21}=\frac{-2}{21}$
then $v=-\frac{21}{2}=-10.5 cm$
but not at infinity

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Question 24 Marks

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40″ deg What is the refractive index of the material of the prism? The refracting angle of the prism is 60 deg If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer

Given that:
$A =60^{\circ}, D _m=40^{\circ}, n=?$
${ }_a n_g=\sin \frac{\left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}$
$=\frac{\sin \left(\frac{60^{\circ}+40^{\circ}}{2}\right)}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}$
${ }_a n_g=\frac{0.766}{0.54}=1.532$
If prism is placed in water
${ }_w n_g=\frac{\sin \left(\frac{ A + D _m^{\prime}}{2}\right)}{\sin \frac{ A }{2}}$
or $\frac{{ }_a n_g}{{ }_a n_w}=\frac{\sin \left(\frac{60^{\circ}+ D _m^{\prime}}{2}\right)}{\sin 30^{\circ}}$
or $\sin \left(\frac{60^{\circ}+ D _m^{\prime}}{2}\right)=\sin 30^{\prime} \times \frac{{ }_a n_g}{{ }_a n_w}$
or $\sin \left(\frac{60^{\circ}+ D _m^{\prime}}{2}\right)=\frac{1}{2} \times \frac{1.532}{1.33}$
or $\sin \left(\frac{60^{\circ}+ D _m^{\prime}}{2}\right)=0.5759$ or $30+\frac{ D _m^{\prime}}{2}=35^{\circ} 10^{\prime}$
or $\frac{ D _m^{\prime}}{2}=35^{\circ} 10^{\prime}-30=5^{\circ} 10^{\prime}$ or $D _m^{\prime}=10^{\circ} 20^{\prime}$

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Question 34 Marks

Figure (a) and (b) show refraction of a ray in air incident at 60 deg with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45 deg with the normal to a water-glass interface [Fig. (c)].

Answer

From fig. (a): Incidence angle = i = 60°
Refraction angle = r = 35°
${ }_a n_g=$ Refractive index of glass with respect to air $=$ ?
Relation, ${ }_a n_g=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ}}{\sin 35^{\circ}}$
$=\frac{0.8660}{0.5736}=1.51$
From fig. (b) : $i=60^{\circ}, r=47^{\circ}$
${ }_a n_w=$ Refractive index of water w.r.t. air $=$ ?
We know that,
${ }_a n_w=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ}}{\sin 47^{\circ}}=\frac{0.8660}{0.7314}=1.184$
From fig. (c) Angle of incidence $=i_1-45^{\circ}$
Angle of refraction $=r=$ ?
${ }_w n_g=$ Refractive index of glass w.r.t. water $=$ ?
We know that :${ }_a n_g={ }_a n_w \times{ }_w n_g$
or ${ }_w n_g=\frac{{ }_a n_g}{{ }_a n_w}=\frac{1.51}{1.184}=1.275$
$\therefore$ Relation, $\quad{ }_w n_g=\frac{\sin i}{\sin r} \Rightarrow 1.275=\frac{\sin 45^{\circ}}{\sin r}$
or $\sin r=\frac{\sin 45}{1.275}=\frac{1 / \sqrt{2}}{1.275}$
$\sin r=\frac{0.707}{1.275}=0.5545=\sin 34^{\circ} ; r=34^{\circ}$

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Question 44 Marks

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer

Given that: Angle of prism = A = 60°
Refractive index of material of prism = n = 1.524.
When the light ray PQ incident on face AB then this is total internal reflection at face AC and at face AC, incident angle will be i, which is equal to the critical angle i of material of prism.
Now 60°+90°− r+90°−i_c = 180°
or $\quad 240^{\circ}-r-i_c=180^{\circ}$
or $r=60^{\circ}-i_c$

Now, $\quad \sin i_c=\frac{1}{n}=\frac{1}{1.524}$
$i_c=\sin ^{-1}\left(\frac{1}{1.524}\right)=\sin ^{-1}(0.6562)=41^{\circ}$
$\therefore \quad r=60^{\circ}-41^{\circ}=19^{\circ}$
Using Snell's law :
$n=\frac{\sin i}{\sin r} \quad \therefore \sin i=\sin 19^{\circ} \times 1.524$
$=0.3256 \times 1.524$
$=0.4962$
or $i=\sin ^{-1}(0.4962)=29.75^{\circ}$
$\therefore \quad i \approx 30^{\circ}$

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Question 54 Marks

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer

Given that:
Distance of object (needle) from convex mirror
u = - 12 cm
Focal length of convex mirror f = 15 cm
Size of object h = 4.5 cm
Position of image = v = ?
Using the formula $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ or $\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$
$\frac{1}{v}=\frac{1}{15}-\left(-\frac{1}{12}\right)=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}$
or $\frac{1}{v}=\frac{9}{60}=\frac{3}{20}$
$\therefore \quad v=\frac{20}{3} cm=6.67 cm \simeq 6.7 cm$.
Here value of v is positive. From this it is clear that image is formed behind the mirror.
And image will be become virtual, small and erect.
Now using magnification $m=-\frac{v}{u}=\frac{h^{\prime}}{h}$
$\therefore \quad h^{\prime}=\left(-\frac{v}{u}\right) \times h$
Put the values $h^{\prime}=\left(\frac{-20 / 3}{-12}\right) \times 4.5$
$=\frac{20}{3} \times \frac{1}{12} \times \frac{9}{2}=+2.5 cm$
Therefore size of image h' = 2.5 cm.
Magnification, $m=\frac{h^{\prime}}{h}=\frac{2.5}{4.5}=\frac{5}{9}$
As the needle is moved away from the mirror the image moves closer to the focus, but becomes smaller and smaller in size upto the focus. That is, as u → ∞ v → f (but does not move beyond the focus) while m → 0

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Question 64 Marks

(a) Figure shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angle of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

(b) What is the answer if there is no outer covering of the pipe?

Answer

(a) $\sin I _c=n=\frac{\sin 1}{\sin r}=\frac{1.44}{1.68}=0.8571$
∴ $\begin{array}{l} I _c=\sin ^{-1}(0.8571)=59^{\circ} \\ I _c=59^{\circ}\end{array}$
When $I > I _c$ i.e. $I >59$ since value of I lies between $59^{\circ}$ to $90^{\circ}$. Here, there will be total internal reflection and value of angle $r$ will be between 0 to $31^{\circ}$.
∴ $r_{\max }=31^{\circ}$
Now refractive index of glass fibre w.r.t. air
$n=\frac{\sin I _{\text {max }}}{\sin r_{\text {max }}}$ From Snell's law
$\Rightarrow 1.68=\frac{\sin I _{\max }}{\sin 31^{\circ}} \Rightarrow \sin I _{\max }=1.68 \sin 31^{\circ}$
$=1.68 \times 0.515=0.8652$
or $I_{\max }=\sin ^{-1}(0.8652)=59.9^{\circ} \simeq 60^{\circ}$
Therefore, $0< I <60^{\circ}$. At this range, the incident rays will be totally reflected in pipe.
(b) If there is no external coating on the pipe, then the refraction from glass to air inside the pipe will be:
$\sin I _c^{\prime}=\frac{1}{1.68}=0.5952$
∴ $I _C^{\prime}=\sin ^{-1}(0.5952) \Rightarrow I _C^{\prime}=36.5^{\circ}$
Hence for total internal reflection $i>36.5^{\circ}$ i.e. should be $36.5^{\circ}$.
Now will be $\quad I =90^{\circ}$ to $r=36.5^{\circ}$
$\because \quad \frac{\sin I}{\sin r}=n=1.68$
$\therefore r_{\max }=90^{\circ}-36.5^{\circ}=53.5^{\circ}$ which is greater than $I _c^{\prime}$.
Hence for total internal reflection $i>36.5^{\circ}$ i.e. should be $36.5^{\circ}$
Now will be $I =90^{\circ}$ to $r=36.5^{\circ}$
$\because \quad \frac{\sin 1}{\sin r}=n=1.68$
$\therefore r_{\max }=90^{\circ}-36.5^{\circ}=53.5^{\circ}$ which is greater than $I _c^{\prime}$.
$\therefore$ The rays incident on the axis on the range 0 to $90^{\circ}$
pass through the pipe, will be completely internal from within.

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Question 74 Marks

(a) 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?
(b) 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$3.48 \times 10^6$ m , and the radius of lunar orbit is $3.8 \times 10^8 m$.

Answer

Given that :
Focal length of objective lens of telescope $=f_o=15 m$
$\therefore \quad$ Focal length of eyepiece $f_e=1.0 cm$
$=10^{-2} m$
(a) Angular magnification $=-\frac{f_o}{f_e}=\frac{-15}{10^{-2}}$
$=-1500=|1500|$
(b) Internal angle by moon $(\alpha)=\frac{\text { diameter of moon }}{\text { radius of moon }}$
$\alpha=\frac{3.48 \times 10^6}{3.8 \times 10^8}$ .... (i)
As second found by objective lens, internal angle by image I of moon :
$\alpha=\frac{\text { size of image }}{\text { focal length of objective lens }}$
$\alpha=\frac{h^{\prime}}{15}...(ii)$
From eqn. (i) and (ii),
$\Rightarrow \quad \frac{h^{\prime}}{15}=\frac{3.48 \times 10^6}{3.8 \times 10^8}$ or $h^{\prime}=\frac{3.48 \times 10^6 \times 15}{3.8 \times 10^8}$
$\Rightarrow \frac{3.48 \times 15}{380}=\frac{52.2}{380} m \Rightarrow \frac{52.2 \times 100}{380} cm=13.736$
$\simeq 13.74 cm$

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Question 84 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 difference 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?

Answer

Given that: Size of the object = h = 2.5 cm
Distance of the object from concave mirror u = - 27 cm.
Radius of curvature of concave mirror = R = - 36 cm.
$\therefore$ Focal length of concave mirror $f=\frac{1}{2} R$
⇒$R =-\frac{36}{2}=-18 cm$.
(i) Distance of object from mirror = v = ?
Using formula $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ or
$\frac{1}{v}=\frac{1}{-18}-\frac{1}{-27}$
$=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}$
$\frac{1}{v}=\frac{1}{54} \therefore v=-54 cm$
The ve sign indicates that the image is formed in front of the mirror and it is formed on the same side on which the object is placed. Hence the screen should be placed in front of the mirror at a distance of 54 cm.
(ii) To find out the nature and size of the image, using the relation m
$=\frac{\text { size of the image }\left(h^{\prime}\right)}{\text { size of the object }(h)}=-\frac{v}{u}$
$\frac{h^{\prime}}{25}=-\left(\frac{-54}{-27}\right)=\frac{-2}{1}$
or $h^{\prime}=-2.5 \times 2=-5 cm$
From this, it is clear that the image is real, inverted, large and magnified.
(iii) If the candle is brought near the mirror then the screen will have to be moved further and further i.e. screen will have to be kept at infinity. But when the distance of the candle is less than the focal length of the mirror like for $u \rightarrow f , v \rightarrow \infty, u \rightarrow f$ then the image will be imaginary (virtual).

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