Ray Optics and Optical Test

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Ray Optics and Optical Test - 66

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Weekly Quiz Competition

Question 1
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An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is :

A
$$1 m$$

B
$$0.5 m$$

C
$$0.24 m$$

D
$$2 m$$

Solution
Question 2
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A sphere of radius $$R$$ is placed in air such that its centre is at origin. A ray of light traveling along $$y = + \dfrac{{\sqrt 3 }}{2}R$$ (in x-y plane) is incident on this sphere.
If ray travels through sphere as shown in figure, then:

A
Refractive index of the sphere is 2

B
Refractive index of the sphere $$\sqrt 3 $$

C
Emergent ray travels along the line $$\sqrt 3 x + y = \sqrt 3 R$$

D
Total deviation suffered by the ray is $${45^o}$$

Solution

$$ \begin{array}{l} \text { equation of sphere } x^{2}+y^{2}=R^{2} \\ \text { at point, } \quad \begin{array}{rl} =R^{2} & y=+\dfrac{\sqrt{3}}{2} R \\ x^{2}+\left(\dfrac{\sqrt{3 R}}{2}\right)^{2}\\ x^{2} & =R^{2}-\dfrac{3 R^{2}}{4} \\ x=\dfrac{R}{2} \end{array} \end{array} $$

considers $$\triangle P H O$$

$$\begin{aligned} \tan \theta=& \dfrac{\dfrac{\sqrt{3} R}{2}}{\dfrac{3 R}{2}} \\ \tan \theta=\dfrac{1}{\sqrt{3}} \\ \theta &=30^{\circ} \end{aligned}$$
Consider $$\triangle \mathrm{PH} \mathrm{c}$$

$$\begin{array}{l}\tan \phi=\dfrac{\frac{\sqrt{3 P}}{2}}{\dfrac{R}{2}}\\\tan\phi=\sqrt{3}\\\phi=60^{\circ}\end{array}$$

So, incident angle at $$P=60^{\circ}$$

refracted angle at $$p=30^{\circ}$$

$$\mu_{1} \sin i=\mu_{2} \sin \gamma \quad$$ at point $$p$$

(1) $$\sin 60=\mu \sin 30$$

$$\qquad \mu=\sqrt{3}]-1$$

$$\mu_{1} \sin i=\mu_{2} \sin \gamma$$ at point 0

$$\begin{aligned} \sqrt{3} \sin 30 &=(1)\sin\alpha\\\alpha &=60^{\circ} \end{aligned}$$

slope of emergent ray $$=\tan (90+\alpha)$$

$$=-\cot 60=-\dfrac{1}{\sqrt{3}}$$

equation of emergent ray $$(x-R)-\dfrac{1}{\sqrt{3}}=y$$

By equation (1), option $$B$$ is correct, option $$c$$ is incorrect
Question 3
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A ray of light enters a spherical drop of water of refractive index $$\mu$$ as shown in the figure. The angle $$\phi$$ for which minimum deviation is produced will be given by :

A
$$\cos^{2}\phi = \dfrac {\mu^{2} + 1}{3}$$

B
$$\cos^{2}\phi = \dfrac {\mu^{2} - 1}{3}$$

C
$$\sin^{2}\phi = \dfrac {\mu^{2} + 1}{3}$$

D
$$\sin^{2}\phi = \dfrac {\mu^{2} - 1}{3}$$
Question 4
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One end of a glass rod of refractive index $$n=1.5$$ is spherical surface of radius of curvature $$R$$. The centre of the spherical surface lies inside the glass. A point object placed in air on the axis of the rod at the point $$P$$ has its real image inside glass at the point Q (see figure). A line joining the points P and Q cuts the surface at O such that $$OP=2OQ$$. The distance PO is

A
$$8R$$

B
$$7R$$

C
$$2R$$

D
None of these

Solution

Use formula for refraction at spherical surface
$$\cfrac { { \mu }_{ 2 } }{ v } -\cfrac { { \mu }_{ 1 } }{ u } =\cfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ R } \\ \cfrac { 1.5 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1.5-1 }{ +R } $$
Also, $$OP=2\left( OQ \right) \\ or\quad OQ=\cfrac { OP }{ 2 } \\ or\quad v=\cfrac { -u }{ 2 } $$
$$\therefore \cfrac { 1.5 }{ \cfrac { -4 }{ 2 } } -\cfrac { 1 }{ u } =\cfrac { 0.5 }{ R } \\ \cfrac { -3 }{ 4 } -\cfrac { 1 }{ u } =\cfrac { 1 }{ 2R } \\ \cfrac { -4 }{ u } =\cfrac { 1 }{ 2R } \\ \therefore u=-8R\\ or\quad PO=8R$$
Question 5
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A light ray passes through a prism with an apex angle $$30^\circ$$ as shown in figure. Incident medium is air. Critical angle for air prism interface is $$42^\circ$$. As angle 'i' increases from zero to $$90^\circ$$ in anti clock wise sense, angle of emergence changes. Which of following options is/are correct, when $$i = 90^\circ$$.
[Given $$\sin \ 42^\circ \cong 0.670; \sin \ 12^\circ \cong 0.21; \sin \ 18^\circ \cong 0.313.$$]

A
Emergent ray will bent towards base of prism

B
Emergent ray will bent away from base of prism

C
Deviation produced by prism is $$42^\circ$$ in clockwise sense

D
Angle of emergence will be $$18^\circ$$
Question 6
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An achromatic combination is made by combining 2 prisms as shown:-
If $$\omega_1 > \omega_2$$, then

A
Average deviation would be clockwise

B
Average deviation would be anticlockwise

C
Average deviation would be zero

D
Cannot be predicted on basis of give information

Solution
Question 7
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A point object $$O$$ is placed in front of a glass rod having a spherical end of radius of curvature $$30\ cm$$. The image would be formed at :

A
$$30\ cm$$ left

B
Infinity

C
$$1\ cm$$ to the right

D
$$18\ cm$$ to the left

Solution

$$\begin{array}{l}{\mu _1}({\rm{air}}) = 1\\{\mu _2}(glass) = 1.5\\u = - 15cm\\R = 30cm\\\dfrac{{{\mu _2} - {\mu _1}}}{R} = \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u}\\\dfrac{{1.5 - 1}}{{30}} = \dfrac{{1.5}}{v} - \dfrac{1}{{15}}\\v = - 30\end{array}$$
Ans. $$30cm$$ towards the left.
Question 8
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A spherical refractive surface of radius $$10$$ cm separates two media of refractive indices $$ \mu = 1$$ and $$ \mu = 3/2$$ respectively. A point object P starts from rest at $$t = 0$$ with an acceleration $$2 \ cm s^{-2}$$. Find the speed of image at $$t =1$$ s.

A
$$ 2 cms^-1 $$

B
$$ 6 cms^-1 $$

C
$$ 3.7 cms^-1 $$

D
none of these

Solution

Velocity of P at $$t=1s$$

$$\dfrac{du}{dt}=-(0+2\times 1)=-2 cm/s^2$$

We have, $$\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$$

$$\dfrac{n_2}{v}=\dfrac{n_2-n_1}{R}+\dfrac{n_1}{u}$$

$$v=\dfrac{n_2Ru}{(n_2-n_1)u+n_1R}$$

$$\dfrac{dv}{dt}==\dfrac { { n }_{ 2 }R\dfrac { du }{ dt } \left( \left( { n }_{ 2 }-{ n }_{ 1 } \right) u+{ n }_{ 1 }R \right) -\left( { n }_{ 2 }-{ n }_{ 1 } \right) \dfrac { du }{ dt } { n }_{ 2 }Ru }{ { \left[ \left( { n }_{ 2 }-{ n }_{ 1 } \right) u+{ n }_{ 1 }R \right] }^{ 2 } } $$

$$=\dfrac { \left( { n }_{ 2 }{ n }_{ 1 } \right) { R }^{ 2 }\dfrac { du }{ dt } }{ { \left[ \left( { n }_{ 2 }-{ n }_{ 1 } \right) u+{ n }_{ 1 }R \right] }^{ 2 } } \\ =\dfrac { 1\times \dfrac { 3 }{ 2 } \times { 10 }^{ 2 }\times (-2) }{ { \left[ 0.5\times (-2)+10 \right] }^{ 2 } } =-3.703\quad cm/s$$
Question 9
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A light ray is incident normally on one of the refracting faces of a prism and just emerges out grazing the second surface. The relation between angle of the prism and its critical angle is

A
$$A=C$$

B
$$A\neq C$$

C
$$A< C$$

D
$$A>C$$

Solution
Question 10
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A double convex lens of focal length $$30cm$$ is made of glass. When it is immersed in a liquid of refractive index $$1.5$$, the focal length is found to be $$120cm$$. The critical angle between glass and the liquid is

A
$$\sin ^{ -1 }{ \left( \cfrac { 1}{ 2 } \right) } \quad $$

B
$$\sin ^{ -1 }{ \left( \cfrac { 1 }{ 1.2} \right) } \quad $$

C
$$\sin ^{ -1 }{ \left( \cfrac { 7 }{ 13 } \right) } \quad $$

D
$$\sin ^{ -1 }{ \left( \cfrac { 7 }{ 8 } \right) } \quad $$

Solution