In Section A, defects of eye are given, in Section B, effects arising due to defects are given and in section C, remedies are given using lenses. Find the correct pair. Section A Section B Section C Myopia (x) Focal length increases (a) Bifocal lens Hypermetropia (y)Focal length decreases (b) Concave lens Presbyopia (z) Power of accommodation decreases (c) Convex lens

(1 - y - b), (2 - x - c), (3 - z - a)

(1 - z - b), (2 - z - b), (3 - y - b)

(1 - x - a), (2 - y - c), (3 - x - a)

(1 - y - a), (2 - x - a), (3 - z - c)

MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 12

A ray of light enters a spherical drop of water of refractive index

$\mu$ as shown in fig. An expression of the angle between incident ray and emergent ray (angle of deviation) as shown in fig. is :

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$0^{\circ}$
0%
$\phi$
0%
$\alpha -\phi$
0%
$\pi - 4\alpha + 2\phi$

Explanation

$\alpha = (\phi - \alpha) + \theta$

$\delta = \pi - 2\theta = \pi - 4\alpha + 2\phi$ .
1589461_160965_ans_fb66f5fb36664f4b81740ff362ca07a7.jpg

Which of the following are defects of vision?

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Myopia
0%
Hypermetropia
0%
Colour blindness
0%
Night blindness

A man can see objects clearly up to

$3 m$ . What type of lens he should use in order to see clearly up to

$12 m$ ?

0%
Convex lens, $f= 4 m$
0%
Concave lens, $f= 4 m$
0%
Convex lens, $f= -4 m$
0%
Concave lens, $f= -4 m$

Explanation

Clearly, the lens used should be the concave lens.

Actual far point

$u=-12m$

Far point of the person,

$v=-3m$

We have lens formula,

$\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f }$

$\Rightarrow \cfrac { 1 }{ -3 } +\cfrac { 1 }{ 12 } =\cfrac { 1 }{ f }$

$\Rightarrow f = - 4m$

Thus a concave lens of focal length

$4\ m$ is used.

Three identical isosceles right-angled prism (ABC, BCD and CDE) are placed as shown in the figure

$\displaystyle \angle ABC,\angle BCD\, and\, \angle CDE$ are

$\displaystyle 90^{\circ}$ in the given prism respectively. A ray is incident along line AB Refraction takes placed at BC and CD and then it emerges along the line DE

$\displaystyle \mu _{1},\mu _{2}$ and

$\displaystyle \mu _{3}$ are the refractive index of prism ABC, BCD and CDE respectively. The relation between

$\displaystyle \mu _{1},\mu _{2}\,and\,\mu _{3}$ is

0%
$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-\mu _{2}^{3}+2=0$
0%
$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-\mu _{3}^{2}+2=0$
0%
$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-2\mu _{3}^{2}+2=0$
0%
$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-5\mu _{3}^{2}+2=0$

Explanation

$c=$ critical angle with respect

$\mu_{1}$ (Prism

$\left.A B C\right)$

$\mu_{1} \sin i_{1}=M_{2} \sin \gamma_{c}$

at surface

$A B$

(1)

$\sin (90)=M, \sin C$

$\frac{1}{M_{1}}=\sin c$

$\mu_{1} \sin (90-c)=\mu_{2} \sin \gamma_{1}$

$\mu_{1} \cos c=\mu_{2} \sin \gamma_{1}$

$\mu_{1} \frac{\sqrt{\mu_{1}^{2}+1}}{\mu_{1}}=\mu_{2} \sin \gamma_{1}$

$\mu_{1}^{2}+1=m_{2}^{2} \sin ^{2} \gamma_{1}-$ (3)

at surface

$x$

$\begin{array}{c}\mu_{2} \sin \left(q 0\gamma_{2}\right)=\mu_{3}\sin\gamma_{2}\\\mu_{2}\cos\gamma_{1}=\mu_{3} \sin \gamma_{2}\end{array}-(1)$

at surface

$D \in$

$\begin{aligned}\mu_{3}\sin\left(90\gamma_{2}\right) &=1 \\ \cos \gamma_{2} &=\frac{1}{\mu_{3}}\end{aligned}-(2)$

$\frac{\mu_{2}^{2} \cos ^{2} \gamma_{1}}{\mu_{3}^{2}}=\sin ^{2} \gamma_{2} \quad$ Squaring equation

$+\frac{1}{\mu_{3}^{2}}=\cos ^{2} \gamma_{2}$ sequaring equation

$\mu_{2}^{2} \cos ^{2} \gamma_{1}+1=\mu_{3}^{2}$

Substitute (3) in this

$\mu_{2}^{2}\left[\frac{\mu_{1}^{2}+1}{\mu_{2}^{2}}\right]+1=\mu_{3}^{2}$

$\mu_{1}^{2}-\mu_{3}^{2}+2=0$

2005491_332952_ans_51edfb4a397c4b03a6bf1c2104176efd.PNG

A spherical lens has a focal length

$2\ cm$ . The lens will be

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Concave lens
0%
Convex lens
0%
Concavo convex lens
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None

As shown in figure, the liquids

$L_1,L_2$ and

$L_3$ have refractive indices.

$1.55,\ 1.50$ and

$1.20$ respectively. Therefore, the arrangement corresponds to

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biconvex lens
0%
biconcave lens
0%
concavo-convex lens
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convexo-concave lens

There is an equiconvex lens of focal length of

$20 cm$ . If the lens is cut into tow equal parts perpendicular to the optical axis, the focal length of each part will be

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20 cm
0%
10 cm
0%
40 cm
0%
15 cm

Explanation

The focal length of equiconvex lens is 20 cm

$\frac { 1 }{ f } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \\ for\quad equiconvex\quad lens\quad { R }_{ 1 }=-{ R }_{ 2 }=R\\ thus,\\ \frac { 1 }{ 20 } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } } +\frac { 1 }{ { R } } \right) \\ \Rightarrow R=40\left( \mu -1 \right)$

when lens is cut into two equal parts perpendicular to optical axis

then,

${ R }_{ 1 }=\infty ,\quad { R }_{ 2 }=R$

thus

$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right)$

$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } } \right)$

by putting value of R we get,

$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { 40\left( \mu -1 \right) } } \right) \\ \Rightarrow f'=40cm$

new focal length of each part is 40 cm , becomes double

Option C is correct.

The ratio of

$\frac {\theta _1}{\theta _2}$ is.

0%
1
0%
$\frac {1}{2}$
0%
$\sqrt 2$
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$\frac {1}{\sqrt 2}$

Explanation

When ray PQ enters in prism, by Snell's law:

$\dfrac{\sin\theta_{1}}{\sin r_{1}}=\mu=2\sin\theta_{1}$ .......eq1

$\sin r_{1}=1/2=\sin30^{o}$

$r_{1}(\angle RQN)=30^{o}$

$\angle RQN'=90-30=60^{o}$

Now BMC is

$45^{o}=45^{o}-90^{o}$ prism (

$\angle M=90$ )

Hence,

$\angle B=30+45=75^{o}$

$\triangle BQR$

$\angle BRQ=180-(75+60)=45^{o}$

Hence,

$\angle QRE=90-45=45^{o}$

In quadrilateral QRSA ,

$\angle Q=90+30=120^{o}$

$\angle A=30+60=90^{o}$

$\angle R=45+45=90^{o}$

Hence,

$\angle ASR=360-(120+90+90)=60^{o}$

It gives

$\angle RSK=90-60=30^{o} =r_{2}$

When ray ST emerges from prism, by Snell's law

$\dfrac{\sin\theta_{2}}{\sin r_{2}}=\dfrac{1}{\mu}$ ................eq2

Dividing eq1 by eq2 ,

$\dfrac{\theta_{1}}{\theta _{2}}=1$ (as

$r_{1}=r_{2}=30^{o}$ )

The total deviation of the incident ray when it emerges out of the prism is.

0%
$30^o$
0%
$60^o$
0%
$90^o$
0%
$45^o$

Explanation

In figure (a), the incident ray

$P$ hits the point Q, R and S and then emerges as

$T$ . The overall scenario is seen in figure (b) which represents the deviation as

$\delta$ .

$\therefore$ Deviation

$\delta = 90- \theta_2 + \theta_1$ .............(1)

In the prism

$DAC$ , prism angle

$= 30^o$

Using prism angle formula

$A = r_2+ r'$ ,

$\therefore$

$30^0 = r_1 + 0^o \implies r_2 = 30^o$

Given: Refractive index of the prism

$n = 2sin\theta_1$

Using Snell's law at point

$S$ ,

$1 \times sin\theta_2 = n \times sinr_2$

$\therefore$

$sin\theta_2 = 2sin\theta_1 \times sin30$

$sin\theta_2 = sin\theta_1$

$\implies$

$\theta_1 = \theta_2$

$\therefore$ From equation (1), deviation of the ray

$\delta = 90^o$

A man can see distinctly from a distance of

$0.5 m$ . If he wants to read a book placed at a distance of

$25 cm$ , then what kind of lens should he use?

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Convex lens with focal length of $-50 cm$
0%
Concave lens with focal length of $-50 cm$
0%
Convex lens with focal length of $50 cm$
0%
Concave lens with focal length of $50 cm$

Explanation

Since given that;

$u = 0.5m = 50cm$

$v = 25cm$

To find

$f$ , we know that

$\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f }$

$\cfrac { 1 }{ 50 } -\cfrac { 1 }{ 25 } =\cfrac { 1 }{ f }$

$\cfrac { 1 }{ f } =\cfrac { -1 }{ 50 }$

$f = 50cm$

Hence a concave lens of focal length

$50 cm$ is used.

The process of re-emission of absorbed light in all directions with different intensities by the atom or molecule is called ____________.

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Scattering of light
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Dispersion of light
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Reflection of light
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Refraction of light

Magnification of lens is:

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$\dfrac {\text {Size of the image}}{\text {Size of the object}}$
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$\dfrac {\text {Image distance}}{\text {Object distance}}$
0%
Both A and B
0%
None

Explanation

For lens, magnification is given by

$m=\dfrac{h_i}{h_o}=\dfrac{v}{u}$

where

$h_i$ = image height

$h_0$ = object height

$v$ =image distance

$u$ = object distance

Answer-(c)

A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45

$^o$ and that of refraction is 30

$^o$ . Find the refractive index of the medium.

0%
2
0%
$\dfrac{1}{\sqrt 2}$
0%
$2 \sqrt 2$
0%
$\sqrt 2$

Explanation

Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by

$\displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r}$

The initial medium is air. Thus

$\mu_1 = 1$

Given incidence angle

$i = 45^o$ and refraction angle is

$30^o$

By Snell's law,

$\mu_2 = \mu_1\times \displaystyle \frac{\sin i}{\sin r} = 1\times \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$

a diverging lens of focal length -

$10$ cm is moving towards right with a velocity

$5$ m/s. An object, placed on Principal axis is moving towards left with a velocity

$3$ m/s. The velocity of the image at the instant when the lateral magnification produced is

$1/2$ is: (All velocities are with respect to ground)

0%
$3$ m/s towards right
0%
$3$ m/s towards left
0%
$7$ m/s towards right
0%
$7$ m/s towards left

Explanation

$\begin{array}{l} f=-10\, cm \\ M=\frac { 1 }{ 2 } =\left( { \frac { v }{ u } } \right) \\ { v_{ IM } }=\left( { \frac { { { v^{ 2 } } } }{ { { u^{ 2 } } } } } \right) { v_{ OM } } \\ \Rightarrow { v_{ I } }-{ v_{ M } }=\frac { 1 }{ 4 } \left( { { v_{ O } }-{ v_{ M } } } \right) \\ { v_{ I } }-5=\frac { 1 }{ 4 } \left[ { -3-5 } \right] \\ { v_{ I } }-5=-2 \\ { v_{ I } }=3\, m/s\, \, \, \, towards\, right \end{array}$

Hence,

option

$(A)$ is correct answer.

Two glass prisms

$P_1$ and

$P_2$ are to be combined together to produce dispersion without deviation. The angles of the prisms

$P_1$ and

$P_2$ are selected as

$4^o$ and

$3^o$ respectively. If the refractive index of prism

$P_1$ is 1.54, then that of

$P_2$ will be

0%
1.48
0%
1.58
0%
1.62
0%
1.72

Explanation

Prism angle of first prism

$A_1 =4^o$

Its refractive index

$\mu_1=1.54$

Prism angle of second prism prism

$A_2 =3^o$

Let refractive index of second prism be

$\mu_2$ .

For dispersion without deviation, net deviation of the ray must be zero i.e.

$\delta_1 - \delta_2 =0$

$\therefore$

$(\mu_1 -1)A_1 -(\mu_2 -1)A_2 =0$

$(1.54 -1)(4) -(\mu_2 -1)3 =0$

$\mu_2 -1 =0.72$

$\implies \mu_2 = 1.72$

A rectangular block is composed of three different glass prisms (with refractive indices

$\mu_1, \mu_2$ and

$\mu_3$ ) as shown in the figure below. A ray of light incident to the left face emerges normal to the right face. Then the refractive indices are related as:

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$\mu_1^2 + \mu_2^2 = 2 \mu_3^2$
0%
$\mu_1^2 + \mu_2^2 = \mu_3^2$
0%
$\mu_1^2 + \mu_3^2 = 2 \mu_2^2$
0%
$\mu_2^2 + \mu_3^2 = 2 \mu_1^2$

Explanation

From the figure,

$i_1 = r_2 = 45^o$

$\Delta AEF,$ ,

$\angle A + \angle E + \angle F = 180^o$

$\implies \angle A = 90^o$

In quadrilateral

$ABCD,$

$\angle A + \angle B + \angle C + \angle D = 180^o$

$\implies \angle C = 90^o$

$\Delta BCD,$

$\angle B + \angle C + \angle D = 180^o$

$r_1 + i_2 = 90^o$

$\sin r_1 = \cos i_2$

$\sin^2 r_1 = \cos^2 i_2$

$\sin^2 r_1 = 1-\sin^2 i_2$

Using snell's law,

$\cfrac{\sin^2 i_1}{(\mu_2/\mu_1)^2} = 1 - (\mu_3/\mu_2)^2 \sin^2 r_2$

$\cfrac{\mu_1^2}{2\mu_2^2} + \cfrac{\mu_3^2}{2\mu_2^2}=1$

$\mu_1^2 + \mu_3^2 = 2\mu_2^2$

A vessel is filled with oil as shown in the diagram. A ray of light from point

$O$ at the bottom of vessel is incident on the oil - air interface at point

$P$ and grazes the surface along

$PQ$ . The refractive index of the oil is close to-

0%
$1.41$
0%
$1.50$
0%
$1.630$
0%
$1.73$

Explanation

$\sin { { i }_{ c } } =\cfrac { 12 }{ \sqrt { { (17) }^{ 2 }+{ (12) }^{ 2 } } } =\cfrac { 12 }{ 20.80 }$

$\quad \mu =\cfrac { 1 }{ \sin { { i }_{ c } } } =\cfrac { 20.80 }{ 12 } =1.73\quad \quad$
1144600_721061_ans_9da872397f2a45c980d106db31aa7d94.png

In Section A, defects of eye are given, in Section B, effects arising due to defects are given and in section C, remedies are given using lenses. Find the correct pair.

Section A	Section B	Section C
Myopia	(x) Focal length increases	(a) Bifocal lens
Hypermetropia	(y)Focal length decreases	(b) Concave lens
Presbyopia	(z) Power of accommodation decreases	(c) Convex lens

0%
(1 - y - b), (2 - x - c), (3 - z - a)
0%
(1 - z - b), (2 - z - b), (3 - y - b)
0%
(1 - x - a), (2 - y - c), (3 - x - a)
0%
(1 - y - a), (2 - x - a), (3 - z - c)

Figure shows an irregular block of material of refractive index

$\sqrt { 2 }$ . A ray of light strikes the face

$AB$ as shown in figure. After refraction, it is incident on a spherical surface

$CD$ of radius of curvature

$0.4 m$ and enters a medium of refractive index

$1.514$ to meet

$PQ$ at

$E$ . Find the distance

$OE$ upto two places of decimal :

0%
$7 m$
0%
$.29 m$
0%
$6.06 m$
0%
$8.55 m$

Explanation

From Snell's law,

${ \mu }_{ 1 }\sin { i } ={ \mu }_{ 2 }\sin { r }$

$\Rightarrow \sin { r } =\dfrac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } \sin { i } =\dfrac { 1 }{ \sqrt { 2 } } \sin { { 45 }^{ o } }$

$\Rightarrow \dfrac { 1 }{ 2 } =\sin { r } \Rightarrow r={ 30 }^{ o }$

This means that the ray becomes parallel to side

$AD$ inside the slab.

This implies for the second face

$CD$ ,

$u=\infty$

Given

$R=0.4 m$

$\dfrac { { \mu }_{ 3 } }{ v } -\dfrac { { \mu }_{ 2 } }{ v } =\dfrac { { \mu }_{ 3 }-{ \mu }_{ 2 } }{ R }$

$\Rightarrow \dfrac { 1.514 }{ v } -\dfrac { \sqrt { 2 } }{ \infty } =\dfrac { 1.514-\sqrt { 2 } }{ R }$

$v=\dfrac { 1.514\times 0.4 }{ 1.514-1.414 } =\dfrac { 1.514\times 0.4 }{ 0.1 }$

$v=6.056=6.06 m$ (upto

$2$ decimal places).

A man is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is

$1.39 \times 10^9 m$ and its mean distance from the earth is

$1.5 \times 10^{11} \ m$ , the diameter of the sun's image on the paper is

0%
$3.1 \times 10^{-4} m$
0%
$6.5 \times 10^{-5} m$
0%
$6.5 \times 10^{-4} m$
0%
$9.2 \times 10^{-4} m$

Explanation

Magnification of convex lens,

$m \, = \, \dfrac{-f}{f \, - \, u}$

Here,

$f \, = \, 10 \, cm \, = \, 10 \, \times \, 10^{-2} \, m, \, \, u \, = \, 1.5 \, \times \, 10^{11} \, m$

$\therefore \, m \, = \, \dfrac{-10 \, \times \, 10^{-2}}{10 \, \times \, 10^{-2} \, - \, 1.5 \, \times \, 10^{11}} \, = \, 6.67 \, \times \, 10^{-13}$

$\therefore$ Diameter of the image,

$d \, = \, m \, \times \, 1.39 \, \times \, 10^9 \, = \, 6.67 \, \times \, 10^{-13} \, \times \, 1.39 \, \times \, 10^9 \, = \, 9.27 \, \times \, 10^{-4} \, m$
Hence, the correct option is

$(D)$

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 isThe focal length of the lens is :

0%
$1 m$
0%
$0.5 m$
0%
$0.24 m$
0%
$2 m$

The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

0%
$1\ D$
0%
$1.5\ D$
0%
$0.75\ D$
0%
$2.67\ D$

Choose the correct answer from alternatives given.
The mixture of a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions <<vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

0%
$- \, \dfrac{1}{d} \, \left(\frac{du}{dy} \right)\mu$
0%
$\dfrac{d}{h} \, \left(\frac{du}{dy} \right)$
0%
$\left(\frac{du}{dy} \right)h$
0%
$- \, \dfrac{1}{\mu} \, \left(\frac{du}{dy} \right)d$

Explanation

Let us consider a light ray entering the column at (x, y) at

$90^o$ to the vertical.

$\theta$ be the angle of incidence at x and it enter the thin column at height y.

$d\theta$ is the deviation of ray between x and (x + dx) emerging at (x +dx, y + dy ) at angle

$(\theta + d \theta)$ .

By snell's law,

$\mu (y) \, sin \, \theta \, =\, \mu (y +dy) \, sin \, ( \theta \, + \, d \theta )$

$= \, \left[\mu (y) + \dfrac{d\mu}{dy} dy \right] \, (sin \, \theta \, cos \, d \theta \, + \, cos \, \theta \, sin \, d \theta)$

$d \theta \, is \, small, \, cos \, d \theta \, =\, 1 \, and \, sin \, d \theta \, \approx \, d\theta$

$\therefore \mu(y) \, cos \, \theta \, d \theta \, =\, \dfrac{d \mu}{dy} dy \, sin \, \theta ; \, d \theta = \dfrac{-1 }{\mu} \dfrac{d \mu dy \, tan \, \theta}{dy}$

$\left[ tan \, \theta \, = \, \dfrac{dx}{dy} and \, raplacing \, \mu (y) \, by \, \mu \right]$

$\theta = -\dfrac{1}{\mu} \dfrac{d \mu}{dy} \int_{o}^{d} dx = -\dfrac{1}{\mu} \left(\dfrac{d \mu}{dy}\right) d$

1028255_943891_ans_3ba2b43df78041ed89e3af84f4f74fb6.png

Ocean appears blue due to _____________________.

0%
Greater depth
0%
Scattering of light by water particles
0%
Blue colour of water
0%
None of these

Which defect of vision of eye is shown respectively?

0%
P: Presbyopia, Q: Near sightedness
0%
P: Far sightedness, Q: Near sightedness
0%
P: Near sightedness, Q: Far sightedness
0%
P: Presbyopia, Q: Far sightedness

Explanation

$P:-$

In this we see that the eye can't see the nearby objects as the lens can't focus the image on the retina but focuses the image beyond the retina.

This defect of the eye is called

$far-sightedness$ as the eye in this defect can't see the nearby objects but can see far-off objects.

$Q:-$

In this we see that the eye can't see the far off objects clearly as the lens can't focus the image on the retina but focuses the image before the retina.

This defect of the eye is called

$near-sightedness$ as the eye in this defect can't see the far away objects but can see nearbyf objects.

Answer-(B)

In the given figure, the radius of curvature of a curved surface for both the piano-convex and plano-concave lens is 10 cm and refractive index for both is 1.The location of the final image after all the refractions through lenses is:

0%
$15 cm$
0%
$20 cm$
0%
$25 cm$
0%
$40 cm$

Explanation

The focal length of the plano - convex lens is

$\dfrac{1}{f} \, = \, (1.5 \, - \,1) \left ( \dfrac{1}{+10} \, - \, \dfrac{1}{\infty }\right ) \, = \, \dfrac{1}{20}$
Focal length of plano - concave lens is

$\dfrac{1}{f} \, = \, (1.5 \, - \,1) \left ( \dfrac{1}{\infty} \, - \, \dfrac{1}{10}\right ) \, = \, \dfrac{-1}{20}$
since parallel beams are incident on the lens, its image from plano - concave lens will be formed at

$+20$ cm from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of

$10$ cm from each other, therefore, for the next lens

$u = +10 cm.$

$\therefore v = \dfrac{uf}{u + f} = \dfrac{10 \, \times \, 20}{10 - 20}$

$= 20 cm$

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:

0%
Refractive index of the sphere is 2
0%
Refractive index of the sphere $\sqrt 3$
0%
Emergent ray travels along the line $\sqrt 3 x + y = \sqrt 3 R$
0%
Total deviation suffered by the ray is ${45^o}$

Explanation

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

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

2005763_970342_ans_6dd9078b840b4cd4980267d6ff249634.PNG

A ray of light travelling in water is incident on its surface open to air. The angle of incidence

$\theta,$ which is less than the critical angle. Then there will be:

0%
only a reflected ray and no refracted ray
0%
only a refracted ray and no reflected ray
0%
a reflected ray and a refracted ray and the angle between them would be less than $180^{o}-2\theta$
0%
a reflected ray and a refracted ray and the angle between them would be greater than $180^{o}-2\theta$

Explanation

The correct answer is option (C).

Hint: Draw the diagram based on the given condition.

Step 1: Explanation

According to the question,

Since the angle of incidence $\theta$ is less than the critical angle, the incident ray will undergo both reflection and refraction. A part of it is reflected while the other part is refracted.
Also, from the diagram, we see that the angle between the reflected and the refracted ray is less than $180^o-2\theta$

Hence, there will be both reflected and refracted ray and the angle between them is less than $180^o-2\theta$

The correct answer is option (C).

A ray of light travels from water to glass as shown below. The refractive index of water is

$1.3$ and the refractive index of glass is

$1.5$ . What is the angle of refraction ?

0%
$30.7^{o}$
0%
$35.3^{o}$
0%
$41.7^{o}$
0%
$48.6^{o}$

An equal convex lens of

$\upsilon =1.5$ has a focal length of 20 cm in air. Now that lens kept in the medium of refractive index 3. An object is kept at distance 30 cm from the lens in the same medium. Find its image distance from the lens is :

0%
24 cm
0%
12 cm
0%
10 cm
0%
None

A point object O is placed at a distance of 20 cm in front of a equiconvex lens

$(^a\mu_g \, = \, 1.5)$ of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in the figure. An image will be formed at a distance h from a lens. The value of h is

0%
5 cm
0%
10 cm
0%
20 cm
0%
40 cm

Explanation

$\dfrac{1}{f} = (\mu - 1) \left[\dfrac{1}{R_1} - \dfrac{1}{R_2}\right]$

$\dfrac{1}{10} = (1.5 - 1) \left(\dfrac{1}{R} + \dfrac{1}{R} \right) = 0.5 \times \dfrac{2}{R} \Rightarrow R = 10 \, cm$

$\therefore \dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$

Refraction from first surface,

$\dfrac{1.5}{v_1} - \dfrac{1}{-20} = \dfrac{1.5 - 1}{+10} \Rightarrow v_1 = \infty$

For the second surface,

$\dfrac{2}{v} - \dfrac{1.5}{\infty} = \dfrac{2 - 1.5}{-10} \Rightarrow v = -40 \, cm \,\, \therefore h = 40 \, cm$

A gain telescope in an observatory has an objective of focal length

$19\ m$ and an eye-piece of focal length

$1.0\ cm$ . What is the diameter of the image of moon formed by the objective in normal adjustment? The diameter of moon is

$3.5\times {10}^{6}\ m$ and the radius of the lunar orbit round the earth is

$3.8\times {10}^{8}\ m.$

0%
$10\ cm$
0%
$12.5\ cm$
0%
$15\ cm$
0%
$17.5\ cm$

Explanation

$f_{0}=19 \mathrm{~m} \quad f_{e}=0.01 \mathrm{~m}$

Lens formula at objective lens

$\begin{aligned}\frac{1}{f} &=\frac{1}{v}-\frac{1}{u} \\\frac{1}{19} &=\frac{1}{v}-\frac{1}{3.8 \times 10^{8}} \\& v \simeq 19 \mathrm{~cm}\end{aligned}$

Given diameter of

$\mathrm{moon}=3.5 \times 10^{6}$

Image diameter

$=\frac{3.5 \times 10^{6}}{3.8 \times 10^{8}} \times 19 \mathrm{~cm}$

$=17.5 \mathrm{~cm}$

2007762_1015184_ans_368281ccc4a24c9eba6370e4498aea98.PNG

A ray incident at a point at an angle of incidence of

${60}^{o}$ enters a glass sphere of

$R.l.n=\sqrt 3$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

0%
${50}^{o}$
0%
${60}^{o}$
0%
${90}^{o}$
0%
${40}^{o}$

Explanation

Refraction at

$P$ ,

$\dfrac { \sin { 60 }^{o} }{ \sin { { r }_{ 1 } } } =\sqrt { 3 }$

$\Rightarrow \sin { { r }_{ 1 } } =1/2 \quad \Rightarrow { r }_{ 1 }=30$
Since

${ { r }_{ 2 } } ={ r }_{ 1 } \quad \therefore \quad{ r }_{ 2 }=30$
Reflection at

$Q$ ,

$\dfrac { \sin { { r }_{ 2 } } }{ \sin { { i }_{ 2 } } } =\dfrac { 1 }{ \sqrt { 3 } }$
Putting

${R}_{2}={30}^{o}$ we obtain

${i}_{2}={60}^{o}$
Reflection at

$Q, {r}^{\prime }_{2}={r}_{2}$

$\therefore \quad \alpha =180^{o}-({ r }^{ \prime }_{ 2 }+{ r }_{ 2 })=180^{o}-(30^{o}+60^{o})=90^{o}$
Hence (C) is correct.
1032847_1014826_ans_9e786de5f0ed4724bf3452c221d557c2.png

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 :

0%
$30\ cm$ left
0%
Infinity
0%
$1\ cm$ to the right
0%
$18\ cm$ to the left

Explanation

$\begin{array}{l}{\mu _1}({\rm{air}}) = 1\\{\mu _2}(glass) = 1.5\\u = - 15cm\\R = 30cm\\\dfrac{{{\mu _2} - {\mu _1}}}{R} = \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u}\\\dfrac{{1.5 - 1}}{{30}} = \dfrac{{1.5}}{v} - \dfrac{1}{{15}}\\v = - 30\end{array}$
Ans.

$30cm$ towards the left.

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 :

0%
$\cos^{2}\phi = \dfrac {\mu^{2} + 1}{3}$
0%
$\cos^{2}\phi = \dfrac {\mu^{2} - 1}{3}$
0%
$\sin^{2}\phi = \dfrac {\mu^{2} + 1}{3}$
0%
$\sin^{2}\phi = \dfrac {\mu^{2} - 1}{3}$

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.

0%
$2 cms^-1$
0%
$6 cms^-1$
0%
$3.7 cms^-1$
0%
none of these

Explanation

Velocity of P at

$t=1s$

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

We have,

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

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

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

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

$=\dfrac { \left( { n }_{ 2 }{ n }_{ 1 } \right) { R }^{ 2 }\dfrac { du }{ dt } }{ { \left[ \left( { n }_{ 2 }-{ n }_{ 1 } \right) u+{ n }_{ 1 }R \right] }^{ 2 } } \\ =\dfrac { 1\times \dfrac { 3 }{ 2 } \times { 10 }^{ 2 }\times (-2) }{ { \left[ 0.5\times (-2)+10 \right] }^{ 2 } } =-3.703\quad cm/s$

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

0%
$8R$
0%
$7R$
0%
$2R$
0%
None of these

Explanation

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,

$OP=2\left( OQ \right) \\ or\quad OQ=\cfrac { OP }{ 2 } \\ or\quad v=\cfrac { -u }{ 2 }$

$\therefore \cfrac { 1.5 }{ \cfrac { -4 }{ 2 } } -\cfrac { 1 }{ u } =\cfrac { 0.5 }{ R } \\ \cfrac { -3 }{ 4 } -\cfrac { 1 }{ u } =\cfrac { 1 }{ 2R } \\ \cfrac { -4 }{ u } =\cfrac { 1 }{ 2R } \\ \therefore u=-8R\\ or\quad PO=8R$

An achromatic combination is made by combining 2 prisms as shown:-
If

$\omega_1 > \omega_2$ , then

0%
Average deviation would be clockwise
0%
Average deviation would be anticlockwise
0%
Average deviation would be zero
0%
Cannot be predicted on basis of give information

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

0%
$A=C$
0%
$A\neq C$
0%
$A< C$
0%
$A>C$

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

0%
$\sin ^{ -1 }{ \left( \cfrac { 1}{ 2 } \right) } \quad$
0%
$\sin ^{ -1 }{ \left( \cfrac { 1 }{ 1.2} \right) } \quad$
0%
$\sin ^{ -1 }{ \left( \cfrac { 7 }{ 13 } \right) } \quad$
0%
$\sin ^{ -1 }{ \left( \cfrac { 7 }{ 8 } \right) } \quad$

After testing the eyes of a child, the optician has prescribed the following lenses for his spectacles:
left eye :

$+ 2\, D$ Right eye:

$+ 2.25\, D$
the child is suffering from the defective vision called:

0%
Short-sightedness
0%
Long-sightedness
0%
Cataract
0%
Presbyopia

Rays from an object immersed in water

$( \mu = 1.33 )$ traverse a spherical air bubble of radius R. If the object is located far away from the bubble, its image as seen by the observer located on the other side of the bubble will be

0%
virtual, erect and diminished
0%
real, inverted and magnified
0%
virtual, erect and magnified
0%
real, inverted and diminished

Explanation

$\textbf{Given:-}$ Radius of Spherical air bubble

$= R$

$\mu = 1.33$

$\textbf{Solution:-}$

Case - 1

$\dfrac{1}{v} - \dfrac{\dfrac{4}{3}}{-\infty } = \dfrac{1 - \dfrac{4}{3}}{R} = - \dfrac{1}{3R}$

$\Rightarrow V = -3R$

$m_{1} = \dfrac{-\dfrac{3R}{1}}{-\dfrac{\dfrac{\infty }{4}}{3}} = +0$

Case - 2

$\dfrac{4}{3v} - \dfrac{1}{-5R} = \dfrac{\dfrac{4}{3} - 1}{-R}$

$\dfrac{4}{3v} + \dfrac{1}{5R} = -\dfrac{1}{3R}$

$\Rightarrow \dfrac{4}{3v} = -\dfrac{1}{3R} - \dfrac{1}{5R} = -\dfrac{8}{15R}$

$v = - \dfrac{5R}{2}$

$m_{2} = \dfrac{\dfrac{5R}{2(4/3)}}{- \dfrac{5R}{1}} = + \dfrac{3}{8}$

$m_{1}m_{2} = +0$

image is virtual, erect and diminished

$\textbf{Hence the correct option is A}$

Two Nicol prisms are first crossed and then one of them is rotated through

$30^{0}$ .The percentage of incident unpolarized light transmitted is

0%
37.5
0%
50
0%
12.5
0%
25.0

A diverging meniscus of radii of curvatures

$25\ cm \ and \ 50 \ cm$ has a refractive index

$1.5$ . Its focal length is :

0%
$-50cm$
0%
$-100cm$
0%
$100cm$
0%
$50cm$

Monochromatic light is incident on a glass prism of angle

$A$ . If the refractive index of the material of the prism is

$\mu$ , a ray, incident at an angle

$\theta$ , on the face

$AB$ would get transmitted through the face

$AC$ of the prism provided.q

0%
$\theta > \sin^{-1} \left [\mu \sin \left (A - \sin^{-1} \left (\dfrac {1}{\mu}\right )\right )\right ]$
0%
$\theta < \sin^{-1} \left [\mu \sin \left (A - \sin^{-1} \left (\dfrac {1}{\mu}\right )\right )\right ]$
0%
$\theta > \cos^{-1} \left [\mu \sin \left (A + \sin \left (\dfrac {1}{\mu}\right )\right )\right ]$
0%
$\theta < \cos^{-1} \left [\mu \sin \left (A + \sin \left (\dfrac {1}{\mu}\right )\right )\right ]$

The angular resolution of a radio telescope is to be

${ 0.100 }^{ 0 }$ when the incident beam of wavelength

$3.00mm$ is used. What is minimum diameter required for the telescope's receiving dish?

0%
2.0 m
0%
4.20 m
0%
2.20 m
0%
3.20 m

A ray of light is incident on the surface of separating two transparent medium at an angle

$45^0$ and is refracted in medium at an angle

$30^0$ . Velocity of light in the medium will be...

0%
$3.8 \times 10^8 \ m/s$
0%
$3.38 \times 10^8 \ m/s$
0%
$2.12 \times 10^8 \ m/s$
0%
$1.5 \times 10^8 \ m/s$

Two identical thin isosceles prism of refracting angle 'A' and refractive index

$\mu$ are placed their bases touching each other. Two parallel rays of light are incident on this system as shown The distance of the point where the rays converge from the prism is:

0%
$\dfrac { h }{ \mu A }$
0%
$\dfrac { h }{ A }$
0%
$\dfrac { h }{ \left( \mu -1 \right) a }$
0%
$\dfrac { \mu h }{ \left( \mu -1 \right) a }$

Explanation

$\begin{array}{l} \text { since } \tan \delta=\frac{h}{f} \Rightarrow f=\frac{h}{\tan \delta} \\ \text { further, } \delta=(\mu-1) A \Rightarrow f=\frac{h}{(\mu-1) A} \\ \therefore f=\frac{h}{(\mu-1) A} \end{array}$

2001058_1234719_ans_421cb6c88aab4e04964f099621cd50bb.PNG

2001058_1234719_ans_421cb6c88aab4e04964f099621cd50bb.PNG

A ray of light makes normal incidence on the diagonal face of a right angled prism as shown in figure. If

$\theta = 37^o$ , then the angle of deviation after second step (from AB) is (sin

$37^o = 3/5$ )

0%
$53^o$
0%
$74^o$
0%
$106^o$
0%
$90^o$

Explanation

When light strike face

$A B$

$\Rightarrow \sin _{i c}=\frac{1}{\mu}\left(\therefore i_{c}=\right.$ critical angle

$)$

$\Rightarrow \sin _{i c}=\frac{3}{5}$

$\Rightarrow \quad i_{c}=37^{\circ}$

But in face incidance angle

$=53^{\circ}$

$\therefore$ reflection takes place

$\angle i\rangle<\angle i_{c}$

9n face BC Incidence angle

$=37^{\circ}$

$\begin{array}{l}\quad i=i c\\\therefore\quad\angle\gamma=90\end{array}$

$c D$ is divergent ray angle blw incident and divergent ray is

$53^{\circ}$ (from figure)

Hence (A) is correct

1999923_1223537_ans_44ee2563a8f149e39e4a33465134d65e.PNG

A convex lens (refractive index

$\mu=1.5$ ) has a power P. If it is immersed in a liquid (

$\mu$ =4/3), then its power will become will become/remain

0%
P
0%
$\dfrac { P }{ 2 }$
0%
$\dfrac { P }{ 4 }$
0%
4P

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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