MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 9

The ray diagram could be correct

0%
If ${ n }_{ 1 }={ n }_{ 2 }={ n }_{ g }$
0%
If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }<{ n }_{ g }$
0%
If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }> { n }_{ g }$
0%
$under\ no\ circumstances$

Explanation

It is to be remembered that to make a lens nature opposite i.e. from convergent to divergent we should start using it in denser medium or if it is in dense medium then use it in rarer medium.

Hence,

${ n }_{ 1 }={ n }_{ 2 }\quad \& \quad { n }_{ 1 }>{ n }_{ g }$

due to surrounding medium.

The relation among

$u,v$ and

$f$ for a mirror is:

0%
$f=uv(u+v)$
0%
$v=fu(u+f)$
0%
$u=fv(f+v)$
0%
None of these

Explanation

from mirror formula

$\dfrac { 1 }{ v } +\dfrac { 1 }{ u } =\dfrac { 1 }{ f }$

$\Rightarrow \quad \boxed { f=\dfrac { uv }{ u+v } }$

and

$1+\dfrac { v }{ u } =\dfrac { v }{ f } \Rightarrow \boxed { u=\dfrac { fv }{ f-v } }$

The sky appears blue, due

0%
To scattering of light by dust particles
0%
To the colour of the sea water
0%
To the presence of water in clouds
0%
None of these

In the given figure a Plane concave lens is placed on paper on which a flower is drawn. How far above its actual position does the flower appear to be?

0%
$10cm$
0%
$15cm$
0%
$50cm$
0%
none of these

Explanation

Since apparent shift =

$t\left( \mu -1 \right)$

here t is thickness of glass slab

$\mu$ is Refractive index of glass

hence

$\delta =20\left( 3/2-1 \right)$

$\delta =10cm$

The refractive index of glass with respect to air is

$\dfrac{3}{2}$ and that of water with respect to air is

$\dfrac{3}{2}$ . What is the refractive index of glass with respect to water?

0%
$9 : 8$
0%
$1 : 2$
0%
$2 : 1$
0%
$8 : 9$

Explanation

Given,

Refractive index of glass=

$\dfrac{3}{2}$

Refractive index of water=

$\dfrac{4}{3}$

Refractive Index of glass with respect to water=

$\dfrac{3/2}{4/3}$

$\dfrac{9}{8}$

A ray of light passes from diamond to air. Critical angle is minimum for.

0%
Green
0%
Red
0%
Blue
0%
Yellow

Explanation

$\mu(\lambda)=B+\cfrac{C}{{\lambda}^{2}}+\cfrac{D}{{\lambda}^{4}}+\dots\dots$ [cauchy's equation]

$\theta c=\sin ^{-1}(\cfrac{1}{\mu})$

$\bullet$ For minimum critical angle,

$\mu$ should be maximum.

$\bullet$ For

$\mu$ to me maximum,

$\lambda$ should be minimum.

hence, out of the following options,

$Blue$ color will have minimum critical angle.

In the given figure, if P is a point of the source of light, R is retina and L is the lens of the eye, the person having such condition of his eyes is suffering from :

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Hypermetropia
0%
Myopia
0%
Cataract
0%
Night Blindness

The phenomenon of bending of light at the surface of separation of two media is called :

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Refraction of light
0%
Reflection of light
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Deflection of light
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Absorption of light

Which of the following forms a virtual and erect image for all positions of the object ?

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Concave lens
0%
Concave mirror
0%
Convex mirror
0%
Both (a) and (c)

Of the following instruments, which one would be used for navigating a ship?

0%
Telescope
0%
Sonometer
0%
Aneroid barometer
0%
Sextant

The minimum distance between an object and its real image formed by a convex lens is:

0%
1.5 f
0%
2 f
0%
2.5 f
0%
4 f

Explanation

Four times the focal length, we have the formula,

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ where,

$u,v$ are the object and image distnaces and f is the focal length.

The minimum distnace is achieved when

$u=v=2f$

Coincidentally the magnification is -1. (The image is same size as the object,but inverted)

An air bubble is inside water. The refractive index of water is

$4/3$ . At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble?

0%
$2R$
0%
$3R$
0%
$4R$
0%
The air bubble cannot form a real image

A glass prism of refractive index

$1.5$ is immersed in water (refractive index

$4/3$ ). A light beam incident normally on the face:

$AB$ as shown in the figure is totally reflected to reach the face

$BC$ if:

0%
$\sin { \theta } >8/9$
0%
$2/3<\sin { \theta } <8/9$
0%
$\sin { \theta } \le 2/3$
0%
$none\ of\ the\ above$

Explanation

$\begin{array}{l} \text { Condition for } \mathbb{Refraction} \quad \mu_{1} \sin \theta_{1}, \quad>\mu_{2} \sin \theta_{2} \\ 1.5 \sin \theta \cdot>\dfrac{4}{3} \sin (90) \\ \sin \theta>\dfrac{8}{9} \\ \text { So option } A \text { is correct } \end{array}$
2005774_1011282_ans_fb8385749fab43be9dfb3051f098420f.PNG

2005774_1011282_ans_fb8385749fab43be9dfb3051f098420f.PNG

In the given situation, a spherical transparent surface of radius

$R$ is placed. The refractive index of sphere is such that image for an object kept at distance

$R$ is formed at the same distance of the opposite side. What is refractive index of sphere?

0%
$1.5$
0%
$1.75$
0%
$1.25$
0%
$2$

Explanation

From the symmetry of the figure, ray inside the sphere is parallel to principal axis. Taking refraction at

$A$ .

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

$\dfrac{\mu}{\infty}-\dfrac{1}{-R}=\dfrac{\mu-1}{R}$

$\dfrac 1R=\dfrac{\mu -1}{R}\implies \mu -1=1$

$\implies \mu=2$

1481387_1032605_ans_41e76cece05a4c2fa1aaf4f3166d78a8.png

Two concave lens and a convex of radii 3 cm and 2 cm are joined. What is the refractive index of the combination ?

0%
$\dfrac{3}{5}$
0%
-6
0%
$-\dfrac{3}{5}$
0%
- 12

Explanation

It is given that,

${{R}_{1}}=-3\,cm$

Focal length of concave lens ${{f}_{1}}=\dfrac{-3}{2}$

${{R}_{2}}=2\,cm$

Focal length of convex lens ${{f}_{2}}=1\,cm$

Combined focal length is

$\dfrac{1}{f}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{f}_{2}}}$

$\dfrac{1}{f}=\dfrac{-2}{3}+1$

$\dfrac{1}{f}=\dfrac{1}{3}........(1)$

The relation between redfractive index and focal length is

$\dfrac{1}{f}=\left( \mu -1 \right)\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$

$\dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{1}{-3}-\dfrac{1}{2} \right]$

$\dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{-5}{6} \right]$

$\mu =\dfrac{3}{5}$

Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are

$\mu_r,$

$\mu_y$ and

$\mu_v$ respectively and those for the flint glass are

$\mu^{ '}_{r}$ ,

$\mu^{'}_{y}$ and

$\mu^{'}_{v}$ respectively. Find the

$\dfrac {A'}{A}$ for which, there is no net angular dispersion.

0%
$\dfrac{\mu_y - 1}{\mu^{'}_{y} - 1}$
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$\dfrac{2\mu_y - 1}{\mu^{'}_{y} - 1}$
0%
$\dfrac{3\mu_y - 1}{\mu^{'}_{y} - 1}$
0%
$\dfrac{4\mu_y - 1}{\mu^{'}_{y} - 1}$

Explanation

Total deviation for yellow ray produced

$\delta y=\delta_{cy}-\delta_{fy}+\delta _{cy}$

$=2\delta_{cy}-\delta_{fy}$

$=2(\mu_{cy-1})A-(\mu_{-cy}-1)A^1$

Angular dispersion,

$\delta_v-\delta_r=[(\mu_{vc}-1)A-(\mu_{vf}-1)A^1+(\mu_{vc}-1)A]$

$-[(\mu_{rc}-1)A-(\mu_{rf}-1)A^1+(\mu_{r}-1)A]$

$2(\mu_{vc}-1)A-(\mu_{vf}-1)A^1$

For net angular dispersion to be zero,

$\delta_r-\delta_r=0$

$2(\mu_{vc}-1)A=(\mu_{vf}-1)A^1$

A light ray is incident a glass sphere at an angle of incidence of

$45^o$ as shown in figure. Find the total deviation suffered by ray of two successive reflection.

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$45^o$
0%
$30^o$
0%
$60^o$
0%
$120^o$

Explanation

Now from a above image

At First reflection

$1 \times Sin 45^\circ = \sqrt{2} Sin\space r$

$r = 30^\circ$

Since

$r = i$ (Both are symmetric)

Show at second reflection

$\sqrt{2} Sin\space i = 1 \times Sin\space r'$

$r' = 45^\circ$

Now consider quadrilateral A O B C

$\angle A+\angle O+\angle B+\angle C$ = 360

$45+120+45+\angle C = 360$

$\angle C = 150$

Now

$\angle C +S = 180$

$\delta = 30^\circ$

Hence (B) is correct answer

961449_1050741_ans_ab046bf7da264e86b1b153ad128853f7.jpeg

If $\mu ,$ represents the refractive index when a light ray goes from medium i to j, then $_2{\mu _1}{\text{ x}}{{\text{ }}_3}{\mu _2}{\text{ x}}{{\text{ }}_4}{\mu _3}$ is equal to

0%
$_3{\mu _1}$
0%
$_3{\mu _2}$
0%
$\dfrac{1}{{_1{\mu _4}}}$
0%
$_4{\mu _2}$

Explanation

According to the question,

$_2\mu_1\times _3\mu_2 \times _4\mu_3 =\dfrac {\mu_1}{\mu_2}\times \dfrac {\mu_2}{\mu_3}\times \dfrac {\mu_3}{\mu_4}=\dfrac {\mu_1}{\mu_4}= _4\mu_1 =\dfrac {1}{_1\mu_4}$

An air bubble is inside water. The refractive index of water is

$4/3$ . At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble:

0%
$2R$
0%
$3R$
0%
$4R$
0%
The air bubble cannot form a real image

Explanation

$\begin{array}{l} { n_{1}=n_{2} } \\ \text { Here, } \\ \text { But here is } \gamma \text { (not possibhe }\\ \text { is this case ) } \\ \text { So the air bubble cannot form a } \\ \text { real inage } \\ \text { Option D is correct. } \end{array}$
2007777_1024886_ans_ec838f23dcc04f179eb996c1c4df2a9c.PNG

2007777_1024886_ans_ec838f23dcc04f179eb996c1c4df2a9c.PNG

A rectangle tank

$6m$ deep is full of water . By how much does the bottom of the tank appear to be raised when viewed from above. Given

$(\mu_{water} = \frac{4}{3})$

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3 m
0%
4.5 m
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1.5 m
0%
6 m

A light ray is incident normally on the surface $AB$ of a prism of refracting angle $60^\circ$ . If the light ray does not emerge from $AC$ , then find the refractive index of the prism.

0%
$\mu \geqslant \cfrac{\sqrt{3}}{2}$
0%
$\mu \leqslant \cfrac{\sqrt{3}}{2}$
0%
$\mu \geqslant \cfrac{2}{\sqrt{3}}$
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$\mu \leqslant \cfrac{2}{\sqrt{3}}$

Explanation

Here, for no emergence through AC, TIR happens at AC.

$\therefore \mu \sin i\ge 1\times \sin 90°$

$\mu \ge \dfrac{1}{\sin i}$

$\mu \ge \dfrac{1}{\sin 60°}$

$\mu \ge \dfrac{2}{\sqrt{3}}$

A man wearing glasses of focal length + 1m cannot clearly see beyond 1 m

0%
if he is farsighted
0%
if he is nearsighted
0%
if his vision is normal
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in each of these cases.

A point object is placed at a distance

$10cm$ and its real image is formed at a distance of

$20cm$ from concave mirror. If the object is moved by

$0.1cm$ towards the mirror, the image will shift by about

0%
$0.4cm$ away from the mirror
0%
$0.4cm$ towards the mirror
0%
$0.8cm$ away from the mirror
0%
$0.8cm$ towards the mirror

Explanation

$We\quad know\quad \dfrac { 1 }{ f } =\dfrac { 1 }{ u } +\dfrac { 1 }{ v } ;\quad \\ where\quad f=focal\quad length=?\\ u=\quad initial\quad distance=20\quad cm\\ v=\quad final\quad distance=10cm\\ \dfrac { 1 }{ f } =\dfrac { 1 }{ -20 } +\dfrac { 1 }{ (-10) } \\ f=\dfrac { 20 }{ 3 } \quad cm\\ Now\quad object\quad moved\quad towards\quad the\quad mirror=0.1\quad cm,so\quad the\quad new\quad value\quad of\quad u=9.9\quad cm\\ Again\quad \quad \quad \\ \dfrac { 1 }{ f } =\dfrac { 1 }{ { u }_{ 1 } } +\dfrac { 1 }{ { v }_{ 1 } } \\ \dfrac { 1 }{ f } -\dfrac { 1 }{ { u }_{ 1 } } =\dfrac { 1 }{ { v }_{ 1 } } \\ \dfrac { 1 }{ { v }_{ 1 } } =\quad \dfrac { 3 }{ 20 } -\dfrac { 1 }{ (-9.9) } \\ { v }_{ 1 }=20.4\quad cm\\ The\quad change\quad in\quad final\quad distance\quad =20.4cm\quad -20\quad cm=0.4\quad cm\\ \\ v=\dfrac { 9\times 1600 }{ 3600 } =\quad 4cm/s$

A thin concavo-concave lens is surrounded by two different liquids A and B as shown in figure. The system is supported by a plane mirror at the bottom. Refractive index of A, lens and B are 9/5 , 7/5 and 8/7 respectively.The radius of curvature of the surfaces of the lens are same and equal to 10 cm. Where should an object be placed infront of this system so that final image is formed on the object itself.

0%
60 cm
0%
75 cm
0%
50 cm
0%
None

A telescope of objective lens diameter

$2m$ uses light of wavelength

$5000 \mathring {A}$ for viewing starts. The minimum angular separation between two stars whose image is just resolved by their telescope is:

0%
$4\times 10^{-4}rad$
0%
$40.25times 10^{-6}rad$
0%
$0.31\times 10^{-6}rad$
0%
$5\times 10^{-3}rad$

Explanation

Given that,

Diameter $d=2\,m$

Wave length $\lambda =5000\,\overset{\circ }{\mathop{A}}\,$

Now, minimum angular separation is

$\Delta \theta =\dfrac{1.22\lambda }{d}$

$\Delta \theta =\dfrac{1.22\times 5000\times {{10}^{-10}}}{2}$

$\Delta \theta =0.3\times {{10}^{-6}}\,rad$

Hence, the resolving power is $0.3\times {{10}^{-6}}\,rad$

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describe best the image formed of an object of height 2 cm placed 30 cm from the lens?

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Real, inverted, height = 1 cm
0%
Virtual, upright, height = 1 cm
0%
Virtual, upright, height = 0.5 cm
0%
Real, inverted, height =4 cm

A Plano convex lens is made of glass of refractive index 1.5.The of curvature of its convex surface is R. Its focal length:

0%
R/2
0%
R
0%
2R
0%
1.5 R

Explanation

Given,

Radius of curvature, $R$

Refractive index = $1.5$

Focal length of lens, $f=\dfrac{R}{(\mu-1)}=\dfrac{R}{(1.5-1)}=2R$

A transparent cube contains a small air bubble. Its apparent is

$2\ cm$ when seen through one face and

$5\ cm$ when seen through other face. If the refractive index of the material of the cube is

$1.5$ , the real length of the edge of cube must be

0%
$7\ cm$
0%
$7.5\ cm$
0%
$10.5\ cm$
0%
$\dfrac{14}{3}\ cm$

Explanation

When viwed through (1),

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

$\dfrac{1}{2}-\dfrac{1.5}{u}=\dfrac{0.5}{\infty}$ (

$R=\infty$ ,plane)

$\dfrac{1}{2}=\dfrac{1.5}{u}$

$u=3cm$

When viewed through (1),

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

$\dfrac{1}{5}=\dfrac{1.5}{L-u}$

$L-u=7.5$

$L=7.5+3=10.5cm$

1034126_1120500_ans_c44479ce0ea2428abfff63eaa5cd700e.JPG

A lens forms a vertical image 14 cm away from it when an object is placed 10 cm away from it. The lens is a....... less of focal length.......

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concave , 6.67 cm
0%
concave , 2.86 cm
0%
convex , 2.86 cm
0%
none

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 cm tall tower at a distance of 1 Km is observed through this telescope in normal setting then the image of the tower is

$\theta$ then

$\theta$ is close to;

0%
$60^\circ$
0%
$86^\circ$
0%
$90^\circ$
0%
$2^\circ$

Explanation

Magnification from mirror

$m = \dfrac { h _ { i } } { h _ { o } } = \dfrac { - v } { u }$

Whereas

$h _ { i } =$ Height of image from principal axis

$h _ { o } =$ Height of object from principal axis

$u =$ Object distance

$v =$ Image distance

Angular magnification $m = \dfrac { f _ { o } } { f _ { e } } = \dfrac { 150 } { 5 } = 30$

$\therefore \quad \dfrac { \tan \beta } { \tan \alpha } = 30 \Rightarrow \tan \beta = \tan \alpha \times 30$

$\tan \beta = \left( \dfrac { 50 } { 1000 } \times 30 \right) = \dfrac { 15 } { 10 } = \dfrac { 3 } { 2 }$

$\tan \beta = \dfrac { 3 } { 2 } \Rightarrow \tan \beta = \tan ^ { - 1 } \left( \dfrac { 3 } { 2 } \right)$

$\theta = \beta \simeq 60 ^ { \circ }$

At what angle will a ray of light be inclined on one face of an equilateral prism, so that the emergen ray may graze the second surface of the prism $\left( {\mu = 2} \right)$

0%
${30^0}\;$
0%
$\,{90^0}$
0%
$\;{45^0}$
0%
$\;{60^0}$

Explanation

REF.Image

According to the question

For the emergent ray

$\dfrac{sin r_{2}}{sin 90^{\circ}}= \dfrac{1}{\mu }$

$sin r_{2}=\dfrac{1}{2}$

$r_{2}= 30^{\circ}$

Also, by the

$eq^{n}$

$r_{1}+r_{2}=A= 60^{\circ}$

$r_{1}+30^{\circ}= 60^{\circ}\Rightarrow r_{1}= 30^{\circ}$

Then,

$\dfrac{sin i}{sin r_{1}}=\mu \Rightarrow sin\, i= \dfrac{2}{2}=1$

$i = 90^{\circ}$

1178299_1142000_ans_50bcfee2588f441da08fb9c46dfdb51e.jpg

A thin equiconvex glass lens of refractive index

$1.5$ has power of

$5D$ . When the lens is immersed in a liquid of refractive index

$\mu$ , it acts as a divergent lens of focal length

$100 \,cm$ . The value of

$\mu$ of liquid is

0%
$4/3$
0%
$3/4$
0%
$15/11$
0%
$8/3$

Explanation

Formula for focal length

$\dfrac{1}{f}=\left(\dfrac{\mu_{2}}{\mu_{1}}=1\right)\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)$

Here,

$\dfrac{1}{f}=50\mu_{2}=1.5\mu_{1}=1$

$5=(1.5-1)\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]$ ........

$(1)$

$\therefore f=100\ cm=1\ m$

$\mu_{2}=1.5$ let

$\mu_{1}=x$

$\therefore \dfrac{1}{1}=\left(\dfrac{1.5}{x}-1\right)\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)$ ........

$(2)$

Dividing

$(1)$ by

$(2)$

$\dfrac{5}{1}=\dfrac{0.5}{\dfrac{1.5}{x}-1}$

$\therefore \dfrac{1.5}{x}-1=\dfrac{0.5}{5}$

$\therefore \dfrac{1.5}{x}=\dfrac{11}{10}$

$\therefore \boxed{x=\dfrac{15}{11}}\therefore \mu_{1}=\dfrac{15}{11}=1.37$

Time taken by the sunlight to pass through a window of thickness 4mm whose refractive index is 1.5 is

0%
$3 \times 10^{-8}s$
0%
$2 \times 10^{-8}s$
0%
$3 \times 10^{-11}s$
0%
$2 \times 10^{-11}s$

Explanation

Given,

Refractive index of medium = 1.5

Thickness of window

$=4 \ mm=4\times 10^{-3} \ m$

Velocity of light in the window

$=\dfrac {3\times 10^8}{1.5}ms^{-1}=2\times 10^8 ms^{-1}$

Hence

$t=\dfrac {4\times 10^{-3}}{2\times 10^8}s=2\times 10^{-11}s$

An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of object on a screen placed at distance 75 cm from the lens. Find : (i) the focal length of lens, and (ii) the magnification.

0%
(i) 15.75 cm (ii) -2
0%
(i) 18.75 cm (ii) -3
0%
(i) 19.75 cm (ii) 2
0%
(i) 17.75 cm (ii) -3

Explanation

$\begin{array}{l} u=25 cm \\ v=75 \mathrm{~cm} \\ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{75}-\frac{1}{25}=\frac{1}{f} \\ \frac{1+3}{75}=\frac{1}{f} \\ f=+\frac{75}{4} \end{array}$

$\begin{array}{l} =18.75 \mathrm{~cm} \\ \text { Magnification }=\frac{v}{u}=\frac{75}{-25}=-3 \text { Ans } \end{array}$

2009056_1169212_ans_bc1755b0b97646f193d49216989a7b17.jpg

Two plano concave lenses of glass of refractive index 1.5 have radii of curvature 20 cm and 30 cm respectively. They are placed in contact with the curved surface towards each other and the space between them is filled with a liquid of refractive index 5/The focal length of the combination is (in cm)

0%
6
0%
-92
0%
108
0%
12

A double convex lens

$( \mu = 1.5 )$ in air has its focal length equal to

$20\ cm$ . When immersed in water

$( \mu = 4 / 3 )$ its focal length will be

0%
$20\ cm$
0%
$80 / 9\ cm$
0%
$360 / 9\ cm$
0%
$80\ cm$

Explanation

Given,

Refractive indexes for, glass ${{\mu }_{g}}=1.5$ , water ${{\mu }_{w}}=\dfrac{4}{3}$ and ${{\mu }_{a}}=1$ .

If f is the focal length of the lens in air then

$\dfrac{1}{{{f}_{a}}}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)\times \left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\,\,.........\,\,(1)$

If f is the focal length of the lens in water then
$\dfrac{1}{{{f}_{w}}}=\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)\times \left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\,\,.........\,\,(2)$

Divide equation (1) by (2)

$\dfrac{\dfrac{1}{{{f}_{a}}}}{\dfrac{1}{{{f}_{w}}}}=\dfrac{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)}{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)}$

${{f}_{w}}=\left( \dfrac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{a}}}\times \dfrac{{{\mu }_{w}}}{{{\mu }_{g}}-{{\mu }_{w}}} \right){{f}_{a}}$

${{f}_{w}}=\left( \dfrac{1.5-1}{1}\times \dfrac{4/3}{1.5-4/3} \right)\times 20=80\,cm$

Hence, focal length in water, ${{f}_{w}}=80\,cm$

The angle of a prism is

$6^0$ and its refractive index for green light is 1.If a green ray passes through it, the deviation will be :-

0%
$30^0$
0%
$15^0$
0%
$9^0$
0%
$3^0$

Explanation

Given,

Angle of prism,

$A={6}^{o}$

Refractive index,

$\mu =1.5$

Deviation,

$\delta =(\mu -1)A$

$=(1.5-1){6}^{o}$

$\delta ={3}^{o}$

The focal point of an equi- convex lens whose refractive index is

$1.5$ is

$10\ cm$ in air. Its focal point inside a liquid of refractive index

$1.25$ is

0%
$20\ cm$
0%
$25\ cm$
0%
$30\ cm$
0%
$50\ cm$

Explanation

Given that,

Focal length $f=10\,cm$

Refractive index ${{\mu }_{1}}=1.5$

Refractive index ${{\mu }_{2}}=1.25$

We know that,

$\dfrac{1}{f}=\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

$\dfrac{1}{10}=\left( 1.5-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(I)$

$\dfrac{1}{f}=\left( 1.25-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(II)$

Now, divided equation (I) by equation (II)

$\dfrac{f}{10}=\dfrac{0.5}{0.25}$

$f=20\,cm$

Hence, the focal length is $20\ cm$

A person can not see the distant objects clearly (though he can see the nearby object clearly). He is suffering from the defect of vision called:

0%
cataract
0%
hypermetropia
0%
myopia
0%
presbyopia

A person cannot see distant objects clearly. his vision can be corrected by using the spectacle containing :

0%
concave lenses
0%
plane lenses
0%
contact lenses
0%
convex lenses

A 2 cm diameter coin rests flat on the bottom of a bowl in which the water is which the water is 20 cm deep

$\left( { \mu }_{ w }=4/3 \right)$ . if the coin is viewed directly from above, what is the apparent diameter:

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2 cm
0%
1.5 cm
0%
2.67 cm
0%
1.67 cm

which instrument is used to check whether the body is charged or not?

0%
Telescope
0%
Lighting conductor
0%
Microscope
0%
Electroscope

An object placed at a distance of 16 cm from first principal focus of convex lens, produces a real image at a distance of 36 cm from its second principal focus. Then the focal length of the lens is :

0%
$9 cm$
0%
$25 cm$
0%
$24 cm$
0%
$17 cm$

Explanation

Let focal length of the lens be

$f$ ,

Given,

Object distance

$=16+f=u$

Image distance

$=36+f=v$

Distance of 1st and 2nd principal focus is same from the optical centre

From, lens formula

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

$\cfrac{1}{f}=\cfrac{1}{36+f}+\cfrac{1}{16+f}$

$\cfrac{1}{f}=\cfrac{16+f+36+f}{(36+f)(16+f)}$

On solving

$f=24cm$

A young man has to hold a book at arm's length to be able to read it clearly. The defect of vision is:

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astigmatism
0%
myopia
0%
presbyopia
0%
hypermetropia

Though a woman can see the distant objects clearly, she cannot see the nearby object clearly. She is suffering from the defect called :

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Long-sight
0%
Short-sight
0%
Hind-sight
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Mid-sight

In a thin spherical fish bowl of radius

$10\ cm$ filled with water of refractive index

$\dfrac{4}{3}$ , there is a fish at a distance of

$4\ cm$ there is a small fish at a distance of

$4\ cm$ form the center

$C$ as shown. The image of the fish appears, when seen from the point

$D$ would be

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$5.2\ cm$
0%
$4.8\ cm$
0%
$2.6\ cm$
0%
$7.4\ cm$

Explanation

$v \to image\,position$

$u \to object\,position$

$R \to Radius\,of\,surface$

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

$\Rightarrow \dfrac{1}{v} - \dfrac{{\left( {4/3} \right)}}{{\left( { - 6} \right)}} = \dfrac{{1 - 4/3}}{{\left( { - 10} \right)}}$

$\Rightarrow \dfrac{1}{v} + \dfrac{4}{{3 \times 6}} = \dfrac{1}{{3 \times 10}}$

$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}} - \dfrac{4}{{18}}$

$\Rightarrow \dfrac{1}{v} = \dfrac{{ - 17}}{{90}}$

$\Rightarrow v = \dfrac{{ - 90}}{{17}} = 5.2\,cm$

Hence,

option

$(A)$ is correct answer.

Choose the wrong statement related to refraction of light

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Twinkling of stars
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Oval shaped of sun in morning and evening
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Object in water appears bigger in size
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Red light undergoes dispersion, while passing through prism

Explanation

* Red light undergoes dispersion, while passing thruugh brism the bhemenon is called dispersion 8 if leads to it is due to the long-wavelength of red light that is bents the least while violet bends the most , that's why red gets segrigated out, its not due to refraction of light. While 'A'B' C' option's are due to refraction of light

Hence abtion

$D$ is correct.

An observer looks at a distant tree of height

$10$ m with a telescope magnifying power of

$20$ . To the observer, the top appears

0%
10 times taller
0%
10 times nearer
0%
20 times taller
0%
20 times nearer

Explanation

$\begin{array}{l} \text { Work of telescope is to give an observer a } \\ \text { closer view on an object, it non-doubtly } \\ \text { magnifies image (but its telescope's internal mechanism) } \\ \text { the actual output which an observer, observes } \\ \text { is an close image. Here the telescope } \\ \text { will bring the object close by 20 times. } \\ \qquad E \text { : option } C^{\prime} \text { is correct. } \end{array}$

A convex lens produces an image of an object on a screen with a magnification of $\frac{1}{2}$ .When the lens is moved 30 cm away from the object, the magnification of the image is 2.The focal length of the lens is

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20 cm
0%
25 cm
0%
30 cm
0%
35 cm

Explanation

We know that

$m=\cfrac{f-v}{f}$

As given,

$\cfrac{1}{2}=\cfrac{f-v}{f}\\f=2(f-v)\\f=2f-2v\\2v=2f-f\\2v=f\\v=(\cfrac{f}{2})\rightarrow (i)$

Now we have

$m=2$

and

$v^1=(v+30)$

$\Rightarrow m=\cfrac{f-v}{f}\\Rightarrow 2=\cfrac{f-(v+30)}{f}\\ \Rightarrow 2=\cfrac{f-(f/2+30)}{f}\quad (from(i))$

$2f=f-f/2+30\\ \Rightarrow 2f=\cfrac{f}{2}+30\\ \Rightarrow 2f-\cfrac{f}{2}=30cm\\ \Rightarrow \cfrac{4f-f}{2}=30cm\\ \Rightarrow \cfrac{3f}{2}=30\\\Rightarrow 3f=60\\ \Rightarrow f=20cm$

Four identical lenses are kept one beside the other on the same optical as shown in the figure. The right surface of rightmost lens is silvered. Focal length of lens is 20 cm and radius of silvered is 20 cm. The focal length of the combined system is

0%
2 cm
0%
-2 cm
0%
5 cm
0%
-5 cm

Explanation

$\begin{array}{l}\text { for net focus of lenses } a, b, c \& d \text { we use formula } \\\frac{1}{F_{n e t}}=\frac{1}{f_{a}}+\frac{1}{f_{b}}+\frac{1}{f_{c}}+\frac{1}{f_{d}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{4}{20}=\frac{1}{5}&={F_{net}}=5cm\end{array}$

for net focus of the systern we use formula

$\frac{1}{F_{\text {net }}}=\frac{-2}{f_{\text {lens }}}+\frac{a_{2}}{R_{\text {mirror }}}=\frac{-2}{5}+\frac{2}{(-20)}$

$F_{\text {net }}=-2 \mathrm{~cm}$

Hence

$B^{\prime}$ is correct

2001977_1280828_ans_e453cf1c76f44e61a3bb317f2709701e.PNG

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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