MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 11

The hyper-metropia is a

0%
Short-side defect
0%
Long-side defect
0%
Bad vision due to old age
0%
None of these

The focal length of a concave mirror is

$50\ cm.$ Where an object be placed so that its image is two times and inverted

0%
$75\ cm$
0%
$72\ cm$
0%
$63\ cm$
0%
$50\ cm$

Explanation

for Real image

$m=-2,$ so by using

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

$\Rightarrow -2=\dfrac{-50}{-50-u}\Rightarrow u=-75\ cm$

Retina of eye acts like ______ of camera.

0%
Shutter
0%
Film
0%
Lens
0%
None of these

The critical angle between an equilateral prism and air is

$45^{o}$ . If the incident ray is perpendicular to the refracting surface, then

0%
After deviation it will emerge from the second refracting surface
0%
It is totally reflected on the second and third emerges out perpendicularly from third surface in air
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It is totally reflected on the second and third refracting surface and finally emerges out from the first surface
0%
It is a totally reflected from all the three sides of prism and never emerges out

If the speed of light in vacuum is

$C\ \ m/sec$ , then the velocity of light in a medium of refractive index

$1.5$ is

0%
$1.5\times C$
0%
$C$
0%
$\dfrac {C}{1.5}$
0%
Can have any velocity

Explanation

Spped of light in vacuum

$=C$ m/s

Refrective index of medium

$\mu=1.5$

Velocity of light in any medium having refrective index

$\mu$ ,

$v=\dfrac C\mu$

On putting the value of

$\mu$ ,

$v=\dfrac C{1.5}$

Option D.

Two parallel pillars are

$11\ km$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

0%
$3.2\ m$
0%
$20.8\ m$
0%
$91.5\ m$
0%
$183\ m$

Explanation

As limit of resolution of eye is

$\left( \dfrac{1}{60}\right)^o$ , the pillars will be seen
distinctly if

$\theta > \left( \dfrac{1}{60}\right)^o$
i.e.

$\dfrac dx > \left( \dfrac{1}{60}\right)\times \dfrac{\pi}{180}$

$\Rightarrow d > \dfrac{\pi \times x}{60\times 180}$

$\Rightarrow d > \dfrac{3.14 \times 11 \times 10^3}{60\times 180}\Rightarrow d > 3.2\ m$
1749372_1824228_ans_b81716e340f14a1097948a7ed02097eb.png

A convex lens makes a real image

$4\ cm$ long on a screen. When the lens is shifted to a new position without disturbing the object, we again get a real image on the screen which is

$16\ cm$ tall. The length of the object must be

0%
$1/4\ cm$
0%
$8\ cm$
0%
$12\ cm$
0%
$20\ cm$

Explanation

As we know,

$O=\sqrt{I_1I_2}=\sqrt{4 \times 16}=8\ cm$

To remove myopia ( short sightedness ) a lens of power

$0.66\ D$ is required. The distant point of the eye is approximately

0%
$100\ cm$
0%
$150\ cm$
0%
$50\ cm$
0%
$25\ cm$

Explanation

Given,

Power of lens

$P=0.66\ D$

Far point of the eye

$=$ focal length of the lens

$=\dfrac{100}{P}=\dfrac{100}{0.66}=151\ cm$

Hence, The distant point of the eye is approximately

$151\ cm$

A prism

$(\mu=1.5)$ has the refracting angle of

$30^{o}$ . The deviation of a monochromatic ray incident normally on its one surface will be

$(\sin 48^{o}36=0.75)$

0%
$18^{o}36'$
0%
$20^{o}30'$
0%
$18^{o}$
0%
$22^{o}1'$

Explanation

For surface

$AC\dfrac{1}{\mu}=\dfrac{\sin 30^{o}}{\sin e}\Rightarrow \sin e=\mu\sin 30^{o}$

$\Rightarrow \sin e=1.5\times \dfrac{1}{2}=0.75$

$\Rightarrow e=\sin^{-1}(0.75)=48^{o}36'$
From figure

$\delta =e-30^{o}$

$=48^{o}36'-30^{o}=18^{o}36'$
1749362_1824218_ans_037f17bca89b403db3761d38028767fe.png

The separation between two microscopic particles is measured

$P_A$ and

$P_B$ by two different lights of wavelength

$2000\ A^o$ and

$3000\ A^o$ respectively, then

0%
$P_A > P_B$
0%
$P_A < P_B$
0%
$P_A < 3/2P_B$
0%
$P_A =P_B$

Explanation

Resolving limit ( minimum separation )

$\propto \lambda$

$\Rightarrow \dfrac{P_A}{P_B}=\dfrac{2000}{3000}\Rightarrow P_A < P_B$

A person is suffering from 'presbyopia' ( myopia and hyper metropia both defects ) should use

0%
A concave lens
0%
A convex lens
0%
A bifocal lens whose lower portion is convex
0%
A bifocal lens whose upper portion is convex

Velocity of light in a medium is

$1.5\times 10^8 m/s$ . Its refractive index will be

0%
$8$
0%
$6$
0%
$4$
0%
$2$

Explanation

$c$ =

$3 \times 10 ^{8} m/s^{-1}$ ........Speed of light in vaccum

$v$ =

$1.5 \times 10 ^{8} m/s^{-1}$ ........Speed of light in medium

$\mu =\dfrac {c}{v}=\dfrac {3\times 10^8}{1.5\times 10^8}=2$

A ray of light strikes a plane mirror

$M$ at an angle of

$45^o$ as shown in the figure. After reflection, the ray passes through a prism of refractive index

$4\$1.5$

$whose apex angle is$

$4^o$ $. The total angle through which the ray is deviated is

0%
$90^o$
0%
$91^o$
0%
$92^o$
0%
$93^o$

Explanation

As we know,

$\delta_{net}=\delta_{mirror}+\delta_{prism}$

$=(180-2i)+(\mu-1)A$

$=(180-2\times 45)+(1.5-1)\times 4=92^{o}$

The image distance of an object placed

$10\ cm%$ in front of a thin lens of focal length

$5cm$ is

0%
$6.5\ cm$
0%
$8.0\ cm$
0%
$9.5\ cm$
0%
$10.0\ cm$

Explanation

As we know,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \Rightarrow \dfrac{1}{+5}=\dfrac{1}{v}-\dfrac{1}{(-10)} \Rightarrow v=10\ cm$

On a glass plate a light wave is incident at an angle of

$60^o$ . If the reflected and the refracted waves are mutually perpendicular, the refractive index of material is

0%
$\dfrac {\sqrt 3}{2}$
0%
$\sqrt 3$
0%
$\dfrac {3}{2}$
0%
$\dfrac {1}{\sqrt 3}$

Explanation

From figure

$< i=60^o, < r=30^o$
so

$\mu =\dfrac {\sin 60}{\sin 30}=\sqrt 3$
1750589_1824265_ans_bacc43e35ef14e26bb8b00d298638a24.png

Refractive index of glass is

$\dfrac {3}{2}$ and refractive index of water is

$\dfrac {4}{3}$ . If the speed of light in glass is

$2.00\times 10^8\ m/s$ , the speed in water will be

0%
$2.67\times 10^8 m/s$
0%
$2.25\times 10^8 m/s$
0%
$1.78\times 10^8 m/s$
0%
$1.50\times 10^8 m/s$

Explanation

As we know,

$\mu=\dfrac{c}{v}$

$\mu\propto \dfrac {1}{v}$ where

$\mu$

is refractive index of medium ,

$c$ speed of light in vaccum and

$v$ is speed of light in medium

$\Rightarrow \dfrac {\mu_g}{\mu_w}=\dfrac {v_w}{v_g}\Rightarrow \dfrac {3/2}{4/3}=\dfrac {v_w}{2\times 10^8}$

$\Rightarrow v_w =2.25\times 10^8 m/s$

The relation between the linear magnification

$m,$ the object distance

$u$ and the focal length

$f$ is

0%
$m=\dfrac{f-u}{f}$
0%
$m=\dfrac{f}{f-u}$
0%
$m=\dfrac{f+u}{f}$
0%
$m=\dfrac{f}{f+u}$

Explanation

$\because m=-\dfrac{v}{u}$ also

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

$\Rightarrow -\dfrac{u}{v}=1-\dfrac{u}{f}\Rightarrow \dfrac{-v}{u}=\dfrac{f}{f-u}$ so

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

A lens which has focal length of

$4\ cm$ and refractive index of

$1.4$ is immersed in a liquid of refractive index

$1.6$ , then the focal length will be

0%
$-12.8\ cm$
0%
$32\ cm$
0%
$12.8\ cm$
0%
$-32\ cm$

Explanation

$\dfrac{f_l}{f_a}=$

$\left(\dfrac{_a \mu_g-1}{_l \mu_g-1} \right)$

$\Rightarrow \dfrac{f_l}{4}=\dfrac{(1.4-1)}{\dfrac{1.4}{1.6}-1} \Rightarrow f_l=-12.8\ cm$

The maximum magnification that can be obtained with a convex lens of focal length

$2.5\ cm$ is ( the least distance of distinct vision is

$25\ cm$ )

0%
$10$
0%
$0.1$
0%
$62.5$
0%
$11$

Explanation

As we know,

$m_{max}=1+\dfrac Df=1+\dfrac{25}{2.5}=11$ .

Which is not a part of a human eye ?

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Ratina
0%
Cornea
0%
Pupil
0%
Mid plane

At sun rise of sunset, the sun looks more red than at mid-day because

0%
The sun is hottest at these times
0%
Of the scattering of light
0%
Of the effects of refraction
0%
Of the effects of diffraction

Explanation

$I\propto\dfrac{1}{{\lambda}^4}$

According to Rayleigh’s law of scattering, intensity scattered is inversely proportional to the forth power of wavelength. So red is least scattered and sun appears Red.

The sky would appear red instead of blue if

0%
Atmospheric particles scatter blue light more than red light
0%
Atmospheric particles scatter all colours equally
0%
Atmospheric particles scatter red light more than the blue light
0%
The sun was much hotter

For a small angled prism, angle of prism

$A$ , the angle of minimum deviation

$(\delta)$ varies with the refractive index of the prism as shown in the graph

0%
Point $P$ corresponds to $\mu =1$
0%
Slope of the line $PQ=A/2$
0%
Slope $=P$
0%
None of the above statement is true

Explanation

$P, \delta=0=A(\mu-1)\Rightarrow \mu=1$
Also

$\delta_{m}=(\mu-1)A=A\mu_{m}-A$
Comparing it with

$y=mx+c$
Slope of the line

$=m=A$

Two point light source are

$24\ cm$ apart. Where should a convex lens of focal length

$9\ cm$ be put in between them from one source so that the images of both the sources are formed at the same place.

0%
$6\ cm$
0%
$9\ cm$
0%
$12\ cm$
0%
$15\ cm$

Explanation

The given condition will be satisfied only if one source

$(S_1)$ placed on one side such that

$u < f$ ( i.e. it lies under the focus).
The other source

$(S_2)$ is placed on the other side of the lens such that

$u > f$ (i.e. it lies beyond the focus).
If

$S_1$ is the object for lens then

$\dfrac 1f=\dfrac {1}{-y}-\dfrac{1}{-x}$

$\Rightarrow \dfrac 1y=\dfrac 1x-\dfrac 1f$ ......(i)
If

$S_2$ is the object for lens then

$\dfrac 1f=\dfrac{1}{+y}-\dfrac{1}{(-24-x)}\Rightarrow \dfrac 1y=\dfrac 1f-\dfrac {1}{(24-x)}$ ....(ii)
From equation (i) and (ii)

$\dfrac 1x-\dfrac 1f=\dfrac 1f-\dfrac {1}{(24-x)}\Rightarrow \dfrac 1x+\dfrac {1}{(24-x)}=\dfrac 2f=\dfrac 29$

$\Rightarrow x^2-24x+108=0$ . After solving the equation

$x=18\ cm, 6\ cm$
1756753_1825248_ans_d6a372e7d72f440ca16648cd61cca841.png

Colour of the sky is blue due to

0%
Scattering of light
0%
Total internal reflection
0%
Total emission
0%
None of the above

Which of the following event is not related to refraction of light ?

0%
The bottom of water filled bowl appears raised.
0%
Appearance of sun before sun rising and after sunset
0%
Formation of image by mirror
0%
Twinkling of stars

When a ray of light passes through the second optical medium with a change in the angle, the phenomenon is known as:

0%
Reflection of light
0%
Absorption of light
0%
Refraction of light
0%
None of these

The length of a simple astronomical telescope is equal to:

0%
Difference between focal length of two lenses
0%
Half of the sum of focal distances
0%
Sum of the focal distances
0%
Multiplication of the focal distances

Explanation

Length of a simple astronomical telescope

$L = f_{0} + f_{e}$

Small hole in Iris is called pupil.

0%
True
0%
False

Shortsightness eye defect persons cannot see nearby objects.

0%
True
0%
False

The apreture of astronomical telescope is large because .

0%
to remove spherical defect
0%
for high limit of revolution
0%
to anlarge observation area
0%
for less dispersion

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

0%
$120^{\circ}$
0%
$90^{\circ}$
0%
$60^{\circ}$
0%
$45^{\circ}$
0%
$30^{\circ}$

Explanation

Angle of incident at the boundary = 60
By snells law

$\Rightarrow \sqrt{3}\sin 60^{\circ}=1.5 \sin r$

$\Rightarrow r=90^{\circ}$
Thus the angle of deviation as seen in the figure is given as

$90^o-60^o=30^o$

A thin convex lens made of glass of

$\mu =$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index

$\dfrac{4}{3}$ . A parallel beam of light is incident on it. The position of the image is :

0%
$16 cm$
0%
$20 cm$
0%
$24cm$
0%
$10 cm$

Explanation

Data:

${ \mu }_{ a }=1,{ \mu }_{ g }=1.5,{ \mu }_{ w }=1.33,{ R }_{ 1 }=10cm,{ R }_{ 2 }=-10cm,u=-\infty$

For left surface,

$\dfrac{\mu_g}{v_1}-\dfrac{\mu_a}{u} = \dfrac{\mu_g-\mu_a}{R_1}$

For right surface,

$\dfrac{\mu_w}{v}-\dfrac{\mu_g}{v_1} = \dfrac{\mu_w-\mu_g}{R_2}$

From above two,

$\dfrac { { \mu }_{ w } }{ v } -\dfrac { { \mu }_{ a } }{ u } =\dfrac { { \mu }_{ g }-{ \mu }_{ a } }{ { R }_{ 1 } } +\dfrac { { \mu }_{ w }-{ \mu }_{ g } }{ { R }_{ 2 } }$

or,

$\dfrac { 1.33 }{ v } =\dfrac { 1.5-1 }{ 10 } -\dfrac { 1.33-1 }{ 10 }$

$v=\dfrac { 1.33\times 10 }{ 0.66 } =20\ cm$

A uniform, horizontal beam of light is incident upon a prism as shown in the fig. The prism is in the shape of a quarter cylinder of radius R

$=$ 5cm, and has a refractive index

$\dfrac{2}{\sqrt{3}}$ . A patch on the table for a distance x from the cylinder is unilluminated. The value of x is

0%
$2.5$ cm
0%
$5$ cm
0%
$5 \sqrt{3}$ cm
0%
$10$ cm

Explanation

Figure-(I) shows how for rays nearer to base are refracted but as rays move away from base, they fall nearer and after a certain distance above base, all rays are internally reflected.

So, we have to find the distance x where there is no light. As on increasing height rays fall nearer to the prism, so we have to find where will the ray for critical angle of incidence will intersect base line. And inside that point there will be no illumination as for larger angle of incidence , rays are internally reflected.

From figure-2, it is clear that

$i=C$ =critical angle and refracted ray for this goes perpendicular to normal which is radius

$OA$ here. Refracted ray is

$AB$ .

$i=C=\sin^{_1}\dfrac{1}{\mu}=\sin^{-1}\dfrac{\sqrt{3}}{2}=60^{o}$

From

$\Delta OAB:-$

$\cos i=\dfrac{OA}{OB}=\dfrac{R}{R+x}$

$\implies\cos 60^{o}= \dfrac{1}{2}=\dfrac{5}{5+x}$

$\implies x=5cm$

Answer-(B)

The refracting angle of a prism

$60^{\mathrm{o}}$ .The refractive index of the material of the prism is

$\sqrt{\dfrac{7}{3}}$ . The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

0%
$30^{0}$
0%
$45^{\mathrm{o}}$
0%
$40^{\mathrm{o}}$
0%
$50^{\mathrm{o}}$

Explanation

$\begin{array}{l}\text { Given } A=60^{\circ}, \mu=\sqrt{\frac{7}{3}} \\\text { we know that } \gamma_{1}+r_{2}=A \\\Rightarrow r_{1}+r_{2}=60^{\circ}\end{array}$

$\begin{array}{l}\text { Snell's law at first refraction } \\\qquad \begin{array}{l}\sin i=\mu \sin \gamma_{1} \\\operatorname{Sin}\gamma_{1}=\sqrt{\frac{3}{7}} \sin i\end{array} \\\text { Snell's law at second refraction }\end{array}$

$\begin{aligned}\mu \sin r_{2} &=1 \\ & \Rightarrow \sin r_{2}=\sqrt{\frac{3}{7}} \\ r_{2}=\sin ^{-1}\left(\sqrt{\frac{3}{7}}\right) \\ r_{2}=& 40.89^{\circ} \Rightarrow r_{1}=19.1066^{\circ}\end{aligned}$

$\begin{aligned}\Rightarrow\operatorname{Sin} r_{1} &=0.327 \\\sin i &=\sqrt{\frac{7}{3}} \sin r_{1} \\\sin i &=1 / 2 \\ i &=30^{\circ}\end{aligned}$

A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices

$\mu_{1}=\sqrt{2}, \mu_{2}=\sqrt{x}$ and

$\mu_{3}=\sqrt{3}$ . If finally emergent light ray also emerges at grazing angle then calculate x :

0%
2
0%
1
0%
5
0%
3

Explanation

Apply Snell's law on various surfaces one by one :

1 sin

$90^{\circ}$

$=\mu_{1}$ sin

$r_{1} \Rightarrow$ sin

$r_{1} = \dfrac{1}{\sqrt{2}} \Rightarrow r_{1}=45^{\circ}$

We have the incident angle for the second interface given using the relation sin

$i_2$ =cos

$r_1$
Thus we get

$\mu_{1}$ cos

$r_{1}$ =

$\mu_{2}$ sin

$r_{2} \Rightarrow$ sin

$r_{2}$ =

$\dfrac{1}{\mu_{2}}$ (cos

$r_2$ =

$\dfrac{1}{\sqrt2}$ )

$\mu_{2}$ cos

$r_{2}=\mu_{3}$ sin

$r_{3}$

$\Rightarrow$ sin

$r_{3}$

$=\dfrac{\mu_{2}\sqrt{1-sin^{2}r_{2}}}{\sqrt{3}}$

$\mu_{3}$ cos

$r_{3}=1$ =

$\dfrac{\mu_{2}^{2}-1}{\sqrt{3}}$

sin

$^{2} r_{3}$ + cos

$^{2} r_{3}=1$

$\Rightarrow \dfrac{\mu_{2}^{2}-1}{3}+\dfrac{1}{3}=1 \Rightarrow {\mu_{2}}^{2}=3\Rightarrow \mu_{2}=\sqrt{3}$

Thus we get

$x=3.$

In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of

$\theta$ so that light incident normally on the face AB does not cross the face BC is (given sin

$^{-1}$ (3/5)

$=$ 37

$^o$ )

0%
$\theta \leq 37^o$
0%
$\theta > 37^o$
0%
$\theta \leq 53^o$
0%
$\theta < 53^o$

Explanation

For no refraction at surface BC

$\dfrac { sin(i) }{ sin(r) } =\dfrac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } \\ for\quad no\quad refraction\quad at\quad surface\quad BC\quad$

$r\ge { 90 }^{ 0 }$

here,

${ \mu }_{ 1 }=\dfrac { 3 }{ 2 } \ (refractive\ index\ of\ ABC)\\$

${ \mu }_{ 2 }=\dfrac { 6 }{ 5 } \ \ \ (refractive\ index\ of\ BCDE)$

thus

$\dfrac { sin(i) }{ sin(90) } =\dfrac { \dfrac { 6 }{ 5 } }{ \dfrac { 3 }{ 2 } } =\dfrac { 4 }{ 5 } \\$

$\Rightarrow sini\ge \dfrac { 4 }{ 5 } \\$

$\Rightarrow i\ge { 53 }^{ 0 }$

here in triangle ABC

$i={ 90 }^{ 0 }-\theta \\ \Rightarrow { 90 }^{ 0 }-\theta \ge { 53 }^{ 0 }\\ \theta \le { 3 }7^{ 0 }$

Rising and setting sun appears to be reddish because:

0%
The sun is colder at sunrise or at sunset
0%
Diffraction sends red rays to the earth at these times
0%
Refraction is responsible for it
0%
Scattering due to dust particles and air molecules is responsible for it

Explanation

According to Rayleigh's law of scattering, amount of scattering

$\alpha \frac {1}{\lambda^4}$ .
Red having maximum wavelength straight to the observer.
Most of the scattered light goes towards the sky, so the sky appears blue.

Two refracting media are separated by a spherical interface as shown in figure. PP' is the principle axis.

$\mu _1$ and

$\mu _2$ are the refractive indices of medium of incidence and medium of refraction respectively. Then,

0%
if $\mu _2>\mu _1$ then there cannot be a real image of real object.
0%
if $\mu _2>\mu _1$ then there cannot be a real image of virtual object.
0%
if $\mu _1>\mu _2$ then there cannot be a virtual image of virtual object
0%
if $\mu _1>\mu _2$ then there cannot be a real image of real object

Explanation

$\dfrac{{\mu}_{2}}{v} - \dfrac{{\mu}_{1}}{u} = \dfrac{{\mu}_{2} - {\mu}_{1}}{R}$

Now, R is always negative (Considering our medium of incidence and direction of incidence is taken as +ve)

Case 1 :

${\mu}_{2} > {\mu}_{1}$ and

$u < 0$ (Real object)

$\underbrace { \dfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ R } }_{ -ve } =\dfrac { { \mu }_{ 2 } }{ v } \underbrace { -\dfrac { { \mu }_{ 1 } }{ -u } }_{ +ve } \Rightarrow v$

has to be negative
Thus, the image can never be real.

Case 2 :

${\mu}_{2} < {\mu}_{1}$ and

$u > 0$ (Virtual object)

$\underbrace { \dfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ R } }_{ +ve } =\dfrac { { \mu }_{ 2 } }{ v } \underbrace { -\dfrac { { \mu }_{ 1 } }{ -u } }_{ -ve } \Rightarrow v$ has to be positive +ve, Thus, the image can never be virtual.

A ray incident at a point at an angle of incidence

$\theta$ enters into a glass sphere placed in air which is reflected and refracted at the farther surface of the sphere as shown in the figure. The angle between reflected and refracted rays at this surface is

$90^{0}$ . If refractive index of the sphere is

$\sqrt{3} ,$ the angle

$\theta$ is :

0%
$\dfrac{\pi }{3}$
0%
$\dfrac{\pi }{4}$
0%
$\dfrac{\pi }{6}$
0%
$\dfrac{2\pi }{3}$

Explanation

Applying Snell's law on the 2 surfaces,

$sin \theta = \sqrt{3} sin r$

Where, r is the angle of refraction.

Applying on the second surface,

$sin r = cos \theta$

Equating the 2 terms,

$sin \theta = \sqrt{3} cos \theta$

Thus,

$tan \theta = \sqrt{3}$

Or,

$\theta = 60^0$

A person who can see the nearer objects clearly but not distant objects is suffering from ....................

0%
Emnetropia
0%
Blindness
0%
Myopia
0%
None of these

A ray of light is incident normally on one face of a prism as shown in figure. The refractive index of the material of the prism is

$\dfrac{5}{3}$ and the prism is immersed in water of refractive index

$\dfrac{4}{3}$ then

0%
The angle of emergence $\Theta_{2}$ of the ray is $sin^{-1}(\dfrac{5}{8})$
0%
The angle of emergence $\Theta_{2}$ of the ray is $sin^{-1}(\dfrac{5}{4\sqrt{3}})$
0%
The angle of emergence $\Theta_{2}$ of the ray is $sin^{-1}(\dfrac{7}{3\sqrt{4}})$
0%
Total internal reflection will not occur at P if the refractive index of water increases to a value greater than $\dfrac{5}{2\sqrt{3}}$ by dissolving some substance

Explanation

Initially at P, there is TIR, There force at Q (snells law)

$\dfrac{5}{3} sin 30^{0} = \dfrac{4}{3} sin \Theta_{2}$

$\Rightarrow sin \Theta_{2} = \dfrac{5}{8} \Rightarrow \Theta_{2} = sin^{-1}(\dfrac{5}{8})$

$\mu_{w} > \dfrac{5}{2 \sqrt{3}} \Rightarrow sin C > \dfrac{5/2\sqrt{3}}{2/3} \Rightarrow C > 60^{0}$

Therefore, now TIR will not occur at P

Consider an optical system placed in water

$\left ( \mu=4/3 \right )$ PQRS is a hollow glass slab filled with air

$\left ( \mu=1 \right )$ This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle

$\theta$ with the axis of the lens.

$\left ( 0<\theta < 180^{\circ} \right )$ , then

0%
for transmission of light takes place through slab $\theta _{min}=\dfrac{\pi }{2}$ ${-}$ sin $^{-1}\left ( 0.75 \right )$
0%
for transmission of light through slab $\theta _{max} =\dfrac{\pi }{2}+$ sin $^{-1}\left ( 0.75 \right )$
0%
for $x = 50$ cm final image is at 3m distance from O
0%
for $x = 100$ cm final image is at 3m distance from O

Explanation

$if \theta < 90^{\circ}$ and If there is no TIR.

$-\theta <$ sin

$^{-1}\left ( \dfrac{3}{4} \right )$

$\theta >\dfrac{\pi }{2}-$ sin

$^{-1}\dfrac{3}{4}$

Similarly if

$\theta > 90^{\circ}$ , then

for No TIR

$\theta <\dfrac{\pi }{2}+$ sin

$^{-1}\dfrac{3}{4}$

$\left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4}<\theta <\dfrac{\pi }{2}+sin ^{-1}\dfrac{3}{4} \right )$

For calculating focal length of lens in air

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

$\dfrac{1}{50}=\left ( \dfrac{3}{2}-1 \right )\left ( k \right )\Rightarrow k=\dfrac{1}{25}$

$f_{w}^{g}=4f_{a}^{G}=4 \times 50cm =200m$

$\dfrac{1}{f_{w}^{o}}=\left ( \dfrac{3}{2}\dfrac{3}{4}-1 \right )\left ( k \right )=\dfrac{k}{8}$

$\Rightarrow f_{w}=200cm$

So that final image will be formed 200 cm in right of lens.

$\Rightarrow Image \ distance,OI = (100 + 200) cm = 300 cm$

Ans. (a) and (b)

$\left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right )$

(d) OI = (100 + 200) cm = 300 cm but this distance is valid only when

$\theta$ is between

$\theta _{min}$ to

$\theta _{max}$

Ans. (a) and (b)

$\left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right )$

The angle of crown glass

$(\mu=1.52)$ prism is

$5^o$ . What should be the angle of flint glass

$(\mu=1.63)$ prism so that the two prisms together may be used in a direct vision spectroscope?

0%
$-2.14^o$
0%
$+2.14^o$
0%
$-4.12^o$
0%
$+4.12^o$

Explanation

For direct vision spectroscope,

$\delta_1+\delta_2=0$

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

$\Rightarrow A_2=-[\dfrac {\mu_1-1}{\mu_2-1}]A_1$

$=-[\dfrac {1.52-1}{1.63-1}]\times 5^0=-4.12^o$

The lens of a simple magnifier has a focal length of 2.5 cm. Calculate the angular magnification produced when the image is at the least distance of distinct vision.

0%
10
0%
5
0%
8
0%
15

Explanation

Focal length of a magnifier

$f=2.5 cm$

(a) At least distance of distant vision, Angular magnification

$=1+\dfrac {D}{f}=1+\dfrac {25 cm}{2.5 cm}$

$=11$

(b) At infinity, angular magnification

$=\dfrac {D}{f}=\dfrac {25 cm}{2.5 cm}=10$

The separation between the lenses :

0%
10 cm
0%
12 cm
0%
18 cm
0%
20 cm

Explanation

The length of the tube is given as,
L =

${f}_{o} + {f}_{e}$

$L = 16 + 2 = 18$

A thin prism

$P_1$ with angle

$6^o$ and made from glass of refractive index 1.54 is combined with another thin prism

$P_2$ of refractive index 1.72 to produce dispersion without deviation. The angle of prism

$P_2$ will be.

0%
$4^o 30'$
0%
$8.5^o$
0%
$6.5^o$
0%
None of these

Explanation

We know that for small angle of prism, deviation ,

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

where

$A$ =angle of prism

Here, net deviaton

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

$\implies A_2=\dfrac{A_1(\mu_1-1)}{\mu_2-1}$

$\implies A_2=\dfrac{6\times 0.54}{0.72}=4.5^{o}$

$\implies A_2=4^{o}30'$

Answer-(A)

A ray

$OP$ of monochromatic light is incident on the face

$AB$ of prism

$ABCD$ near vertex

$B$ at an incident angle of

$60^{\circ}$ (see figure). If the refractive index of the material of the prism is

$\sqrt {3}$ , which of the following is (are) correct?

0%
The ray gets totally internally reflected at face $CD$
0%
The ray comes out through face $AD$
0%
The angle between the incident ray and the emergent ray is $90^{\circ}$
0%
The angle between the incident ray and the emergent ray is $120^{\circ}$

Explanation

Ray OP goes parallel to the side BC. Hence the angle of incidence of this ray on the side CD becomes

$45^{\circ}$ .

Applying Snell's Law at the point Q,

$\sqrt{3} \sin 45^{\circ}=\sin r>1$

Hence, this is a case of total internal reflection.

Thus the angle of incidence on Q is equal to the angle of reflection.

This forms

$30^{\circ}$ as angle of incidence on side AD.

By Snell's Law,

$\sqrt{3} \sin 30^{\circ}=\sin r=\dfrac{\sqrt{3}}{2}$

$\implies r=60^{\circ}$

Thus the angle between rays OP and RS is

$90^{\circ}$ .

Three right angled prisms of refractive indices

$n_1, n_2$ and

$n_3$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

0%
$n_1 = n_2 = n_3$
0%
$n_1 = n_2 \neq n_3$
0%
$1 + n_1 = n_2 + n_3$
0%
$1 + n_2^2 = n_1^2 + n_3^2$

Explanation

Applying Snell's Law at each surface,

AT B,

$sin i=n_1sin r_1 \implies sin^2 i=n_1^2 sin^2r_1$

At C,

$n_1sin(90-r_1)=n_2 sin r_2 \implies n_1^2cos^2 r_1=n_2^2sin^2 r_2$

At D,

$n_2 sin(90-r_2)=n_3 sinr_3 \implies n_2^2 cos^2r_2=n_3^2 sin^2r_3$

At E,

$n_3 sin(90-r_3)=sin(90-i) \implies n_3^2cos^2r_3=cos^2i$

Adding all above equations gives,

$1+n_2^2=n_1^2+n_3^2$

A parallel beam of light falls on a solid transparent sphere. Which of the following options is/are correct?

0%
If the beam is thick, then whole beam can be focussed at A.
0%
The whole beam can be focussed at A only if the beam is thin enough.
0%
If the beam is thin, then the beam can't be focussed before A.
0%
None of these

Explanation

Sphere is considered as a solid transparent with two

Spherical refractive surface of radius of curvature

$r_1=r_2=r$

The image is formed at

$2r$ distance from the pole of the 1st surface and so no refraction takes place in the second surface.

effectively refraction takes place in single surface and image is formed inside the refractive medium.

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

$n_1 = \text{refractive index of air}$

$n_2 = \text{refractive index of medium}=\mu$

$u=∞$

$v=2r$

$\dfrac{1}{\infty}+\dfrac{\mu}{2r}=\dfrac{(\mu-1)}{r}$

$\mu =2 \mu -2$

$\mu=2$

Hence b is the correct answer.

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 11 - 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 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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