MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 13

Tyndall effect is applicable when?

0%
The diameter of the dispersed particle is not much smaller than the wavelength of the light used
0%
The diameter of the dispersed particles is much smaller than the wavelength of the light used
0%
The refractive indices of the dispersed phase and the dispersion medium must be same
0%
The refractive indices of the dispersed phase and the dispersion medium must differ greatly in magnitude

Explanation

$\begin{array}{l} \text { Tyndall effect is observed only when the following two } \\ \text { conditions are satisfied. } \\ \text { i) The diameter of the dispersed particles is not much } \\ \text { smaller then the wavelength of the light used } \\ \text { ii) The refractive indices of the dispersed phase and the } \\ \text { dispersion medium differ greatly in magnitude. } \\ \text { Hence option (A) and (D) are correct } \end{array}$

Consider a hemisphere of radius R with centre of curvature at origin O, as shown. Refractive index of material of the hemisphere varies as

$\mu = \frac{2R}{2R-x}$ . Where x is x-coordinate of material point. A ray travelling in air in xy-plane is grazingly incident at O, as shown in the figure. Now answer the following question.
Deviation suffered by the ray just before it comes out of the hemispherical surface

0%
lies between $0^o and 30^o$
0%
lies between $30^o and 45^o$
0%
is greater than $60^o$
0%
is equal to $60^o$

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

$60^0$ . The percentage of incident light transmitted is

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

The refractive indices of glass and water are 3/ 2 and 4/ 3 respectively. The refractive index of glass with respect to water is

0%
2
0%
8/9
0%
9/8
0%
5/3

Explanation

Given

$n_{glass}=3/2$

$n_{water}=4/3$

Now refractive index of glass with respect to water is given by

$_{water}n_{glass}$ =

$\dfrac{n_{glass}}{n_{water}}$ =

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

$\dfrac{3\times 3}{4 \times 2}$ =

$\dfrac{9}{8}$

A ray incident at an angle

$53^{\circ}C$ on prism emerges at an angle

$37^{\circ}C$ as shown in the figure. If the angle of incidence is made

$50^{\circ}C$ , which of the following is a possible value of angle of emergence

0%
$35^{\circ}C$
0%
$42^{\circ}C$
0%
$40^{\circ}C$
0%
none

A ray incident at an angle

$53^{\circ}$ on prism emerges at an angle

$37^{\circ}$ as shown in the figure. If the angle of incidence is made

$50^{\circ}$ , which of the following is a possible value of angle of emergence

0%
$35^{\circ}$
0%
$42^{\circ}$
0%
$40^{\circ}$
0%
$38^{\circ}$

A convex lens of focal length

$f$ produces a real image 3 times the size of an object, then distance of the object from the lens is

0%
$-\dfrac{2f} { 3}$
0%
$-\dfrac{\mathrm { f }} { 2}$
0%
$-\dfrac{\mathrm { f } }{ 4}$
0%
$-\dfrac{\mathrm {4 f }} { 3}$

The angular resolution of a

$10cm$ diameter telescope at a wavelength of

$5000A$ is of the order of

0%
$10^{6} \mathrm{rad}$
0%
$10^{-2} \mathrm{rad}$
0%
$10^{-4} \mathrm{rad}$
0%
$10^{-6} \mathrm{rad}$

Explanation

Given

$\lambda = 5000{A^ \circ } = 5 \times {10^{ - 7}}m$

$d = 10cm = 0.1m$

Angular resolution of a telescope,

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

$\Rightarrow \theta = \dfrac{{1.22 \times 5 \times {{10}^{ - 7}}}}{{0.1}}$

$\Rightarrow \theta = 6.1 \times {10^{ - 6}}rad$

Hence the angular resolution is of the order

${10^{ - 6}}rad$ .

The correct answer is option (D).

What kind of lens is used in the spectacles of a person suffering from Hypermetropia?

0%
Convex.
0%
Concave.
0%
Plano -convex
0%
A combination of cocave and convex.

A ray of light is incident on a glass at an angle of

$60^o$ . What would be the refractive index of glass, if reflected and refracted rays are perpendicular to each other ?

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

Explanation

The incident ray $1$ will be reflected at an equal angle as reflected ray $2.$

By laws of reflection, $i=r=60^o$

The refracted ray $3$ is perpendicular to reflected ray $2.$

So by geometry, the angle of refraction $r'=30^o$

Now by Snell's law,

$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$

$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$

$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$

$\mu_{glass}=\sqrt3$

Option A is the answer.

1996023_1441036_ans_e852a520e2664e748aa585b02a1735a0.JPG

A right-angled prism is set to deviate a ray of light through

$90^o$ . If the same prism is to be used for deviating through

$180^o$ , the prism must be turned from the previous set position through the angle :

0%
$45^o$
0%
$90^o$
0%
$135^o$
0%
$180^o$

Explanation

From the figure we can see that the right angled prism needs to be rotated by ${45^\circ }$ .

Hence the correct answer is option (A).

1996607_1442334_ans_9ecc77d5cdd8465290c671375d53c249.JPG

The focal lengths of the objective and eye plece of astronomical telescope are

$100cm$ and

$2cm$ respectively. The magnitude of magnifying power of telescope for adjustment is

0%
$54$
0%
$0.02$
0%
$50$
0%
$100$

Explanation

The correct option is (C).

Given,

The focal length of the objective lens ${f_o} = 100\;cm$

The focal lens of the eyepiece ${f_e} = 2\;cm$

The magnifying power of a telescope can be given as the ratio of the focal length of the objective lens ${f_o}$ to the focal length of the eyepiece ${f_e}$ fo the telescope, hence

$M = \frac{{{f_o}}}{{{f_e}}}$

$M = \frac{{100}}{2}$

$M = 50$ .

Two thin prisms are combined to form an achromatic combination.For prism I,

$A=4^{0}, \mu_{R}=1.35, \mu_{Y}=1.40, \mu_{\nu}=1.42 .$ For prism II

$\mu_{R}=1.7, \mu_{\gamma}=1.8$ and

$\mu_{R}=1.9 .$ Find the prism angle of prism II and the net mean deviation.

0%
$0.44^{\circ}$
0%
$0.45^{\circ}$
0%
$0.48^{\circ}$
0%
$0.46^{\circ}$

Explanation

The correct option is (C).

Given,

The reflecting angle of prism I is

$A = 4^\circ$

Now for aromatic combination

$\left( {{\mu _{v1}} - {\mu _{R1}}} \right){A_1} = \left( {{\mu _{v2}} - {\mu _{R2}}} \right){A_2}$

$\left( {1.42 - 1.35} \right) \times 4^\circ = \left( {1.9 - 1.7} \right){A_2}$

${A_2} = \dfrac{{\left( {1.42 - 1.35} \right) \times 4^\circ }}{{\left( {1.9 - 1.7} \right)}}$

${A_2} = 1.4^\circ$

Now the net minimum deviation is given as

${\delta _1} - {\delta _2} = \left( {{\mu _{Y1}} - 1} \right){A_1} - \left( {{\mu _{Y2}} - 1} \right){A_2}$

${\delta _1} - {\delta _2} = \left( {1.40 - 1} \right) \times 4^\circ - \left( {1.8 - 1} \right) \times 1.4^\circ$

${\delta _1} - {\delta _2} = 0.48^\circ$ .

A student measures the focal length of a convex lens by putting an object pin at a distance 'u' from the lens and measured the distance 'v' of the image pin. The graph between 'u' and 'v' plotted by the student should look like

Distances measured perpendicular to and below the optical center are taken as ________.

0%
Zero
0%
Negative
0%
Positive
0%
None of the above

Distances measured perpendicular to and above the optical center are taken as ________.

0%
Negative
0%
Positive
0%
Zero
0%
None of the above

The angle of minimum deviation produced by an equilateral prism is

$46^0$ . The refractive index of material of the prism.

0%
1.6
0%
1.5
0%
1.4
0%
1.8

Explanation

Given, minimum deviation (

$\delta_m$ ) for an equilateral prism is

$46^\circ$ .

We know, for a equilateral prism, the index angle

$A$ is always

$60^\circ$ .

Let,

$i$ ,

$r_1$ ,

$r_2$ and

$e$ be the incident angle at AB face of prism, refracted angle at AB face, incident angle at AC face and refracted angle at AC face respectively.

Now, as there is a minimum deviation,

So, we can say,

$i=e$ .

As we know,

$\delta_m=i+e+A$ ,

Then,

$\delta_m=2i-A$

$\therefore i=\dfrac{\delta_m+A}{2}=\dfrac{46^\circ+60^\circ}{2}=53^\circ$

Now, from Snell's law,

$sini=\mu sinr_1$ ......... (i)

And,

$sin e=\mu sinr_2$ ............ (ii)

Where,

$\mu$ is the refractive index of the prism.

From (i) and (ii) we get

$r_1=r_2$ because

$i=e$ .

We know,

$r_1+r_2=A$

So,

$r_1=r_2=\dfrac{A}{2}=30^\circ$

$\therefore \mu=\dfrac{sini}{sinr_1}=\dfrac{sin53^\circ}{sin30^\circ}=1.6$

Hence, the correct option is (A).

The objective of a telescope has a focal length of

$1.2\ m$ . It is used to view a

$10.0\ m$ tall tower

$2\ km$ away. What is the height of the image of the tower formed by the objective.

0%
$2\ mm$
0%
$4\ mm$
0%
$6\ mm$
0%
$8\ mm$

A thin Prism

$P_1$ with angle

$4^o$ and made from glass of refractive index 1.54 is combined with another thin prism

$P_2$ made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of Prism

$P_2$ is

0%
$5.33^0$
0%
$4^o$
0%
$3^o$
0%
$2.6^o$

For path

$A \rightarrow B$ optical path is

0%
$2(a + b)$
0%
$2 (a - b)$
0%
$a + b$
0%
$a - b$

Explanation

$AC = \dfrac{a}{\cos \, 60^o} ; CB = \dfrac{b}{\cos 30^o}$

$\sin 60^o = \mu \sin 30^o$

$\mu = \sqrt{3}$
optical path

$= AC + \mu CB$

$= \dfrac{a}{cos 60^o} + \mu \dfrac{\beta}{cos \, 30^o} = 2a + \sqrt{3} \dfrac{6}{\dfrac{\sqrt{3}}{2}} = 2 (a + b)$
1245099_1612354_ans_b7ed6ad978984b5db1f5fd0e9b69e652.png

To focus the image of a nearby object on the retina of an eye

0%
The distance between eye-lens and retina is increased
0%
The distance of the eye-lens and retina is decreased
0%
The thickness of eye-lens is decreased
0%
The thickness of eye-lens is. increased

Range the values of tip angle

$\theta$ for the advice to work as anamorphic de-magnifier is

0%
$\dfrac {1}{3}\cos^{-1}\left(\dfrac {1}{\mu}\right) < \theta < \dfrac {1}{2}\sin^{-1} \left(\dfrac {1}{\mu}\right)$
0%
$\dfrac {1}{3}\sin^{-1}\left(\dfrac {1}{\mu}\right) < \theta < \dfrac {1}{2}\cos^{-1} \left(\dfrac {1}{\mu}\right)$
0%
$\dfrac {1}{3}\sin^{-1}\left(\dfrac {1}{\mu}\right) < \theta < \dfrac {1}{2}\sin^{-1} \left(\dfrac {1}{\mu}\right)$
0%
$\dfrac {1}{3}\cos^{-1}\left(\dfrac {1}{\mu}\right) < \theta < \dfrac {1}{2}\cos^{-1} \left(\dfrac {1}{\mu}\right)$

A light ray is incident ton face AB of prism ABC as shown in figure. The second prism is kept in such a manner that emergent ray from prism ABC is falling normally on face A'B' of prism A'B'C'. The net deviation produced by the optical' system of the two prisms is :

0%
$45^{\circ}$
0%
$112^{\circ}$
0%
$32^{\circ}$
0%
$18^{\circ}$

The ratio of thickness of plates of two transparent medium

$A$ and

$B$ is

$6:4$ . If the takes equal time in passing through them, then refractive index of

$B$ with respect to

$A$ will be

0%
$1.4$
0%
$1.5$
0%
$1.75$
0%
$1.33$

Explanation

Time taken by light to travel distance

$x$ through a medium of refractive index

$\mu$ is

$t=\dfrac {\mu x}{c}\Rightarrow \dfrac {\mu_B}{\mu_A}=\dfrac {x_A}{x_B}=\dfrac {6}{4}\Rightarrow _A\mu_B =\dfrac {3}{2}=1.5$

A medium shows the relation between the angle of incidence

$i$ and the angle of refraction

$r$ . If speed of light in the medium is

$nc$ then value of

$n$ is

0%
$1.5$
0%
$2$
0%
$2^{-1}$
0%
$3^{-1/2}$

Explanation

Let

$\mu$ be the refractive index of the medium.

From Snell's law:

$\mu=\dfrac{sin\ i}{sin\ r}$

From graph it is clear that

$\tan 30^{o}=\dfrac{\sin r}{\sin i}$

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{\sin r}{\sin i}=\dfrac{1}{\mu}\Rightarrow \mu=\sqrt{3}$
Also, speed of light in the medium

$v=\dfrac{c}{\mu}=nc\Rightarrow n=\dfrac{1}{\mu}=\dfrac{1}{\sqrt{3}}=(3)^{-1/2}$

Which of the following ray diagram show physically possible refraction

0%
(i)
0%
(ii)
0%
(iii)
0%
None of these

Explanation

When light ray goes from denser to rarer medium (i.e. more

$\mu$ to less

$\mu$ ) it deviates away from the normal

On the other hand, if light ray goes from rarer to denser medium (i.e. less

$\mu$ to more

$\mu$ ) it bends towards the normal.

It is followed only in option A.

In option B, the refracted ray is on the same side of the normal as the incident ray, which never happens in case of any refraction. So it is incorrect.

Focal length of convex lens is _________

0%
Negative
0%
Zero
0%
Unity
0%
Positive

Ray EF is ____________.

0%
Incident ray
0%
Emergent ray
0%
Refracted ray
0%
None of the above

Focal length of concave lens is ________.

0%
positive
0%
negative
0%
zero
0%
none of the above

A lens has focal length of -30 cm . Identify the lens.

0%
Convex
0%
Cylindrical
0%
Concave
0%
None of the above

In lenses, all the distances measured to the ________ of the origin are taken as positive.

0%
Left
0%
Right
0%
Below
0%
None of the above

What do ray PE denote?

0%
Angle of incidence
0%
Incident ray
0%
Refracted ray
0%
Reflected ray

Explanation

Ray

$PE$ is the incident ray which is getting refracted at the face

$AB$ of the prism.

Ray

$EF$ is the refracted ray which is getting refracted again at the face

$AC$ of the prism.

Ray

$FS$ is the emergent ray coming out the face

$AC$ of the prism.

So option B is the answer.

$\angle$

$N'EF$ is __________.

0%
angle of incidence
0%
angle of emergence
0%
angle of refraction
0%
None of the above

Explanation

$\angle$

$N'EF$ is angle of refraction.

A lens has focal length of

$+20 cm$ . Identify the lens.

0%
Convex
0%
Concave
0%
Cylindrical
0%
None of the above

In lenses, the distances measured to the left of the origin are taken as _________.

0%
Zero
0%
One
0%
Positive
0%
Negative

In case of lenses, object distance is taken to be_______.

0%
Positive
0%
Zero
0%
Negative
0%
One

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 13 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.