MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 3

A parallel beam of light falling on the eye gets focused on the retina because of refraction at

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The cornea
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The crystalline lens
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The vitreous humor
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Various surfaces in the eye

The human eye part which can be used again and again for forming different images is called

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Iris
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Pupil
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Cornea
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Retina

State whether true or false :

The number of cones in the human eye are more than the number of rods.

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True
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False

Large aperture objective is used in telescopes as it helps in

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increasing the brightness of image
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reducing image size
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increasing field of view
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increasing intensity by gathering more light

Which of the following statements is correct?

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A person with myopia can see nearby objects clearly
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A person with hypermetropia can see nearby objects clearly
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A person with myopia can see distant objects clearly
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A person with hypermetropia cannot see distant objects clearly

A ray of light travels from a denser to a rarer medium then, the ray :

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doesn't bend at all
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bends towards the normal
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bends away from the normal
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goes along the normal

Large astronomical telescopes always use as objective

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lens
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mirror
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combinations of lenses
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none of the above

The radius curvature for a convex lens is

$40\ cm$ , for each surface. Its refractive index is

$1.5$ . The local length will be

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$40\ cm$
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$20\ cm$
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$80\ cm$
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$30\ cm$

Explanation

By formula

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

$=(1.5-1) \left(\dfrac{1}{40}+ \dfrac{1}{40} \right) =0.5 \times \dfrac{1}{20} =\dfrac{1}{40}$

$\therefore f=40\ cm$

Microscope is an optical instrument which

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Enlarges the object
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Increases the visual angle formed by the object at the eye
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Decreases the visual angle formed by the object at the eye
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Brings the object nearer

The relation between

$n_{1}$ and

$n_{2}$ , if the behaviour of light ray is as shown in the figure.

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$n_{2}>n_{1}$
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$n_{1}>>n_{2}$
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$n_{1}>n_{2}$
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$n_{1}=n_{2}$

Explanation

Nature of lens changes, if

$\mu_{medium} > \mu_{lens}$

hence option A is correct.

An ellipsometer is an instrument for.

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Measuring stellar distance
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Measuring the path of celestial bodies
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Ending curvature of elliptical surfaces
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Studying thin films on a solid surface

The sensation of vision in the retina is carries to the brain by

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Ciliary muscle
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Blind spot
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Cylindrical lens
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Optic nerve

Finger prints are observed by the use of

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Gallilean telescope
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Telescope
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Microscope
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Concave lens

In order to increase the angular magnification of a simple microscope, one should increase

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The object size
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The aperture of the lens
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The focal length of the lens
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The power of the lens

A person using a lens as a simple microscope sees an

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Inverted virtual image
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Inverted real magnified image
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Upright virtual image
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Upright real magnified image

In the case of refraction of light :
a) Frequency changes
b) Speed changes
c) Wavelength changes

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a is correct
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b and c are correct
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a, b, c are correct
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a and b are correct

Explanation

In the process of refraction, speed of light and wavelength of light changes.

In a medium, speed of light is given by

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

$\mu$ is refractive index of the medium. Hence, for medium with different refractive index , speed of light is different.

Frequency will not change because at the boundary/interface of the medium, the number of waves you send is the number of waves you receive at the other side, almost instantly.

Since, for a travelling wave we have a relation

$\nu = \dfrac{v}{\lambda}$ .

Hence wavelength also changes in the process of refraction.

So, option B is correct.

In the diagram shown

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$\mu ^{1}=\mu$
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$\mu ^{1}<\mu$
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$\mu ^{1}>\mu$
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$\mu ^{1}\geq \mu$

Explanation

Since, the given lens is a convex lens hence it should converge a parallel beam of light falling on it to a single focal point. However, since it is showing the exactly opposite behaviour, thus it is definitely placed in a medium of higher refractive index.
Thus, we conclude that

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

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If

$\mu$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

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$sin^{-1}\left ( \dfrac{1}{\mu } \right )$
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$tan^{-1}\left ( \dfrac{1}{\mu } \right )$
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$sin^{-1}\left ( \dfrac{\mu -1}{\mu } \right )$
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$cos^{-1}\left ( \dfrac{1}{\mu } \right )$

Explanation

$\textbf {Hint :}$ Use critical angle

$C=\sin^{-1}\left(\dfrac{1}{\mu}\right)$

$\textbf{Step 1: }$ Calculating the maximum value of the base angle

$\alpha$

$\alpha$ is maximum value of base angle for which light is totally reflected from the hypotenuse and

$\theta$ is the minimum value of angle of incidence at hypotenuse for total internal reflection.

Then ,

$\theta =sin^{-1}\left ( \dfrac{1}{\mu}\right )$

From diagram

$\alpha +\theta =90^o$

$\Rightarrow \alpha = 90^o-sin^{-1}\left ( 1/\mu \right )=cos^{-1}\left ( 1/\mu \right )$

${\textbf{Correct option: D}}$

A double convex lens, made of a material of refractive index

$\mu _{1}$ , is placed inside two liquids of refractive indices

$\mu _{2}$ and

$\mu _{3}$ , where

$\mu _{2}>\mu _{1}>\mu _{3}$ . A wide , parallel beam of light is incident on the lens from the left .The lens will give rise to

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a single convergent beam
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two different convergent beams
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two different divergent beams
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a convergent and a divergent beam

Explanation

$\therefore$ analyzing the rays it gives a divergent ray above the axis and
convergent ray below

$\mu_2 > \mu_1 > \mu_3$

from medium 2

$\Rightarrow$ when light ray enters into medium

$1 (\mu_1)$ then moves away from normal

from medium 3

$\Rightarrow$ when light ray enters into medium

$1 (\mu_1)$ the moves towards the normal.

A ray of light falls on the surface of a spherical glass paperweight making an angle

$a$ with the normal and is refracted in the medium at angle

$b$ . The angle of deviation of the emergent ray from the direction of incident ray is :

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$a-b$
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$2(a-b)$
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$\dfrac{(a-b)}{2}$
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$b-a$

Explanation

From the above figure we can see that

$\theta =2(a-b)$

Exterior angle=sum of interior opposite angles. So answer is option B.

Match the elements of table I and II.

Table I	Table II
a) Myopia	i) Bifocal len
b) Hypermetropia	ii) Cylindrical lens
c) Presbyopia	iii) Concave lens
d) Astigmation	iv) Convex lens

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a - iii, b - iv, c- i, d -ii
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a - iv, b - iii, c -i, d - ii
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a - i, b - ii, c - iii, d - iv
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a - ii, b - iv, c -i, d - iii

Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/The combination works as :

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converging lens of focal length 50 cm.
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diverging lens of focal length 15 cm
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converging lens of focal length 15 cm
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diverging lens of focal length 50 cm

Explanation

$\begin{array}{l}\text { Ths is a combination of convex, concave and convex } \\\text { lens. Number them } 1,2,3 \\\qquad \begin{aligned}\text { given } f_{1} &=f_{3}=20 \mathrm{~cm} \\\mu_{1} &=\mu_{3}=1.5 \\\mu_{2} &=4/3\end{aligned}\end{array}$

$\begin{array}{l}\quad \mu_{2}=4 / 3 \\\text { Now densmaker formula } \frac{1}{f}=\left(\mu_{r}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\text { for (1) } \quad \frac{1}{f_{1}}=(1.5-1)\left(\frac{1}{R_{1}}-\left(-\frac{1}{R}\right)\right)\end{array}$

$\text { (equi convex lenS }\left.R_{1}=R_{2}\right)$

$\begin{array}{l}\frac{1}{20}=\frac{1}{2}\times \frac{2}{R} \\ R=20 \mathrm{~cm} \\\text { for (2) } \\\qquad \begin{array}{l}\frac{1}{f_{2}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\frac{1}{f_{2}}=\frac{1}{3}\left(\frac{-2}{R}\right) \\ f_{2}=-30\mathrm{~cm}\end{array}\end{array}$

$\begin{array}{l}\text { Now we know that } \\\qquad \begin{aligned}\text { ieffective } &=P_{1}+P_{2}+P_{3} \\\frac{1}{f_{e q}} &=\frac{1}{20}-\frac{1}{30}+\frac{1}{20} \\ f_{\text {eq }}=15\mathrm{~cm}\end{aligned} \\\text { So it is a converging lens of focal length } 15 \mathrm{~cm} .\end{array}$

The refractive index of the material of a concave lens is

$\mu$ and is placed in the medium of refractive index

$\mu ^{1}$ . If

$\mu ^{1}>\mu$ , then which of the following ray diagram is correct:

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None of these

A ray incident at a point at an angle of incidence

$60$

$^{\circ}$ enters a glass sphere of

$\mu =\sqrt{3}$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

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$50$ $^{\circ}$
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$90$ $^{\circ}$
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$60$ $^{\circ}$
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$40$ $^{\circ}$

Explanation

Using Snell's Law at point

$P$ we get:-

$\dfrac{sin\ 60^o}{sin\ r_1}=\sqrt3$

$sin \ r_1= \dfrac12\implies r_1= 30^o$

As we know that

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

Using Snell's Law at point

$Q$ we get:-

$\dfrac{sin \ 30^o}{sin\ i_2}=\dfrac1{\sqrt3}\implies i_2= 60^o$

Since reflection at point

$Q$ occurs

$\implies r_2'=r_2= 60^o$

Let angle between the refracted ray and reflected ray be

$\alpha$

$\alpha= 180^o- (r_2'+r_2)= 90^o$

Hence option

$(B)$ is correct.

A defective eye cannot see close objects clearly because their image is formed:

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On the eye lens
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Between eye lens and retina
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On the retina
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Beyond retina

ACB is a right-angled glass prism with refractive index 1.Angle A, B, C are 60

$^{\circ}$ ,30

$^{\circ}$ and 90

$^{\circ}$ respectively. A thin layer of liquid is on the AB. For a ray of light which is incident normally on AC to ;be totally reflected at AB, the refractive index of the liquid on AB should be

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1.5
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1.4
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1.3
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1.2

Explanation

Given,

$\mu$ =refractive index of liquid.

From figure it is itself clear that , angle of incidence,

$i=60^{o}$

Now, for internal reflection , angle of refraction

$r=90^{o}$

Now,

$\dfrac{\sin i}{\sin r}=\dfrac{\mu}{\mu_1}$

$\implies \sin 60^{o}=\dfrac{\mu}{1.5}$

$\implies \mu=\dfrac{\sqrt{3}}{2}\times 1.5$

$\implies \mu=1.3$

Hence, answer is option-(C)

A light ray is incident perpendicularly to one face of a 90

$^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45

$^{\circ}$ , we conclude that the refractive index

$n$ :

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$n>\dfrac{1}{\sqrt{2}}$
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$n>\sqrt{2}$
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$n<\dfrac{1}{\sqrt{2}}$
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$n<\sqrt{2}$

Explanation

We know

$\theta _{cric}=sin^{-1}(\dfrac{1}{n })$

$45^{o}> \theta$

$sin(45^o)> \dfrac{1}{n}$

$\Rightarrow n > \sqrt{2}$

In the figure, light is incident on the thin lens as shown. The radius of curvature for both the surface is R. The focal length of this system is

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$f=\dfrac{\mu_3R}{\mu_3-\mu_1}$
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$f=\dfrac{\mu_2R}{\mu_1-\mu_2}$
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$f=\dfrac{\mu_3}{\mu_2-\mu_1}$
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$f=\dfrac{\mu_1R}{\mu_1-\mu_2}$

Explanation

Consider refraction from the two spherical surfaces.

At surface 1,

$\dfrac{\mu_2}{v_1}-\dfrac{\mu_1}{\infty}=\dfrac{\mu_2-\mu_1}{R}$

At surface 2,

$\dfrac{\mu_3}{v_2}-\dfrac{\mu_2}{v_1}=\dfrac{\mu_3-\mu_2}{R}$

$\implies v_2=\dfrac{\mu_3R}{\mu_3-\mu_1}=f$

A thin prism of angle

$15^{o}$ made of glass of refractive index

$\mu _{1} = 1.5$ is combined with another prism of glass of refractive index

$\mu _{2} = 1.75$ . The combination of the prism produced dispersion without deviation. The angle of the second prism should be :

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$5^{o}$
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$7^{o}$
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$10^{o}$
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$12^{o}$

Explanation

For without deviation

$\frac{A}{A'}=-\frac{\mu ' -1}{\mu -1}$

$\frac{15^{o}}{A'}=-\frac{1.75-1}{1.50-1}$

$\frac{15^{o}}{A'}=-\frac{0.75}{0.50}$

$A'=-10^{o}$

If the behavior of light rays through a convex looking lens is shown in the figure below , then

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$\mu > \mu _{2}$
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$\mu \leq \mu _{2}$
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$\mu = \mu _{2}$
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$\mu < \mu _{2}$

Explanation

A medium inverts its behaviour if placed in a medium of refractive index higher than the material of the lens.
Thus, a converging lens diverges light if placed in a medium of higher refractive index.
Thus,

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

The angle at which the ray is incident on the second mirror is___?

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$10^{\circ}$
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$20^{\circ}$
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$30^{\circ}$
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$40^{\circ}$

Explanation

$\theta =40^{\circ}$

$\phi = 180 - 60 40$

$=80^{\circ}$

$\therefore i =90^{\circ}-80^{\circ}$

$=10$

The angular resolution of a 10cm diameter telescope at a wavelength of

$5000 A^0$ is of the order of -

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$10^{4} rad$
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$10^{-6} rad$
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$10^{6} rad$
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$10^{2} rad$

Explanation

Given parameters are,

Wavelength $\lambda = 5000\ A^{\circ}$

= $5000 \times 10^{-10}$

Diameter of telescope, $D = 10 cm = 0.1 m$

Now, Angular resolution (d\theta) formula for telescope is,

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

Substituting the values, we get

$\Rightarrow d\theta = \dfrac{1.22 \times 5000 \times 10^{-10}}{0.1}= 6.1 \times 10^{-6}\\$

Clearly it is having significance order of $10^{-6}$ .

Thus option B is correct.

A student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his textbook. Which of the following statements is correct?

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The near point of his eyes has receded away.
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The near point of his eyes has come closer to him.
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The far point of his eyes has come closer to him.
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The far point of his eyes has receded away.

A person needs a lens of power -4.5 D for correction of her vision. What kind of defect in vision is she suffering from?

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Myopia
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Ptosis
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Presbyopia
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Diplopia.

At noon the sun appears white because:

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light is least scattered.
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all the colours of the white light are scattered away.
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blue colour is scattered the most.
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red colour is scattered the most.

Which of the following phenomena contributes significantly to the reddish appearance of the sun at sunrise or sunset?

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Dispersion of light
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Scattering of light
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Total internal reflection of light
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Reflection of light from the earth

The bluish colour of water in deep sea is due to:

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the presence of algae and other plants found in water.
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reflection of sky in water.
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scattering of light.
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absorption of light by the sea.

The clear sky appears blue because:

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blue light gets absorbed in the atmosphere.
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ultraviolet radiations are absorbed in the atmosphere.
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violet and blue lights get scattered more than lights of all other colours by the atmosphere.
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light of all other colours is scattered more than the violet and blue colour lights by the atmosphere.

The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light:

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is scattered the most by smoke or fog.
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is scattered the least by smoke or fog.
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is absorbed the most by smoke or fog.
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moves fastest in air.

Which of the following statement is correct?

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A person with myopia can see distant objects clearly
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A person with hypermetropia can see nearby objects clearly
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A person with myopia can see nearby objects clearly
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A person with hypermetropia cannot see distant objects clearly

A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her?

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Doctor advises her to use a convex lens of appropriate power.
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Doctor advises her to use a concave lens of appropriate power.
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Doctor advises her to use a plane mirror of appropriate power.
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Doctor advises her to use a plano convex lens of appropriate power.

Which of the following lens is used to correct hypermetropia?

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Convex lens
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Concave lens
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Cylindrical lens
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None of these

What kind of defect in vision is she suffering from?

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Myopia
0%
Hypermetropia
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Astigmatism
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None of these

Explanation

Power of

$P=4.5D \Rightarrow f=\cfrac{1}{4.5}$

$f=0.22m =22cm$

Positive focal length means that the lens is a convex lens.

Hence person should have trouble in seeing things close to his eyes.

He is suffering from long-sighted-ness that is

$hypermetropia$ .

Answer-(B)

Why is the colour of the clear sky blue?

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Blue colour gets scattered the maximum.
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water of sea is blue.
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Blue colour gets scattered the minimum.
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none of these.

Which of the following statements is correct regarding rods and cones in the human eye?

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Cones are sensitive to dim light.
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Cones are sensitive to bright light.
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Rods are sensitive to bright light and dim light.
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Rods can sense colour.

How can we explain the reddish appearance of sun at sunrise or sunset?

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Scattering of blue light is more than the scattering of red light
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Scattering of red light is more than the scattering of blue light
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Intensity of sun reduces during sunrise and sunset
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Due to the view angle, it appears blue

Which is the path of rays of light when it enters with oblique incidence from water into air ?

When do we consider a person to be myopic?

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He has poor near vision or blurred vision at all distances
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He can not identify colors
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He has poor night vision
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He can see fine upto close distance, but distant objects are a blur

When the length of a microscope tube increases

its magnifying power:

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decreases
0%
increases
0%
does not change
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may decrease or increase depending on observer and place of observation

What is the magnifying power of a telescope whose objective and eye - piece have focal length 180 cm and 3 cm respectively?

0%
50
0%
40
0%
30
0%
60

Explanation

Here

$F=180 \ cm$ and

$f=30 \ cm$

$\therefore$ Magnifying power M

$=\dfrac {F}{f}=\dfrac {180}{3}=60$

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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