MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 1

Monochromatic light is refracted from air into the glass of refractive index

$\mu$ . The ratio of the wavelength of incident and refracted waves is

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1: $\mu$
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1: $\mu$ $^{2}$
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$\mu$ :1
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1:1

The refractive index of the material of a concave lens is

$\mu.$ It is immersed in a medium of refractive index

$\mu_1$ . A parallel beam of light is incident on the lens. The path of the emergent rays when

$\mu_1>\mu$ is :

What are the colours of the Sun observed most during sunrise/sunset and noon?

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white and red
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reddish and orange
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yellow and reddish
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orange and blue

The magnifying power of a telescope with tube length

$60cm$ is

$5$ . What is the focal length of its eye piece?

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$30cm$
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$20cm$
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$10cm$
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$40cm$

Explanation

The magnifying power of a telescope is given by

$m = \dfrac{f_0}{f_e}, \, 5 = \dfrac{f_0}{f_e}$

$f_0 + f_e = 60$ [The total length of the telescope is

$60cm$ ]

And

$f_0 = 5f_e$

$\therefore f_0 + f_e = 5f_0, \, 6f_e = 60, f_e = 10cm$

$f_0 = 5f_e = 50cm$

$\therefore$ The focal length of the eyepiece is

$f_e = 10cm$

The aperture diameter of a telescope is

$5m$ . The separation between the moon and the earth is

$4\times 10^5$ km. With light of wavelength of

$5500\overset{o}{A}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to?

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$60$ m
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$20$ m
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$600$ m
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$200$ m

Explanation

Refer above image.

Let the aperture

$=d$

$D=4\times 10^8m$

$wavelength =\lambda=5500\times 10^{-10}m$

for two objects to be resolved

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

also

$d=D\theta$

$\Rightarrow \theta=\dfrac{d}{D}$

$\Rightarrow \dfrac{d}{4\times 10^8}=\dfrac{1.22\times 5500\times 10^{-10}}{5}$

$d=53.68\approx 60m$

Option A is correct.

1479410_1697279_ans_5cba9a0bf938456da7a4b7acfece96cc.png

A ray of light is incident on a prism

$\mathrm{A}BC$ of refractive index

$\sqrt{3}$ as shown in figure.
(a) Find the angle of incidence for which the deviation of light ray by the prism

$\mathrm{A}BC$ is minimum.
(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

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$30^o$ , $60^o$
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$60^o$ , $30^o$
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$60^o$ , $60^o$
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None of these

Explanation

Minimum deviation

$i_{1}=i_{2}=i$ and

$r_{1}=r_{2}=r$

and

$r_{1}+r_{2}=A$

$r=A/2=30^{\circ}$

Using

Snell's law,

$\dfrac{sini_{1}}{sinr_{1}}=\mu$

$sini_{1}=\sqrt{3}sin30^{\circ}$

$sini_{1}=\dfrac{\sqrt{3}}{2}$

$i_{1}=60^{\circ}$

When the prism is rotated anticlockwise by

$60^{\circ}$ the combined prism will become a slab as shown in the figure 2. In this case the net deviation by the prism will be zero as the deviation by the two prism will be equal and opposite.

Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is

$\mu$ , a ray incident at an angle

$\theta$ , on the face AB would get transmitted through the face AC of the prism provided

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

Explanation

Consider refraction though AB surface

Using Snell's law,

$1\sin\theta=\mu \sin r$ ...(i)

Now for ray to transmit through AC the angle of incidence should be less the

critical angle

i.e.

$r'<\sin^{-1}\dfrac{1}{\mu}$

From triangle APQ

$A+90^\circ{}-r+90^\circ{}-r'=180$

$r'=A-r;$

$A-r<\sin^{-1}\dfrac{1}{\mu}$

From equation (i)

$\theta > sin^{-1}\left[\mu \, sin \left(A - sin^{-1}\left(\dfrac{1}{\mu}\right) \right)\right]$

Option A is correct.

A ray

$OP$ monochromatic light is incident on the face

$AB$ of prism

$ABCD$ near vertex

$B$ at an incident angle of

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

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

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

Explanation

Using snell's law at surface AB, CD and AD

$\displaystyle sin^{-1}\frac{1}{\sqrt{3}} < sin^{-1} \frac{1}{\sqrt{2}}$
Net deviation is

$90^{o}$

An astronomical refracting telescope will have large angular magnification and high resolution, when it has an objective lens of :

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Large focal length and large diameter
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Small focal length and large diameter
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Small focal length and small diameter
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Large focal length and small diameter

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect

Explanation

When a ray of light enters as spherical glass sphere, it is first refracted at first interface and then it strike the inner surface of sphere (second interface) and get totally internally reflected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle

$(42^{\circ}$ , in this case).

A convex lens of refractive index

$\dfrac{3}{2}$ has a power of 2.5 D in air. If it is placed in a liquid of refractive index 2, then the new power of the lens is:

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- 1.25 D
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- 1.5 D
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1.25 D
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1.5 D

Explanation

According to Lens Maker's Formula,

Power of lens

$=P=(\mu_r-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

where

$\mu_r$ is the refractive index of material of lens relative to the surrounding medium

$=\dfrac{\mu_{material}}{\mu_{medium}}$

For convex lens in air,

$\mu_r=\dfrac{3/2}{1}=\dfrac{3}{2}$

For convex lens in the liquid,

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

Thus

$\dfrac{P_2}{P_1}=\dfrac{\mu_{r_2}-1}{\mu_{r_1}-1}$

$=\dfrac{\dfrac{3}{4}-1}{\dfrac{3}{2}-1}=-\dfrac{1}{2}$

$\implies P_2=-\dfrac{1}{2}\times 2.5D=-1.25D$

When Astronauts fly at higher altitude, the sky appears dark because ;

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Scattering of light does not take place.
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Scattering of light takes place.
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Refraction of light takes place.
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Dispersion of light takes place.

Red light is used as danger signal because it is:

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Accepted as symbol
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Pleasing to the eye
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Sensitive to the eye
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Has high wavelength

Two light rays

$P$ and

$Q$ are incident an optical device

$'X'$ which finally goes along

$'R'$ and

$'S'$ . Identify optical device

$'X'$ .

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Concave lens
0%
Concave mirror
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Convex lens
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Convex mirror

In an astronomical telescope the focal lengths of objective and eyepiece should respectively be :

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large and small
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small and large
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equal
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too small are too large

The image formed by the telescope in normal adjustment position is at :

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$D$
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$2D$
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$F$
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$\text{Infinity}$

A myopic person cannot see distinctly , objects that are :

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near
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far
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Neither
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horizontal and vertical signs

Myopia occurs due to:

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Increase in the focal length of eye lens
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Decrease in the distance between retina and lens
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Decrease in the focal length of the eye lens
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Increase in the distance between retina and lens

The optical instrument with zero power is :

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microscope
0%
telescope
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eyepiece
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all of the above

Explanation

A telescope has a larger value of its focal length. we know that Power

$P = \dfrac{1}{f}$

so because telescope has a larger focal length so it's power which is inversely related to its focal length is approximately equal to zero.

Thus the optical instrument with zero power is telescope because it is a system with large focal length.

Inability of the eye to focus on both far and near objects with advancing age is:

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Astigmatism
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Presbyopia
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Myopia
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Hypermetropia

For a myopic (short-sighted) eye, rays from far distant objects are brought to focus at a point:

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on the retina
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behind the retina
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in between eye lens and retina
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at any position

Hypermetropia is a synonym for:

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Far-sightedness
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Near-sightedness
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Bad vision due to old age
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None of these

In the case of hypermetropia:

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the image of a near object is formed behind the retina
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the image of a distant object is formed in front of the retina
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a concave lens should be used for correction
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a bifocal lens should be used for correction

A convex glass lens is immersed in water. Compared to the power in air, its power in water will :

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increase
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decrease
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not change
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decrease for red light and increase for violet light

Explanation

Power of a lens $=\dfrac{1}{f}$

Where $f$ is the focal length of the lens.

So, by lens maker formula

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

As lens is dipped in water, $\mu$ of the medium is increased from 1 to $\dfrac{4}{3}$ (i.e. refractive index of water), so, $\dfrac{\mu_{2}}{\mu_{1}}$ term decreases, so, $\left ( \dfrac{\mu_{2}}{\mu_{1}}-1 \right )$ term decrease which leads to decrease of $\dfrac{1}{f}.$ so, power $\left ( =\dfrac{1}{f} \right )$ decreases.

The focal length of a lens of refractive index

$\mu$ is

$f$ . If the lens is immersed in a liquid of refractive index

$\mu$ then the focal length of lens is

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$\dfrac{f}{2}$
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$2f$
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$\text{infinite}$
0%
$\text{zero}$

Explanation

$\dfrac{1}{f} = (n-1) (\dfrac{1}{R_1}- \dfrac{1}{R_2})$

where,

$n$ is refractive index of lens material with respect to its surrounding medium

If lens is immersed in the liquid of refractive index equal to its own then

$n=1$ and hence above equation becomes:

$\dfrac{1}{f} = 0$

$\Rightarrow f = \infty$

In which of the following instruments is final image erect?

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Simple mircoscope
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Compound mircoscope
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Astronimical telescope
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None of the above

Large aperture objective is used in telescope. Which of the following is not a function of the objective?

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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.

Cornea is a transparent spherical structure which :

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Reflects light
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Scatters light
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Refracts light
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None of these

Explanation

The cornea is the transparent front part of the eye that covers the iris, pupil and anterior chamber. The cornea, with the anterior chamber and lens, refracts light, with the cornea accounting for approximately two-thirds of the eye's total optical power. Thus, option

$C$ is correct.

First adjustment of a spectrometer is :

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focusing the distant object
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focusing the cross wires
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focusing the slit
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focusing near objects

In astronomical telescope, the final image is formed at:

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the least distance of distinct vision
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the focus of objective lens
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the focus of the eye lens
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beyond the focal length of eyepiece.

Explanation

In astronomical telescope, the image formed by objective

$I_1$ is real and inverted. It is formed at focus of objective lens.

$I_1$ is placed between the focal length of eyepiece and the lens. Hence, it forms a virtual image beyond the focal length of eyepiece.

While viewing a distant object with telescope a housefly sits on objective lens. Then which of the following is the correct statement :

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housefly will be seen enlarged in image
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housefly will be seen reduced in image
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intensity of image will be decreased
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intensity of image will be increased

Name the part of the eye which gives colour to the eyes.

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Retina
0%
Iris
0%
Cornea
0%
Fovea

Magnification of an optical instrument is expressed in:

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m
0%
$m^{-1}$
0%
D
0%
It has no unit

Explanation

$\text{Magnification} = \cfrac{\text{Image Height}}{\text{Object Height}}$

because unit of image height and object height is same hence unit of magnification is none because it is a constant number. So, it has got to unit.

When the light is very bright, then:

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Iris makes pupil expand
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Iris makes pupil contract
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Iris and pupil remain as they are
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None of these

If a ray of light goes from a rarer medium to a denser medium , will it bend towards the normal or away from it ?

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bends away from the normal
0%
bends towards the normal
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goes undeviated
0%
is reflected back

The focal length of a lens is

$-0.4\ m$ . The lens is:

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Convex only
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Concave only
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It may be convex or conacve
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None of these

A printed page is seen through a glass slab place on it. The printed words appear raised. This is due to:

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Refraction at the upper surface of the slab
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Refraction at the lower surface of the slab
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Partial reflection at the upper surface of the slab
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Partial reflection at the lower surface of the slab

The eye consists of a screen called ______ on which the light rays are focused.

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Retina
0%
Iris
0%
Pupil
0%
None of these

Explanation

The eye consists of a screen called retina on which the light rays are focussed and an image is formed. Thus, option

$A$ is correct.

The refraction of light is commonly known as :

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Bending
0%
Scattering
0%
Reflection
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Interference

Long-sightedness : Hypermetropia

Short-sightedness :

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Myopia
0%
Focussing
0%
Astigmatism
0%
Accommodation

The instrument that is based on the principle that when an object is placed between first principal focus and the optic centre of convex lens, an upright, virtual and enlarged image on the same side of the object is formed, is:

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Telescope
0%
Projector
0%
Camera
0%
Simple microscope

Blue colour of sky is due to:

0%
scattering.
0%
absorption.
0%
reflection.
0%
refraction.

Fill in the blanks:

The phenomenon of change in the _____of light when it passes from one transparent medium to another is refraction.

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speed
0%
direction
0%
both A & B
0%
None of the above

State whether true or false:

Blind spot is found in the eyes of a blind person only.

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True
0%
False

The phenomenon of light passing through the object is called :

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Reflection
0%
Refraction
0%
Dispersion
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Total internal reflection

In which of the following the final image is erect?

0%
Simple microscope
0%
Compound microscope
0%
Astronomical telescope
0%
none of these

For reading small letters with a lens

0%
one has to keep a convex lens at a distance between F and 2F from the book.
0%
one has to keep a concave lens at a distance less than the focal length from the book
0%
one has to keep concave lens at a distance between F and 2F from the book
0%
One has to keep a convex lens at a distance less than the an focal length from the book
0%
None of these

Explanation

$\bf{concave\ lens}$ always produces virtual, erect and diminished images and the decrease in the size of the image depends on the position of the object.

$\therefore$ Concave lens will shrink the size of the already small letters.

A

$\bf{convex\ lens}$ produces real and virtual, erect and inverted, diminished, same sized and magnified image of the object, depending upon the position of the object on the principal axis.

When the object is placed between F and 2F of convex lens, an enlarged but inverted image of the object is formed. The magnified image makes it easier to read small letters but the inverted image is undesirable.

When the object is placed at a distance less than the focal length of convex lens, an enlarged and erect image of the object is formed, which makes it easier to read small letters.

Hence, the correct answer is OPTION D.

In the white light of sun, maximum scattering by the air molecules present in the earth's atmosphere is for:

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red colour
0%
yellow colour
0%
green colour
0%
blue colour

Explanation

${\Large\underline{\textbf{Concept Used:}}}$

${\large{\textbf{Rayleigh Scattering}}}$

In the process of scattering, light rays are absorbed by the particles of a medium and remitted in random directions. When white light passes through a medium, different wavelengths are scattered differently.

Shorter the wavelength, the higher the chances of the light being scattered due to its high frequency. According to Rayleigh's Theory, the intensity of scattered light in a medium is related to the wavelength as,

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

${\Large\underline{\textbf{Correct Answer: D}}}$

${\Large {\textbf{Explanation}}}$

$\bullet$ By Rayleigh's Criterion intensity of scattered light is given by,

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

$\Rightarrow$ Intensity of scattered light is more for smaller wavelengths of light.

$\bullet$ In the visible spectrum, the

$blue$ side of the spectrum has a

$smaller \ wavelength$ compared to the red side. Hence blue light is

$scattered \ more$ by air molecules present in earth's atmosphere.

The phenomenon due to which a ray of light .......... from its path while travelling from one optical medium to another optical medium is called refraction.

0%
reflected
0%
deviates
0%
normally
0%
less

Which of the following would you prefer to use while reading small letters found in a dictionary ?

0%
A concave mirror
0%
A concave lens
0%
A convex mirror
0%
A convex lens

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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