MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 10

The focal lengths of the objective and eye lens of telescope are respectively

$200\ cm$ and

$5\ cm$ . The maximum magnifying power of the telescope will be.

0%
$-40$
0%
$-48$
0%
$-60$
0%
$-100$

Explanation

Magnifying power of astronomical telescope

$m=-\dfrac{f_o}{f_e}\left( 1+\dfrac{f_e}{D}\right)=-\dfrac{200}{5}\left( 1+\dfrac{5}{25}\right)=-48$ .

Return of light after striking a surface is called the refraction of light.

0%
True
0%
False

A convex lens of focal length f produces a real image 3 times as that size of the object, the distance between the object and the lens is .

0%
$-\left ( \cfrac {2f}3 \right)$
0%
$-\left ( \cfrac {3f}4 \right)$
0%
$-\left ( \cfrac {4f}3 \right)$
0%
$-\left ( \cfrac {3f}2 \right)$

The eye can be regarded as a single refracting surface. The radius of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

0%
2 cm
0%
1 cm
0%
3.1 cm
0%
4.0 cm

Explanation

$\dfrac{1.34}{V} - \dfrac{1}{\infty} = \dfrac{1.34 - 1}{7.8}$

$\dfrac{1.34}{v}=\dfrac{34}{780}$

$v=\dfrac{1.34\times 780}{34}$

$\therefore V = 30.7 mm$

1144220_1333752_ans_918415422d2f4ae5b584eddeac99bfbe.PNG

An observer looks at a distant tree
of height $10$ m with a
telescope of magnifying power of $20$ . To the
observer the tree appears :

0%
$10$ times taller.
0%
$10$ times nearer
0%
$20$ times taller.
0%
$20$ times nearer.

Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence?

0%
$n > \surd 2$
0%
$n = 1$
0%
$n = 1.1$
0%
$n = 1.3$

Explanation

From the following figure

$r+i=90^o \Rightarrow i=90^o -r$
For ray not to emerge from curved surface

$i > C$

$\Rightarrow \sin i > \sin C \Rightarrow \sin (90^o -r)> \sin C \Rightarrow \cos r > \sin C$

$\Rightarrow \sqrt {1-\sin^2 r} > \dfrac 1n\quad \left\{\because \sin C=\dfrac 1n \right\}$

$\Rightarrow 1-\dfrac {\sin^2 \alpha}{n^2} > \dfrac {1}{n^2} \Rightarrow 1 > \dfrac {1}{n^2} (1+\sin^2 \alpha)$

$\Rightarrow n^2 > 1 \sin^2 \alpha \Rightarrow n > \sqrt 2\quad \left\{\sin i \to 1\right\}$

$\Rightarrow$ Least value

$=\sqrt 2$
1745239_1329808_ans_8c19e74954c74a60a935cc5d62c74d76.png

Three glass prisms A, B and C of same refractive indices are placed in contact with each other as shown in fig. 13.77 with no air gap between the prisms. A monochromatic ray of light OP passes through the prism assemble and emerges as QR. The condition of minimum deviation are satisfied in the prisms:

0%
A and C
0%
B and C
0%
A and B
0%
in all prisms A,B and C

Explanation

In both

$A$ and

$B$ , the refracted ray is parallel to the base of prism.

Hypermetropia is possible in.........

0%
children
0%
aged person's
0%
high intensity fall on eye
0%
low intensity fall on eye

Three glass prisms A, B and C of same refractive are placed in contact with each other as shown on Fig. 13.77 with no air gap between the prisms. A monochromatic ray of light OP passes through the prism assembly and emerges as QR. The conditions of minimum deviation are satisfied in the prisms:

0%
A and C
0%
B and C
0%
A and B
0%
in all prisms A , B and C

Explanation

The condition for minimum deviation in a prism is given as

1. $i = e$ where $i$ is the angle of incidence and $e$ is the angle of emergence.

2. ${r_1} = {r_2}$ where ${r_1}$ and ${r_2}$ are the angle of refraction of the two refracting surfaces of the prism.

Hence the first refracted ray inside the prism should be parallel to the base of the prism.

In the figure we can see that the above condition is satisfied by the prisms $A$ and $B$ .

Hence the correct answer is option (C).

The focal length of objective and eye lens of a microscope are

$4\ cm$ and

$8\ cm$ respectively. If the least distance of distinct vision is

$24\ cm$ and object distance is

$4.5\ cm$ from the objective lens, then the magnifying power of the microscope will be

0%
$18$
0%
$32$
0%
$64$
0%
$20$

Explanation

For objective lens

$\dfrac{1}{f_0}=\dfrac{1}{v_0}-\dfrac{1}{u_0}$

$\Rightarrow \dfrac{1}{(+4)}=\dfrac {1}{v_0}-\dfrac{1}{(-4.5)}\Rightarrow v_0=36\ cm$

$\therefore | m_D| =\dfrac {v_0}{u_0}\left( 1+\dfrac{D}{f_e}\right) =\dfrac{36}{4.5}\left( 1+\dfrac{24}{8}\right) =32$

The principle section of a glass prism is isosceles triangle ABC with AB=AC, the face AC is silvered. A ray of light is incident normally on the face AB and after two reflections, it emerges from the base BC perpendicular to the base. Angle of BAC of the prism is

0%
30 $^{ \circ }$
0%
36 $^{ \circ }$
0%
18 $^{ \circ }$
0%
72 $^{ \circ }$

When a glass rod is immersed in a liquid of same refractive index, it will _____.

0%
Appear bent
0%
Disappear
0%
Appear longer
0%
Not be affected

The image of an object is formed due to the _________ of light rays by the eye lens on the retina.

0%
reflection
0%
dispersion
0%
scattering
0%
refraction

A convex lens forms a real image of a point object placed on its principle axis.If the upper half of the lens is painted black,the image will-

0%
be shifted downwards
0%
be shifted upwards
0%
not be shifted
0%
shift on the principal axis

In a terrestrial telescope, the telescope, the focal length of objective is 90 cm of inverting lens is 5 cm and of eye lens is 6 cm. If the final image is at 30 cm, then the magnification will be

0%
21
0%
12
0%
15
0%
18

Explanation

As we know,

$m=\dfrac{f_o}{f_o}\left( 1+\dfrac{f_e}{D}\right)\Rightarrow m=\dfrac{90}{6}\left( 1+\dfrac{6}{30}\right)\Rightarrow m=18$

The focal length of an objective of a telescope is 3 meter and diameter 15 cm. Assuming for a normal eye, the diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must be

0%
$6 cm$
0%
$6.3 cm$
0%
$20 cm$
0%
$60 cm$

Explanation

Full use of resolving power means whole aperture of objective in use. And for relaxed vision.

$\dfrac{f_o}{f_e}=\dfrac{D}{d}\Rightarrow \dfrac{300}{f_e}=\dfrac{f_o}{0.3}\Rightarrow f_e=6\ cm$
1745498_1474925_ans_d008b0f7aa07405b84d03af1d2ad2710.png

The angle of minimum deviation measured with a prism is

$30^o$ and the angle of prism

$60^o$ . The refractive index of prism material is -

0%
$\sqrt{2}$
0%
$2$
0%
$\dfrac{3}{2}$
0%
$\dfrac{4}{3}$

Explanation

As we know,

$\mu=\dfrac{\sin \left(\dfrac{A+\delta_{m}}{2}\right)}{\sin (A/2)}=\dfrac{\sin 45^{o}}{\sin 30^{o}}=\sqrt{2}$

A thick plane mirror shows a number of images of the filament of an electric bulb. Of these, the brightest image is the:

0%
First
0%
Second
0%
Last
0%
Fourth

Explanation

A thick plane mirror forms a number of images of the filament of an electric bulb. Images formed by the front surface which is unpolished reflect only a small part of the light (say

$10\%$ ), while the second image is formed by the polished surface which reflects most of the intensity ($$90\%). Hence the second image is the brightest. The intensity of the next reflections will keep on decreasing.
2082923_1509226_ans_856cf05ffa254b75991d63c525bacdcb.png

2082923_1509226_ans_856cf05ffa254b75991d63c525bacdcb.png

Three right angled prisms of refractive indices

$\mu_1$ ,

$\mu_2$ and

$\mu_3$ are joined together ao that the faces of the middle prism are each in contact with one of the outside prisms. If the ray passes through the composite block undeviated, then

0%
$1 + \mu _2^2 = \mu _1^2 + \mu _3^2$
0%
${\mu_1}^2-{\mu_3}^2+{\mu_2}^3 = 1$
0%
${\mu_1}^2-{\mu_3}^2-{\mu_2}^3 = 1$
0%
${\mu_2}^2+{\mu_3}^3-{\mu_2}^1 = 1$

Explanation

$\alpha = 90-r_i$

$\beta = 90-r_2$

$\gamma = 90-r_3$

$\begin{array}{l} Applying\, snell's\, Law\, at\, each\, surface \\ AtB, \\ \sin i={ \mu _{ 1 } }\sin { r_{ 1 } } \Rightarrow { \sin ^{ 2 } }i=\mu _{ 1 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 1 } } \\ At\, C \\ { \mu _{ 1 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 2 } }\sin { r_{ 2 } } \Rightarrow \mu _{ 1 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 2 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 2 } } \\ At\, D \\ { \mu _{ 2 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 3 } }\sin { r_{ 3 } } \Rightarrow \mu _{ 2 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 3 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 3 } } \\ At\, E, \\ { n_{ 3 } }\sin \left( { 90-{ r_{ 3 } } } \right) =\sin \left( { 90-i } \right) \Rightarrow \mu _{ 3 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 3 } }={ \cos ^{ 2 } }i \\ Adding\, all\, above\, equations\, \, gives, \\ 1+\mu _{ 2 }^{ 2 }=\mu _{ 1 }^{ 2 }+\mu _{ 3 }^{ 2 } \\ Hence,\, the\, option\, \, A\, is\, the\, correct\, answer. \end{array}$
1235288_1502318_ans_4bfd079c5735446781f12545268fdbc2.png

1235288_1502318_ans_4bfd079c5735446781f12545268fdbc2.png

A thin lens produce an image of the same size as the object.Then first the optical center of the lens, distance of the object is

0%
Zero
0%
4f
0%
2f
0%
f/2

A coordinate axis as shown in the the figure is kept in front of a converging lens at a distance 2f from it, where f is the focal length of the lens. Which of the following shows the approximate shape of the image? Assume that X-axis is the principal axis of the lens.

Which colour shows maximum deviation when light is dispersed through prism?

0%
Violet
0%
red
0%
yellow
0%
blue

Find the relation between critical angle "C" of a glass prism and its Angle of Prism "A", such that all the rays will reflect back while hitting second surface of prism

0%
$A<2C$
0%
$2A < C$
0%
$A>2C$
0%
$2A>C$

Explanation

Relation between prism angle A & critical angle C such
that ray will always show T I R at BC :
For this

$(r_2) min > C$ ...(i)
For

$(r_2)$ min,

$r1$ should be maximum and
for (r_1)max imax = 90

$(r_1)max =C$

$(r_2)max =AC$
Now from eq. (i)

$A C > C$

$A > 2C$
Thus for

$A > 2C$ , all rays are reflected back from the second surface
1250672_1615505_ans_bb5f90689c6a4e18a6f904ee31b65dc4.png

A ray of light is incident on the surface of transparent medium at an angle of

$45^o$ and is refracted in the medium at an angle of

$30^o$ . What will be the velocity of light in the transparent medium ?

0%
$1.96 \times 10^8 m/s$
0%
$2.12 \times 10^8 m/s$
0%
$2.65 \times 10^8 m/s$
0%
$1.25 \times 10^8 m/s$

Explanation

From Snell's Law,

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

$\Rightarrow v = c\dfrac{\sin r}{\sin i} = 3\times 10^8 \times \dfrac{\sin 30^o}{\sin 45^o}$

$\Rightarrow v=3\times 10^8\times \dfrac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = 2.12 \times 10^8 m/s$

If the tube length of astronomical telescope is

$96$ cm and magnifying power is

$15$ for normal setting, then the focal length of the objective is _________cm.

0%
$100$
0%
$90$
0%
$105$
0%
$92$

Explanation

$|m|=\dfrac{f_0}{f_e}=15$

$f_0=15f_e$
Tube length

$=f_0+f_e=96$

$15f_e+f_e=96$

$f_e=6$ cm

$f_0=15\times 6=90$ cm.

In a hologram, image is stored in the form of

0%
Diffraction pattern
0%
Photograph
0%
Interference pattern
0%
Intensity recording

A Point source of Light is placed

$4 \ m$ below the surface of water of refractive index

$\dfrac{5}{3}$ . the minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is: (μ=5/3)

0%
$2 \ m$
0%
$6 \ m$
0%
$4\ m$
0%
$$3\ m$$

$n$ simillar thin prisms of same material and refractive index are arranged in series as shown :

0%
if $n$ is even number, no net deviation and no net dispersion
0%
if $n$ is odd number, no net deviation and no net dispersion
0%
it depends upon angle of prism
0%
no sufficient information

Explanation

In given arrangement, if number of prism is even:
Then emerging ray from

$n^{th}$ prism have no net deviation and no net dispersion. Because of every sets of two prisms diminish their effect.

Hence option A

Four similar prism of same material having same angle of prism are arranged. Which of the following arrangements given no net angular deviation ?

The limit of resolution of microscope, if the numerical aperture of microscope is

$0.12$ , and the wavelength of light used is

$600\ nm$ , is:

0%
$0.3\mu m$
0%
$1.2\mu m$
0%
$2.3\mu m$
0%
$3\mu m$

A thin prism of angle

$7^o$ made of glass of refractive index

$1.5$ is combined with another prism made of glass of

$\mu =1.75$ to produce dispersion without deviation. The angle of second prism is :

0%
$7^o$
0%
$4.67^o$
0%
$9^o$
0%
$5^o$

One can not see through fog because:

0%
fog absorbed light
0%
light is scattered by the droplets in fog
0%
light surfers total reflection by the droplets in the fog
0%
the refractive index of fog is in infinity

The rising and setting of sun appear red because of :

0%
refraction
0%
reflection
0%
defraction
0%
scattering

A beam of white light is incident on a combinations of thin prism with equal angles. In the out going beam the rainbow colours are observed as shown

$\mu$ stands for refractive index and

$\theta$ for angle of dispersion. Then:

0%
$\mu_1 < \mu_2$
0%
$\mu_2 = \mu_1$
0%
$\theta_1 > \theta_2$
0%
$\mu_1 > \mu_2$

Consider a system of two thin lenses as shown in figure. An object of height

$1\, cm$ is placed at

$40\,cm$ from convex lens. Mark the correct option related to final image formed by the two-lens-system :

0%
Final image is formed at $32\,cm$ on right of concave lens and is $\dfrac{1}{2.2} \,cm$ in size .
0%
Final image is formed at $32\,cm$ on left side of convex lens and is $1\,cm$ in size .
0%
Final image is formed at $14.5\,cm$ on the left side of of concave lens and is $\dfrac{1}{2.2} \,cm$ in size .
0%
None of the above

The image formed by a convex mirror is only one-third of the size of the object. If the focal length of the mirror is

$12 \,cm,$ the image formed will be

0%
$8 \,cm$ behind the mirror
0%
$10 \,cm$ behind the mirror
0%
$8 \,cm$ in front of the mirror
0%
$10 \,cm$ in front of the mirror

Explanation

Given,
Height of image

$(h_1)$ =

$\dfrac{height \,of \,object (h_0)}{3}$ Focal length of mirror (f) = 12 cm

Magnification =

$m = \dfrac{h_f}{h_o} = \dfrac{-v}{u} = \dfrac{1}{3} \Rightarrow u = -3v$

Using Mirror formula ,

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

$\Rightarrow \dfrac{1}{v} + \dfrac{1}{-3v} = \dfrac{1}{12}$

$\Rightarrow \dfrac{2}{3v} = \dfrac{1}{12}$

$\Rightarrow v = 8 \,cm$ As v is positive, so image is formed at

$8 \,cm$ behind the mirror.

For a person suffering from myopia, the image of the object is formed in front of the retina.

0%
True
0%
False

Abnormalities in the normal vision of the eye are called defects of vision.

0%
True
0%
False

Short-sightedness is caused due to elongation of the eyeball.

0%
True
0%
False

Explanation

Myopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.

This defect may arise due to:

$i)$ excessive curvature of the eye lens

$ii)$ elongation of the eyeball

Write true or false for the following statements :
Presbyopia is a type of far-sightedness.

0%
True
0%
False

Explanation

Presbyopia is an age-related condition in which the lens of the eye becomes less flexible. Seeing details like words in a book or an online article, or adjusting focus between far away and nearby objects is difficult this condition is most common in people between the ages of

$40$ and

$50$ .
While in far-sightedness it's hard to see things close up. People of any age, including babies, can be farsighted.

Write true or false for the following statements :
A near sighted person can see only nearer objects clearly.

0%
True
0%
False

A $10 \mathrm{mm}$ long awl pin is placed vertically in front of a concave mirror. A $5 \mathrm{mm}$ long image of the awl pin is formed at $30 \mathrm{cm}$ in front of the mirror. The focal length of this mirror is:

0%
$-30 \mathrm{cm}$
0%
$-20 c m$
0%
$-40 \mathrm{cm}$
0%
$-60 \mathrm{cm}$

Explanation

Given: object size $\mathrm{h}=10 \mathrm{mm},$ image size $\mathrm{h}^{\prime}=5 \mathrm{mm},$ image distance
$\mathrm{v}=-30 \mathrm{cm},$

To find object distance=? $\mathrm{f}=?$

from the magnification formula $=\dfrac{h^{\prime}}{h}=\dfrac{-v}{u}$

or $\dfrac{5 m m}{10 m m}=\dfrac{-30 c m}{u}$ or $\dfrac{-30 c m}{u}=\dfrac{1}{2}$

or $u=-60$

Now, it can be calculated as follows:

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

or $\dfrac{1}{-30 \mathrm{cm}}+\dfrac{1}{-60 \mathrm{cm}}=\dfrac{1}{f}$ or $\dfrac{-2-1}{60}=\dfrac{1}{f}$

$\dfrac{1}{f}=-\dfrac{1}{20 \mathrm{cm}}$ or $f=-20 \mathrm{cm}$ (the negative sign confirms that the calculated focal length is of a concavemirror)

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

0%
a larger angle to be subtended by the object at the eye and hence viewed in greater detail
0%
the formation of a virtual erect image
0%
increase in the field of view
0%
infinite magnification at the near point

Explanation

In magnifying glass, the object is placed within the focal length and the image formed is magnified and erect. As

$(A'B')$ image is magnified so it subtends larger angle at the eye than object

$(AB)$ , so can be seen more clearly
1685573_1796703_ans_bb54466f482e42e9928f4a3f3f203e89.png

When a ray of light from air enters a denser medium, it:

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

Explanation

The ray of light bends towards the normal.

$\text{Reason:}$ As the speed of light decreases in the denser medium, it bends towards the normal.

1709052_1804912_ans_e5538d918fd4462babd23c27a4325792.png

A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from :

0%
its first focus
0%
its optical centre
0%
its second focus
0%
the centre of curvation of its second surface

The eye defect represented by the figure is

0%
Myopia
0%
Hypermetropia
0%
Cataract
0%
Presbyopia

A student of class

$10,$ is not able to see clearly the blackboard question when seated at a distance of

$5\ \mathrm{m}$ from the board, the defect he is suffering from is

0%
Myopia
0%
Hypermetropia
0%
Presbyopia
0%
Astigmatism

Explanation

Myopia is a vision condition in which you can see nearby objects clearly but far away objects appear blurry. In this defect, the light from far away objects focusses in front of the retina.

Since, the student is not able to see blackboard, which is located at a far away(

$5 \ m$ ) distance, clearly. Hence, he is suffering from Myopia.

Long-sightedness or hypermetropia can be corrected by

0%
Planar lens
0%
Concave lens
0%
Convex lens
0%
Bifocal lens

A thin convex lens of focal length

$10\ cm$ is placed in contact with a concave lens of same material and of same focal length. he focal length of combination will be

0%
Zero
0%
Infinity
0%
$10\ cm$
0%
$20\ cm$

Explanation

As we know,

$f=\dfrac{f_1f_2}{f_1+f_2}=\dfrac{10(-10)}{10+(-100)}=\dfrac{-100}{10-10}=\infty$

The refractive indices of glass and water w.r.t air are

$\dfrac{3}{2}$ and

$\dfrac{4}{3}$ respectively. The refractive index of glass w.r.t water will be

0%
$\dfrac{8}{9}$
0%
$\dfrac{9}{8}$
0%
$\dfrac{7}{6}$
0%
None of these

Explanation

As we know,

$_a\mu_{g}=\dfrac {3}{2}, _a\mu_{w}=\dfrac {4}{3}$

$\therefore \ _w\mu_{g}=\dfrac {_a\mu_{g}}{_a\mu_{w}}=\dfrac {3/2}{4/3}=\dfrac 98$

CBSE Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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