M.C.Q (1 Marks)

MCQ 11 Mark

Consider following statements regarding a simple microscope:
(P) It allows us to keep the object within the least distance of distinct vision.
(Q) Image appears to be biggest if the object is at the focus.
(R) It is simply a convex lens.

A
Only (P) is correct
B
Only (P) and (Q) are correct
C
Only (Q) and (R) are correct
✓
Only (P) and (R) are correct

Answer

Correct option: D.

Only (P) and (R) are correct

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MCQ 21 Mark

Which of the following is not involved in formation of a rainbow?

A
refraction
B
angular dispersion
C
angular deviation
✓
total internal reflection

Answer

Correct option: D.

total internal reflection

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MCQ 31 Mark

Angles of deviation for extreme colours are given for different prisms. Select the one having maximum dispersive power of its material.

✓
7°, 10°
B
8°, 11°
C
12°, 16°
D
10°, 14°

Answer

Correct option: A.

7°, 10°

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MCQ 41 Mark

Select the WRONG statement.

✓
Smaller angle of prism is recommended for greater angular dispersion.
B
Right angled isosceles glass prism is commonly used for total internal reflection.
C
Angle of deviation is practically constant for thin prisms.
D
For emergent ray to be possible from the second refracting surface, certain minimum angle of incidence is necessary from the first surface.

Answer

Correct option: A.

Smaller angle of prism is recommended for greater angular dispersion.

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MCQ 51 Mark

A spherical marble, of refractive index 1.5 and curvature 1.5 cm, contains a tiny air bubble at its centre. Where will it appear when seen from outside?

A
1 cm inside
✓
at the centre
C
$5 / 3 cm$ inside
D
2 cm inside

Answer

Correct option: B.

at the centre

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MCQ 61 Mark

A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are

✓
bi-concave
B
piano concave
C
concavo-convex
D
convexo-concave

Answer

Correct option: A.

bi-concave

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MCQ 71 Mark

Consider following phenomena/applications: P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in

✓
Only R and S
B
Only R
C
Only P, R and S
D
all the four

Answer

Correct option: A.

Only R and S

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MCQ 81 Mark

There are different fish, monkeys and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at 2.5 m on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, height of the tree and the depth at which the fish is swimming are exactly same. Refractive index of that water must be

A
$6 / 5$
✓
$5 / 4$
C
$4 / 3$
D
$7 / 5$

Answer

Correct option: B.

$5 / 4$

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MCQ 91 Mark

Which of the following aberrations will NOT occur for spherical mirrors?

✓
Chromatic aberration
B
Coma
C
Distortion
D
Spherical aberration

Answer

Correct option: A.

Chromatic aberration

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MCQ 101 Mark

A concave mirror of curvature 40 cm, used for shaving purpose produces image of double size as that of the object. Object distance must be

A
10 cm only
B
20 cm only
C
30 cm only
✓
10 cm or 30 cm

Answer

Correct option: D.

10 cm or 30 cm

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MCQ 111 Mark

Two plane mirrors are inclined at angle 40° between them. Number of images seen of a tiny object kept between them is

A
Only 8
B
Only 9
✓
8 or 9
D
9 or 10

Answer

Correct option: C.

8 or 9

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MCQ 121 Mark

Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention which of these have positive focal length?

A
OnlyP
✓
Only P and Q
C
Only P and R
D
Only Q and S

Answer

Correct option: B.

Only P and Q

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MCQ 131 Mark

As per recent understanding light consists of

A
rays
B
waves
C
corpuscles
✓
photons obeying the rules of waves

Answer

Correct option: D.

photons obeying the rules of waves

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MCQ 141 Mark

A transparent glass cube of length $24 cm$ has a small air bubble trapped inside. When seen normally through one surface from air outside, its apparent distance is $10 cm$ from the surface. When seen normally from opposite surface, its apparent distance is $6 cm$. The distance of the air bubble from first surface is

A
$15 cm$
B
$14 cm$
C
$12 cm$
D
$8 cm$

Answer

(a) : $\mu=3 / 2$
Let the real depth from face I is $t$.
$\begin{aligned}
\mu & =\frac{\text { Real depth }}{\text { Apparent depth }} \\
\mu & =\frac{t}{10}...(i) \\
\mu & =\frac{24-t}{6}...(ii)
\end{aligned}$
From (i) and (ii),
$\frac{t}{10}=\frac{24-t}{6} \Rightarrow 3 t=24 \times 5-5 t \Rightarrow t=15 cm$

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MCQ 151 Mark

If a ray of light in denser medium strikes a rarer medium at an angle of incidence $i$, the angles of reflection and refraction are $r$ and $r^{\prime}$ respectively. If the reflected and refractive rays are at right angles to each other, the critical angle for the given pair of media is

A
$\sin ^{-1}\left(\tan r^{\prime}\right)$
B
$\sin ^{-1}(\tan r)$
C
$\tan ^{-1}(\sin i)$
D
$\cos ^{-1}(\tan i)$

Answer

(b) : From figure, $i=r$ and $r^{\prime}=90-r$
$n=\frac{\sin r^{\prime}}{\sin r}=\frac{\sin (90-r)}{\sin r}=\frac{1}{\tan r}$...(i)
$n=\frac{1}{\sin C}$...(ii)
From (i) and (ii),
$
\frac{1}{\sin C}=\frac{1}{\tan r} \Rightarrow C=\sin ^{-1}(\tan r)
$

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MCQ 161 Mark

The rays of different colours fail to converge at a point after passing through a thick converging lens. The defect is called

A
spherical aberration
B
distortion
C
coma aberration.
✓
chromatic

Answer

Correct option: D.

chromatic

(d) : Due to chromatic aberration, the rays of different colours fails to converge at a point after passing through a thick converging lens.

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MCQ 171 Mark

The angle of prism is $A$ and one of its refracting surface is silvered. Light rays falling at an angle of incidence ' $2 A$ ' on the first surface return back through the same path after suffering reflection at the silvered surface. The refractive index of the material of the prism is

A
$2 \sin \left(\frac{A}{2}\right)$
B
$2 \tan A$
✓
$2 \cos A$
D
$2 \sin A$

Answer

Correct option: C.

$2 \cos A$

(c) : $\alpha=90-A$
So, $r=A$
$
\begin{aligned}
& \mu=\frac{\sin i}{\sin r}=\frac{\sin 2 A}{\sin A} \\
& \mu=\frac{2 \sin A \cos A}{\sin A}=2 \cos A
\end{aligned}
$

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MCQ 181 Mark

When a monochromatic ray of light is passed through an equilateral glass prism, it is found that the refracted ray in glass is parallel to the base of the prism. If $i$ and $e$ denote the angles of incidence and emergence respectively, then

A
$i<e$
B
$i<e$
✓
$i=e$
D
$i+e=90^{\circ}$

Answer

Correct option: C.

$i=e$

(c): The refracted ray is parallel to the base.
It means minimum deviation, so, $i=e$.

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MCQ 191 Mark

Refractive index of a glass convex lens is $1.5.$ The radius of curvature of each of the two surfaces of the lens is $20 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is

A
$2: 3$
✓
$2: 5$
C
$3: 2$
D
$5: 2$

Answer

Correct option: B.

$2: 5$

$ \mu_g=1.5, R_1=20 \ cm , R_2=-20 \ cm ; \mu_l=1.25$
$P_g=\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(i)$
$P_l=\left({ }^l \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(ii)$
$\frac{P_l}{P_g}=\frac{l_{\mu_g-1}}{a_{\mu_g-1}}=\frac{\frac{\mu_g}{\mu_l}-1}{a_{\mu_g-1}}$
$\frac{P_l}{P_g}=\frac{\frac{1.5}{1.25}-1}{1.5-1}=\frac{0.25}{1.25 \times 0.5} ; \frac{P_l}{P_g}=\frac{2}{5}$

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MCQ 201 Mark

The angle of deviation produced by a thin prism when placed in air is $' \delta\ ' $ and that when immersed in water is $' \delta_2\ '.$ The refractive index of glass and water are $\frac{3}{2}$ and $\frac{4}{3}$ respectively. The ratio $\delta_1: \delta_2$ is

A
$1: 2$
B
$2: 1$
C
$1: 4$
✓
$4: 1$

Answer

Correct option: D.

$4: 1$

$ \delta=(\mu-1) A$
$\delta_1=\left(\mu_g-1\right) A ...(i)$
$\delta_2=\left({ }^w \mu_g-1\right) A ....(ii)$
$\frac{\delta_1}{\delta_2}=\frac{\frac{3}{2}-1}{\frac{3 \times 3}{2 \times 4}-1}=\frac{\frac{1}{2}}{\frac{1}{8}}=4: 1$

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MCQ 211 Mark

A combination of two thin lenses in contact have power $+10 D$. The power reduces to $+6 D$ when the lenses are $0.25 m$ apart. The power of individual lens is

A
$5 D , 5 D$
B
$6 D , 4 D$
C
$7 D , 3 D$
✓
$8 D , 2 D$

Answer

Correct option: D.

$8 D , 2 D$

(d) : $P_1+P_2=10 D$ ...(i)
$
\begin{aligned}
& P_1+P_2-d P_1 P_2=6 D \\
& 10-0.25 P_1 P_2=6 ..(ii)
\end{aligned}
$
Using equation (ii), we have
$
\begin{aligned}
& \therefore \frac{10-6}{0.25}=P_1 P_2 ; P_1 P_2=16 \\
& P_1-P_2=\sqrt{\left(P_1+P_2\right)^2-4 P_1 P_2}=\sqrt{10^2-4 \times 16} \\
& P_1-P_2=6
...(iii)\end{aligned}
$
Using equation (i) and (iii), $P_1=8 D , P_2=2 D$

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MCQ 221 Mark

The refracting angle of a glass prism is $30^{\circ}$. A ray is incident on one of the faces perpendicular to it. The angle of deviation $\delta$ between the incident ray and that leaves the prism is. $($ Refractive index of glass $=1.5)$. $\left(\sin 30^{\circ}=0.5, \sin (48.6)^{\circ}=0.75\right)$

A
$17^{\circ}$
✓
$18.6^{\circ}$
C
$12.6^{\circ}$
D
$16^{\circ}$

Answer

Correct option: B.

$18.6^{\circ}$

(b) :Given, $A=30^{\circ}, m =1.5 ; i_1=0^{\circ}$
In a prism, $r_1+r_2=A \quad \therefore r_2=A=30^{\circ}$
from snell's law, $\mu=\frac{\sin i_2}{\sin r_2}$
$
\begin{aligned}
& \Rightarrow 1.5=\frac{\sin i_2}{\sin 30^{\circ}} \Rightarrow \sin i_2=1.5 \sin 30^{\circ} \\
& =1.5 \times \frac{1}{2}=0.75 \\
& i_2=\sin ^{-1}(0.75)=48.6^{\circ}
\end{aligned}
$
Deviation, $\delta=\left(i_1+i_2\right)-A=\left(0+48.6^{\circ}\right)-30^{\circ}=18.6^{\circ}$

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MCQ 231 Mark

A compound microscope is a magnification of 24 . The focal length of the eyepiece is $5 cm$.The final image is formed at the least distance of distinct vision. The magnification produced by the objective is

A
5
B
7
✓
4
D
6

Answer

Correct option: C.

(c):For a compound microscope, $M=m_o \times m_e$.
Also, the final image is formed at $D$.
So, $M=m_0\left(1+\frac{D}{f_e}\right)$ $m_0=\frac{M}{1+\frac{D}{f_e}}=\frac{24}{1+\frac{25}{5}}=\frac{24}{6} ; m_0=4$

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MCQ 241 Mark

A glass convex lens is of refractive index 1.55 with both faces of same radius of curvature. What is the radius of curvature required if focal length is to be $20 cm$ ?

A
$18 cm$
B
$21 cm$
C
$20 cm$
✓
$22 cm$

Answer

Correct option: D.

$22 cm$

(d): Lens maker's formula,
$
\begin{aligned}
& \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \text { Here, } f=20 cm , \mu=1.55 ; R_1=R ; R_2=-R \\
& \Rightarrow \frac{1}{20}=(1.55-1)\left(\frac{1}{R}+\frac{1}{R}\right) \\
& \Rightarrow \frac{1}{20}=0.55 \times \frac{2}{R} \Rightarrow R=22 cm
\end{aligned}
$

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MCQ 251 Mark

A ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is

A
zero
✓
$30^{\circ}$
C
$45^{\circ}$
D
$60^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

(b) : Angle of prism, $A=r_1+r_2$
For minimum deviation
$
r_1=r_2=r \quad \therefore A=2 r
$
Given, $A=60^{\circ}$
Hence, $r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}$

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MCQ 261 Mark

A ray of light travelling in a transparent medium of refractive index $\mu$, falls on a surface separating the medium from air at an angle of incidence of $45^{\circ}$. For which of the following value of $\mu$, the ray can undergo total internal reflection?

A
$\mu=1.33$
B
$\mu=1.40$
✓
$\mu=1.50$
D
$\mu=1.25$

Answer

Correct option: C.

$\mu=1.50$

(c) : For total internal reflection, $\sin i>\sin C$
where, $i=$ angle of incidence
$C=$ critical angle
But, $\sin C=\frac{1}{\mu} \quad \therefore \sin i>\frac{1}{\mu}$ or $\mu>\frac{1}{\sin i}$
$
\begin{aligned}
& \mu>\frac{1}{\sin 45^{\circ}} \quad \quad\left(i=45^{\circ} \text { (Given) }\right) \\
& \mu>\sqrt{2}
\end{aligned}
$
Hence, option (c) is correct.

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MCQ 271 Mark

A convex lens of focal length $f$ is placed in contact with a concave lens of the same focal length. The equivalent focal length of the combination will be

A
$f$
✓
infinity
C
$f / 2$
D
zero

Answer

Correct option: B.

infinity

(b) : Here, equivalent focal length of the combination is,
$
\frac{1}{F_{e q}}=\frac{1}{F_1}+\frac{1}{F_2}-\frac{d}{F_1 F_2}
$
Now, as the lenses are in contact, so, $d=0$
$
\therefore \frac{1}{F_{e q}}=\frac{1}{F_1}+\frac{1}{F_2}=\frac{1}{-f}+\frac{1}{f}=0 \Rightarrow F_{e q}=\infty \text { (infinity) }
$

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MCQ 281 Mark

The equiconvex lens has a focal length ' $f$ '. If the lens is cut along the line perpendicular to principal axis and passing through the pole, what will be the focal length of any half part?

A
$\frac{f}{2}$
✓
$2 f$
C
$\frac{3 f}{2}$
D
$f$

Answer

Correct option: B.

$2 f$

(b) : Using lens maker's formula,
$
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$
Since, $R_1=+R$ and $R_2=-R$ for equiconvex lens.
$
\therefore \frac{1}{f}=(\mu-1)\left(\frac{1}{R}+\frac{1}{R}\right)=(\mu-1)\left(\frac{2}{R}\right)
$
Now, when the lens is cut along the line perpendicular to principal axis then,
$
\begin{aligned}
& R_1=R \text { and } R_2=\infty \\
& \therefore \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}+\frac{1}{\infty}\right) \\
& =(\mu-1)\left(\frac{1}{R}\right)=\frac{1}{2}(\mu-1)\left(\frac{2}{R}\right)=\frac{1}{2 f} \quad \therefore f^{\prime}=2 f
\end{aligned}
$

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MCQ 291 Mark

The angle made by incident ray of light with the reflecting surface is called

✓
glancing angle
B
angle of incidence
C
angle of deviation
D
angle of refraction

Answer

Correct option: A.

glancing angle

(a) : The angle between incident ray and the reflecting surface is called glancing angle.

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Physics M.C.Q (1 Marks)

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