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Geometrical Optics

NEET Physics · 99 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance x in front of the second wall. The required focal length of the lens will be :
A less than x4\dfrac{x}{4}
B more than x4\dfrac{x}{4} but less than x2\dfrac{x}{2}
C x2\dfrac{x}{2}
D x4\dfrac{x}{4}
Correct Answer
Option C
Solution
x=2f\mathrm{x=2f}
f=x/2\mathrm{f=x/2}
Q12
Light travels a distance x\mathrm{x} in time t1t_{1} in air and 10x10 \mathrm{x} in time t2t_{2} in another denser medium. What is the critical angle for this medium?
A sin⁡−1(10t2t1)\sin ^{-1}\left(\dfrac{10 \mathrm{t}_{2}}{\mathrm{t}_{1}}\right)
B sin⁡−1(t110t2)\sin ^{-1}\left(\dfrac{t_{1}}{10 t_{2}}\right)
C sin⁡−1(10t1t2)\sin ^{-1}\left(\dfrac{10 t_{1}}{t_{2}}\right)
D sin⁡−1(t2t1)\sin ^{-1}\left(\dfrac{t_{2}}{t_{1}}\right)
Correct Answer
Option C
Solution

The critical angle, denoted as

θc\theta_c

, is the angle of incidence beyond which light is totally internally reflected within a denser medium when it hits the boundary with a less dense medium.

To find the critical angle for the medium in question, first, we need to understand the relationship between the speed of light in different media and their refractive indices.

Let's denote the speed of light in air as

V1V_1

and in the denser medium as

V2V_2

. From the given information:

V1=xt1V_1 = \frac{x}{t_1}

(speed of light in air)

V2=10xt2V_2 = \frac{10x}{t_2}

(speed of light in the denser medium) The refractive index of a medium (n) is inversely proportional to the speed of light in that medium (

n=cVn = \frac{c}{V}

, where c is the speed of light in vacuum). Thus, the refractive index of the denser medium (

n2n_2

) relative to air (

n1n_1

) can be obtained by taking the ratio of the speed of light in air to that in the denser medium:

n2n1=V1V2\frac{n_2}{n_1} = \frac{V_1}{V_2}

Substituting the expressions for

V1V_1

and

V2V_2

:

n2n1=xt110xt2=t210t1\frac{n_2}{n_1} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} = \frac{t_2}{10t_1}

For the critical angle

θc\theta_c

, the light in the denser medium (index

n2n_2

) strikes the boundary with the less dense medium (index

n1n_1

) such that it refracts at 90 degrees (escapes along the boundary). Snell's Law at this boundary is:

n2sin⁡θc=n1n_2 \sin \theta_c = n_1

Given that

n1=1n_1 = 1

(approximating the refractive index of air), we simplify to:

sin⁡θc=1n2\sin \theta_c = \frac{1}{n_2}

Substituting

n2n_2

from the earlier relation and simplifying:

sin⁡θc=10t1t2\sin \theta_c = \frac{10t_1}{t_2}

Hence, the critical angle

θc\theta_c

is given by:

θc=sin⁡−1(10t1t2)\theta_c = \sin^{-1}\left(\frac{10 t_1}{t_2}\right)
Q13
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin) ?
A −40 cm-40 \mathrm{~cm}
B −100 cm-100 \mathrm{~cm}
C −50 cm-50 \mathrm{~cm}
D 40 cm40 \mathrm{~cm}
Correct Answer
Option B
Solution

Use

1f=[μ−1][1R1−1R2]\frac{1}{\mathrm{f}}=[\mu-1]\left[\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right]
1f1=[1.6−1][1∞−120]=−3100\frac{1}{\mathrm{f}_{1}}=[1.6-1]\left[\frac{1}{\infty}-\frac{1}{20}\right]=\frac{-3}{100}
1f2=[1.5−1][120+120]=120\frac{1}{\mathrm{f}_{2}}=[1.5-1]\left[\frac{1}{20}+\frac{1}{20}\right]=\frac{1}{20}
1f3=−3100\frac{1}{\mathrm{f}_{3}}=\frac{-3}{100}
1feq=1f1+1f2+1f3\frac{1}{f_{e q}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}
1feq=−3100+120−3100=−1100\frac{1}{f_{e q}}=-\frac{3}{100}+\frac{1}{20}-\frac{3}{100}=\frac{-1}{100}
Q14
Two thin lenses are of same focal lengths (f)(f), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be :
A f/4f / 4
B f/2f / 2
C Infinite
D Zero
Correct Answer
Option C
Solution
1feq=1f1+1f2\frac{1}{f_{e q}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}
1feq=1f−1f\frac{1}{f_{e q}}=\frac{1}{f}-\frac{1}{f}
feq=∞\mathrm{f}_{\mathrm{eq}}=\infty
Q15
During a cloudy day, a primary and a secondary rainbow may be created, then the :
A secondary rainbow is due to single internal reflection and is formed above the primary one.
B primary rainbow is due to double internal reflection and is formed above the secondary one.
C primary rainbow is due to double internal reflection and is formed below the secondary one.
D secondary rainbow is due to double internal reflection and is formed above the primary one.
Correct Answer
Option D
Solution

Primary rainbow is result of three-step process, that is, refraction, internal reflection and refraction.

Secondary rainbow is result of four step process, that is, refraction, internal reflection, internal reflection and refraction.

Secondary rainbow appears above the primary rainbow.

Q16
An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eye piece used in the construction of the telescope are 20 m and 2 cm respectively. Consider the following statements about the telescope : (a) The distance between the objective and eye piece is 20.02 m (b) The magnification of the telescope is (−-) 1000 (c) The image of the planet is erect and diminished (d) The aperture of eye piece is smaller than that of objective The correct statements are :
A (a), (b) and (d)
B (a), (b) and (c)
C (b), (c) and (d)
D (c), (d) and (a)
Correct Answer
Option A
Solution

Given f 0 = 20 m = 2000 cm, f e = 2 cm ∙\bullet Distance between objective and eye piece l = f 0 + f e = 20 m + 2 cm = 20.02 cm ∙\bullet Magnification of telescope

m=−f0fe=−2000 cm2 cm=−1000m = - {{{f_0}} \over {{f_e}}} = - {{2000\,cm} \over {2\,cm}} = - 1000

∙\bullet Image formed by telescope is inverted.

∙\bullet Aperture of eye piece is smaller than that of objective.

So, statements (a), (b) and (c) are correct.

Q17
A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is
A +2 D
B +20 D
C +5 D
D Infinity
Correct Answer
Option C
Solution

Power of lens is given by

P=1f(m)P = {1 \over {f(m)}}
1f=(μ−1)(1R1−1R2){1 \over f} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
1f={32−1}(120+120){1 \over f} = \left\{ {{3 \over 2} - 1} \right\}\left( {{1 \over {20}} + {1 \over {20}}} \right)
f=20f = 20

cm

P=120×10−2P = {1 \over {20 \times {{10}^{ - 2}}}}
=5= 5

D

Q18
A light ray falls on a glass surface of refractive index 3\sqrt3, at an angle 60∘^\circ. The angle between the refracted and reflected rays would be
A 30∘^\circ
B 60∘^\circ
C 90∘^\circ
D 120∘^\circ
Correct Answer
Option C
Solution

Given i = 60

∘^\circ

and μ\mu =

3\sqrt3

⇒\Rightarrow Here, angle of incidence ⇒\Rightarrow i = tan −-1 (μ\mu) Hence, reflected and refracted rays would be perpendicular to each other.

Q19
Two transparent media A and B are separated by a plane boundary. The speed of light in those media are 1.5 ×\times 10 8 m/s and 2.0 ×\times 10 8 m/s, respectively. The critical angle for a ray of light for these two media is
A sin −-1 (0.500)
B sin −-1 (0.750)
C tan −-1 (0.500)
D tan −-1 (0.750)
Correct Answer
Option B
Solution
μA=3×1081.5×108=2{\mu _A} = {{3 \times {{10}^8}} \over {1.5 \times {{10}^8}}} = 2
μB=3×1082×108=1.5{\mu _B} = {{3 \times {{10}^8}} \over {2 \times {{10}^8}}} = 1.5

For TIR, ray of light should travel from denser to rarer medium

μAsin⁡θC=μBsin⁡90∘{\mu _A}\sin {\theta _C} = {\mu _B}\sin 90^\circ
2sin⁡θC=1.5sin⁡90∘2\sin {\theta _C} = 1.5\sin 90^\circ
sin⁡θC=0.75\sin {\theta _C} = 0.75
θC=sin⁡−1(0.75){\theta _C} = {\sin ^{ - 1}}(0.75)
Q20
Find the value of the angle of emergence from the prism. Refractive index of the glass is 3\sqrt 3
A 90∘^\circ
B 60∘^\circ
C 30∘^\circ
D 45∘^\circ
Correct Answer
Option B
Solution

from Snell's law

3\sqrt 3

sin 30 o = 1 ×\times sin r ⇒\Rightarrow r = 60 o

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