Optics(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

OPTICS MCQs

Wave Optics, Ray Optics Refraction, Ray Optics Fundamentals, Ray Optics Curved Surface Refraction, Ray Optics Curved Mirrors

Total Questions : 111 | Page 2 of 12 pages

Question 11. A ray of light passes from vacuum into a medium of refractive index

μ

,the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is

cos−1(μ2)
2cos−1(μ2)
2sin−1(μ)
2sin−1(μ2)

Discuss Question

Answer: Option B. -> 2cos−1(μ2)
:
B
By using

μ = \frac{s i n i}{s i n r} \Rightarrow μ = \frac{s i n 2 r}{s i n r} = \frac{2 s i n r c o s r}{s i n r}

\Rightarrow r = c o s^{- 1} (\frac{μ}{2}) . S o, i = 2 r = 2 c o s^{- 1} (\frac{μ}{2})

Question 12. Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of

θ

at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is

Two Plane Mirrors P And Q Are Aligned Parallel To Each Other...

ldtanθ
dltanθ
ldtanθ
None of these

Discuss Question

Answer: Option A. -> ldtanθ
:
A
Suppose n = Total number of reflection light ray undergoes before exist out.
x = Horizontal distance travelled by light ray in one reflection.
So nx = l also

t a n θ = \frac{x}{d}

\Rightarrow n = \frac{l}{d t a n θ}

Question 13. A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a deviation of

300^{\circ}

. The number of images observable are

Discuss Question

Answer: Option B. -> 11
:
B
By using

δ = (360 - 2 θ) \Rightarrow 300 = 360 - 2 θ

\Rightarrow θ = 30^{\circ}

.Hence number of images =

\frac{360}{30} - 1 = 11

Question 14. A point object is placed mid-way between two plane mirrors distance 'a' apart. The plane mirror forms an infinite number of images due to multiple reflection. The distance between the nth order image formed in the two mirrors is

Discuss Question

Answer: Option B. -> 2na
:
B

A Point Object Is Placed Mid-way Between Two Plane Mirrors D...

From above figure it can be proved that seperation between nth order image formed in the two mirrors = 2na

Question 15. A thin equiconvex lens is made of glass of R.I 1.5 and its focal length is 0.2 If it acts a concave lens of 0.5 m focal length when dipped in a liquid, the R.I of liquid is

Discuss Question

Answer: Option B. -> 158
:
B

- \frac{f_{1}}{f_{a}} = \frac{(μ_{5} - 1)}{(\frac{μ_{s}}{μ_{l}} - 1)} - 2.5 = \frac{0.5}{(\frac{1.5}{μ_{l}} - 1)} - \frac{0.5}{0.2} = \frac{1.5 - 1}{(\frac{1.5}{μ_{1}} - 1)} \frac{1.5}{μ_{l}} - 1 = - \frac{1}{5} \frac{1.5}{μ_{l}} = 1 - \frac{1}{5} = \frac{4}{5} μ_{l} = \frac{7}{4} = \frac{75}{40} = \frac{15}{8}

Question 16. There is an equiconvex glass lens with radius of each face as R and

_{a} μ_{g} = \frac{3}{2} a n d_{a} μ_{w} = \frac{4}{3}

. If there is water in object space and air in image space, then the focal length is

Discuss Question

Answer: Option C. -> 3R2
:
C
Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium

\frac{μ_{2} - μ_{1}}{R} = \frac{μ_{1}}{- u} + \frac{μ_{2}}{v_{1}} \Rightarrow \frac{(\frac{3}{2}) - (\frac{4}{3})}{R} = \frac{\frac{4}{3}}{\infty} + \frac{\frac{3}{2}}{v_{1}} \Rightarrow v_{1} = 9 R

Now consider the refraction at the second surface of the lens i.e. refraction from denser medium to rarer medium

\frac{1 - \frac{3}{2}}{- R} = - \frac{\frac{3}{2}}{9 R} + \frac{1}{v_{2}} \Rightarrow v_{2} = (\frac{3}{2}) R

The image will be formed at a distance of

\frac{3}{2} R

.This is equal to the focal length of the lens.

There Is An Equiconvex Glass Lens With Radius Of Each Face...

Question 17. Shown in the figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length + 20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is

+ 80 cm
– 80 cm
– 24 cm
–100 cm

Discuss Question

Answer: Option D. -> –100 cm
:
D

\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}

\frac{1}{f_{1}} = (1.6 - 1) (\frac{1}{\infty} - \frac{1}{20}) = \frac{0.6}{20} = - \frac{3}{100} . . . (i)

\frac{1}{f_{2}} = (1.5 - 1) (\frac{1}{20} - \frac{1}{- 20}) = \frac{1}{20} . . . (i i)

\frac{1}{f_{3}} = (1.6 - 1) (\frac{1}{- 20} - \frac{1}{\infty}) = - \frac{3}{100} . . . (i i i)

\Rightarrow \frac{1}{F} = - \frac{3}{100} + \frac{1}{20} - \frac{3}{100} \Rightarrow F = - 100_{c m}

Question 18. A ray of light falls on a transparent sphere with centre at C as shown in figure. The ray emerges from the sphere parallel to line AB. The refractive index of the sphere is:

√2
√3
32
12

Discuss Question

Answer: Option B. -> √3
:
B
Deviation by a aphere is 2(i -r) Here, deviation

δ = 60^{\circ} = 2 (i - r)

(or)

i - r = 30^{\circ}

∴

r = i - 30^{\circ} = 60^{\circ} - 30^{\circ} = 30^{\circ}

∴

μ = \frac{s i n i}{s i n r} = \frac{s i n 60^{\circ}}{s i n 30^{\circ}} = \sqrt{3}

Question 19. An air bubble in sphere having 4 cm diameter appears 1 cm from surface nearest to eye when looked along diameter. If _am_g = 1.5, the distance of bubble from refracting surface is
[CPMT 2002]