CBSE Board Exam 2020: Important MCQs (with Answers) for Class 12 Physics - Chapter 9 - Ray Optics and Optical Instruments; Also useful for JEE Main, UPSEE, WBJEE& Other Engineering Entrance Exams

Check important MCQs (with Answers) for Class 12 Physics Board Exam 2020 (Chapter 9 - Ray Optics and Optical Instruments). Students preparing for CBSE Class 12th Physics Board Exam usually ask about important MCQs (Multiple Choice Questions) & in this article, we are going to provide important questions, based on Chapter 9 (Ray Optics and Optical Instruments) of Class 12 Physics NCERT textbook. Here you will also get important links to access some important articles for the preparation of CBSE 12th board exams 2020.

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NCERT Exemplar Solutions for CBSE Class 12 Physics

Chapter 9

Ray Optics and Optical Instruments

Q1. In the figure given below there are two convex lenses L_A and L_B having focal lengths F_A and F_B respectively. The distance between L_A and L_B is

(a) F_A

(b) F_B

(c) F_A + F_B

(d) F_A ‒ F_B

Sol:

The light ray after falling on lens L_A will converge to focus. The light ray will become parallel to the axis only when the refracted ray starts from the focus of L_B.

Hence, distance between L_A and L_B is F_A + F_B.

Q2. If the critical angle for total internal reflection from a medium to vacuum is 30^o, the velocity of light in the medium is

(a) 3 X 10⁸ m/s

(b) 1.5 X 10⁸ m/s

(c) 0.5 X 10⁸ m/s

(d) 0.2 X 10⁸ m/s

Sol: (b)

Given, critical angle = C = 30^o.

Refractive index of a medium = μ = 1/(sin C) = 1/(sin 30^o) = 2.

Q3. Rainbow is formed due to the combination of

(a) refraction and absorption

(b) dispersion and focusing

(c) dispersion and total internal reflection

(d) refraction and scattering

Sol: (c)

A rainbow is formed due to a combination of dispersion and total internal reflection.

Q4. Refractive index of material of a prism is √2. The angle of prism is 60^o. What is the angle of minimum deviation?

(a) 30^o

(b) 45^o

(c) 60^o

(d) 90^o

Sol: (a)

Here, A = 60^o and μ = √2, putting values in …(i) we have,

Q5. The radius of curvature of a double concave lens is 20 cm. If the refractive index of glass with respect to air is 5/3 then, the focal length of this lens is

(a) 5 cm

(b) 15 cm

(c) ‒ 15 cm

(d) 20 cm

Sol:

Given, R₁ = ‒ 20 cm, R₂ = 20 cm, μ = 5/3 and f =?

Q6. What is the correct relation between the refractive indices n, n₁ and n₂ if the behavior of light rays is as shown in the figure given below?

(a) n₁<n

(b) n₁>n

(c) n₁ = n

(d) None of these

Sol: (a)

In the figure shown, the convex lens is behaving as concave lens, it is possible only when refractive index of lens is less than the surroundings i.e. n₁<n.

Q7. A convex lens of focal length f = 20 cm is combined with a diverging lens of power 65 D. The power and the focal length of the combination is

(a) ‒1.5 D, 66.7 cm

(b) 1.5 D, 33.7 cm

(c) 5 D, 66.7 cm

(d) 5 D, 33.6 cm

Sol: (c)

Given, focal length of the convex lens = f = 20 cm = 0.2 m.

Therefore, its power is: P = 1/f(in meters) = 1/0.2 = 5D.

As, the power of diverging lens is ‒6.5 D therefore, the power of the combination is = 5D + (-6.5 D) = -1.5 D.

The focal length of the system of lens = 1/P = 1/(‒1.5 D) = ‒0.667 m = 66.7 cm.

Q8. The power of a thin convex lens of glass is 5 D. When it is immersed in a liquid, then it behaves like a divergent lens of focal length 100 cm. The refractive index of the liquid is:

(Given, refractive index of glass = μ = 1.5)

(a) 3/5

(b) 1/2

(c) 2/3

(d) 5/3

Sol: (d)

Focal length of the convex lens in air is given by: 1/f_air = 1/P_a = 1/5D = 0.2 m = 20 cm.

Inside the liquid, the lens is behaving as divergent lens and its focal length is given by f_l = ‒100 cm.

Now, f_a/f_l = 20 cm/-100cm = -1/5

From Len’s makers formula,

Q9. An equiconvex lens of focal length f and power P is cut into two halves in thickness. The focal length and power of each half is

(a) f/2

(b) 2f

(c) f

(d) zero

Sol: (b)

Q10. The magnifying power of an astronomical telescope in normal adjustment is 100. The distance between the objective and the eyepiece is 101 cm. The focal length of the objectives and eyepiece is

(a) 10 cm and 1 cm respectively

(b) 100 cm and 1 cm respectively

(c) 1 cm and 100 cm respectively

(d) 1 cm and 10 cm respectively

Sol: (b)

In normal adjustment, the object and final image are both at infinity and the separation between the objective and the eye piece is f_o + f_e.

Therefore, f_o + f_e = 101 cm   …(i)

The magnifying power of the telescope in normal adjustment is: M = -f₀/f_e = -100

⇒ f₀ = 100f_e     ….(ii)

Solving equation …(i) and …(ii), we get f_o = 100 cm and f_e = 1 cm.