# Physics Questions and Answers – Ray Optics – Refraction at Spherical Surfaces and by Lenses

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This set of Physics Written Test Questions and Answers for Class 12 focuses on “Ray Optics – Refraction at Spherical Surfaces and by Lenses”.

1. How many types of spherical refracting surfaces are there?
a) 2
b) 3
c) 4
d) 5

Explanation: The portion of a refracting medium, whose curved surface forms the part of a sphere, is known as a spherical refracting surface. Spherical refracting surfaces are of two types, namely, convex refracting spherical surface and concave refracting spherical surface.

2. Which among the following is a portion of a transparent refracting medium bound by one spherical surface and the other plane surface?
a) Concave mirror
b) Plane mirror
c) Lens
d)Prism

Explanation: Lens is a portion of a transparent refracting medium bound by two spherical surfaces or one spherical surface and the other plane surface. Lenses are divided into two classes, namely, convex lens or converging lens, and concave lens or diverging lens.

3. X is thicker in the middle than at the edges, whereas, Y is thicker at the edges than in the middle. Identify ‘X’ and ‘Y’.
a) X = concave lens; Y = convex lens
b) X = convex lens; Y = concave lens
c) X = plane lens; Y = convex lens
d) X = concave lens; Y = plane lens

Explanation: When the lens is thicker in the middle than at the edges, then it is convex lens. When the lens is thicker at the edges than in the middle, then it is concave lens. Therefore, X = convex lens, and Y = concave lens.
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4. Sign conventions for spherical refracting surface are the same as those for spherical mirrors.
a) True
b) False

Explanation: Yes, this is a true statement. Sign conventions for spherical refracting surface are the same as those for spherical mirrors. The only difference is that instead of the pole of the mirror, we use the optical center of a lens.

5. Identify the Lens Maker’s Formula.
a) f=(μ-1)$$( \frac {1}{R_1} – \frac {1}{R_2} )$$
b) $$\frac {1}{f}$$=(μ-1)$$( \frac {1}{R_1} – \frac {1}{R_2} )$$
c) $$\frac {1}{f}$$=(μ+1)$$( \frac {1}{R_1} – \frac {1}{R_2} )$$
d) $$\frac {1}{f}$$=(μ-1)$$( \frac {1}{R_1} + \frac {1}{R_2} )$$

Explanation: The lens maker formula may be defined as the formula which gives the relationship for calculating the focal length in terms of the radii of the two curvatures of the lens. It is given as:
$$\frac {1}{f}$$=(μ-1)$$( \frac {1}{R_1} – \frac {1}{R_2} )$$
Where R1 and R2 are radii of curvature of the two surfaces of the lens and μ is the refractive index of the material of the lens w.r.t. medium in which the lens is placed.

6. According to the thin lens formula, which one of the following is true regarding the focal length of the lens?
a) f is positive for concave lens
b) f is negative for convex lens
c) f is positive for a diverging lens
d) f is negative for concave lens

Explanation: Thin lens formula is given as:
$$\frac {1}{f}=\frac {1}{v}-\frac {1}{u}$$
Where u = distance of the object from the optical center of the lens; v = distance of the image from the optical center of the lens; f = focal length of a lens. According to the thin lens formula, f is positive for converging or convex lens and f is negative for a diverging or concave lens.

7. What is the SI unit of power of a lens?
a) Watts
b) Unit less
c) Diopter
d) Joule

Explanation: The power of a lens is defined as the reciprocal of its focal length in meters. The expression is given as:
P = $$\frac {1}{f}$$
Where P is the power and f is the focal length in meters. Lens surface power can be found with the index of refraction and the radius of curvature. The SI unit of power of a lens is diopter.

8. The power of a convex lens is negative.
a) True
b) False

Explanation: No, this is a false statement. The power of a convex lens is positive as a convex lens has a positive focal length, while the power of a concave lens is negative as a concave lens has a negative focal length.

9. Calculate the focal length of a biconvex lens if the radii of its surfaces are 50 cm and 20 cm, and index of refraction of the lens glass = 1.2.
a) 0.014 cm
b) 0.715 cm
c) 0.14 cm
d) 71.5 cm

Explanation: Given: R1 = 50 cm; R2 = 20 cm; μ2 = 1.2
Required equation ➔ $$\frac {1}{f}$$=(μ-1)$$( \frac {1}{R_1} – \frac {1}{R_2} )$$
$$\frac {1}{f}=( \frac {\mu 2}{\mu 1} -1 ) ( \frac {1}{R_1} – \frac {1}{R_2} )$$
$$\frac {1}{f}=( \frac {1.2}{1} -1 ) ( \frac {1}{50} – \frac {1}{-20} )$$
$$\frac {1}{f}$$=(0.2)$$( \frac {1}{50} + \frac {1}{20} )$$
$$\frac {1}{f}$$=(0.2)$$( \frac {20}{1000} + \frac {50}{1000} )$$
$$\frac {1}{f}$$=(0.2)$$( \frac {7}{100} )$$
$$\frac {1}{f}$$=0.014
f=+71.5 cm

10. A lens has a focal length of 10 cm. Where the object should be placed if the image is to be 40 cm in the positive direction from the lens?
a) 12 cm
b) 40 cm
c) 13 cm
d) 0.075

Explanation: Given: f = + 20 cm; v = + 40 cm
The required equation ➔
$$\frac {1}{f}=\frac {1}{v}-\frac {1}{u}$$
$$\frac {1}{10}=\frac {1}{10}-\frac {1}{u}$$
$$\frac {1}{u}=\frac {1}{10}-\frac {1}{40} = \frac {40-10}{400}$$
$$\frac {1}{u}=\frac {30}{400}=\frac {3}{40}$$
u=$$\frac {40}{3}$$=13.3 cm

11. Find the magnification of the lens if the focal length of the lens is 10 cm and the size of the image is -30 cm.
a) 2
b) 3
c) 4
d) 5

Explanation: Given: f = +10 cm; v = -30 cm
Required equations ➔
$$\frac {1}{f}=\frac {1}{v}-\frac {1}{u}$$
m=$$\frac {size \, of \, the \, image}{size \, of \, the \, object}=\frac {v}{u}$$
$$\frac {1}{u}=\frac {1}{v}-\frac {1}{f}=\frac {1}{-30}-\frac {1}{10} = \frac {-(10+30)}{300} = \frac {-40}{300}= \frac {-2}{15}$$
u=$$\frac {-15}{2}$$ = -7.5 cm
m=$$\frac {v}{u}= – \frac {30}{-7.5}=\frac {-300}{-75}$$= +4
Therefore, the object should be placed at a distance of 7.5 cm from the lens to get the image at a distance of 30 cm from the lens. It is four times enlarged and is erect.

12. A lens has a power of +3 diopters in air. What will be the power of the lens if it is completely immersed in water? Given, μg = $$\frac {3}{2}$$ and μw = $$\frac {4}{3}$$.
a) 5 D
b) 3 D
c) 1 D
d) $$\frac {3}{4}$$ D

Explanation: Pw=$$\frac {\mu_w}{f_w}$$=μw$$[ \frac {\mu_g}{\mu_w} -1 ] [ \frac {1}{R1} – \frac {1}{R2} ]$$……………………………..1
Pa=$$\frac {1}{f_a}$$=[μg-1]$$[ \frac {1}{R1} – \frac {1}{R2} ]$$……………………………………2
Dividing 1 and 2
We get ➔ $$\frac {P_w}{P_a} = \frac {\mu_g-\mu_w}{μ_g-1}$$
Pw=$$\frac {\frac {3}{2}-\frac {4}{3}}{\frac {3}{2}-1}$$ × 3
Pw=$$\frac {\frac {1}{6}}{\frac {1}{2}}$$ × 3
Pw=$$\frac {2\times 3}{6}$$=1 D
Therefore, the power of the lens gets altered inside water.

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