We have provided Physics Light chapter MCQs with Solutions for Class 10 CBSE & NTSE, to prepare the student for any eventuality and lays a strong foundation for higher level exams.
Introduction of Light
Light is a form of radiant energy that causes the sensation of sight in our eye. In fact, light is an indispensable sensation of sight without which we cannot explore and enjoy colourful beauty of nature.
When light reflected/scattered from objects enters into our eyes then its image is formed in our eyes. The sparkling of diamonds and pearls, shining colours of gems and shells, blue colour of the sky, the red colour of sunrise and sunset, the rainbow and twinkling of stars are some examples of the phenomenon in nature due to light.
Light also provides us means of communication. It is also used for transmitting signals through optical fibres and this is the fastest growing mode of communication. The fibre-optic cables consist of many glass fibres that transmit hundreds of telephone conversations over long distances.
The sun is a natural source of light and is responsible for life on the earth. Sunlight is used by plants to manufacture their food through photosynthesis. The man-made sources of light are electric lamps, oil-lamps, candles, discharge tubes, etc.
Multiple Choice Questions with Solutions of Light
1. Astigmatism can be corrected by using :
(A) Bifocal lens
(B) Cylindrical lens
(C) Concave lens
(D) Convex lens
Solution
Sol. Astigmatism can be corrected by using cylindrical lens.
2. Wavelength of light of frequency 100 Hz is :
(A) 2 × 106 m
(B) 3 × 106 m
(C) 4 × 106 m
(D) 5 × 106 m
Solution
Sol. \lambda = \frac{v}{f} = \frac{{3 \times {{10}^8}}}{{100}}= 3 × 106m
3. A thin lens has more focal length, then its power will be :
(A) More
(B) Less
(C) Zero
(D) Infinity
Solution
Sol. Power of len’s is given by, P = \frac{1}{f}So, as thin len’s has more focal length so its power will be less.
4. A stone lies at the bottom of a lake 8m deep. To an observer directly above the stone, it appears to be at a depth of only 6.0 m below the surface. Then the refractive index of water is :
(A) \frac{3}{4}
(B) \frac{4}{3}
(C) \frac{2}{3}
(D) \frac{3}{5}
Solution
Sol. We know,
apparent depth = \frac{{{\rm{actual}}\,\,{\rm{depth}}}}{\mu } \Rightarrow \,\,\,\mu = \frac{8}{6} = \frac{4}{3}
5. Monochromatic light is refracted from air into a medium of refractive index n. Then the ratio of wavelengths of the incident and refracted waves is :
(A) 1 : 1
(B) 1 : n
(C) n : 1
(D) n2 : 1
Solution
Sol. We know, \mu = \frac{c}{v} = \frac{{{\lambda _v}}}{{{\lambda _m}}} \Rightarrow \,\,\frac{{{\lambda _v}}}{{{\lambda _m}}} = n
6. Angle of deviation experienced by a light ray when incident on a prism of refracting angle A = 60° at an angle of incidence i = 60° and emerging at an angle e = 60° is :
(A) 0°
(B) 30°
(C) 60°
(D) 90°
Solution
Sol. We know angle of deviation is given as, \delta = i + e-A \Rightarrow \,\,\delta = 60^\circ
7. A ray of light enters from a denser medium into a rarer medium. The speed of light in rarer medium is twice that in denser medium. Then critical angle for TIR to take place is :
(A) 60°
(B) 45°
(C) 30°
(D) None of these
Solution
Sol. We know critical angle is given as,
c = {\sin ^{ - 1}}\left( {\frac{{{\mu _r}}}{{{\mu _d}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{c/{v_r}}}{{c/{v_d}}}} \right)\, = \,{\sin ^{ - 1}}\left( {\frac{{{v_d}}}{{{v_r}}}} \right)
\Rightarrow ï€ c = sin–1 \left( {\frac{1}{2}} \right) = 30°
8. An object is placed on the principal axis of a lens at a distance of 20 cm from it. If the power of the lens is –5 D, then the distance of image from the len’s will be :
(A) 10 cm
(B) 15 cm
(C) 5 cm
(D) 20 cm
Solution
Sol. f = \frac{1}{P}\, = \,\frac{1}{{ - 5}}\, = \,0.20\,m\, = \, - 20\,cmu = – 20 cm
Applying len’s formula,
\frac{1}{v}-\frac{1}{u} = \frac{1}{f} \Rightarrow  v = – 10 cm
9. For a concave mirror of focal length 10 cm to form twice magnified image, object distance from its pole is/are :
(A) – 5 cm
(B) – 15 cm
(C) – 10 cm
(D) Both (A) and (B)
Solution
Sol. Magnification is given by,
m = - \frac{v}{u}According to question,
m = ± 2 (as one image would be real and other would be virtual)
\Rightarrow  v = ± 2u
and f = – 10 cm
Thus, applying mirror formula,
\Rightarrow  \frac{1}{v} + \frac{1}{u} = \frac{1}{f}u = – 5 cm, – 15 cm
10. A convex lens of focal length f0 produces an real image \frac{1}{n} times than that of the size of the object. Then the distance of the object from the lens is :
(A) nf0
(B) \frac{{{f_0}}}{n}
(C) (n+1)f0Â Â Â Â Â
(D) (n – 1)f0
Solution
Sol. We know magnification is,
m = \frac{v}{u}\, = \, - \frac{1}{n} \Rightarrow  v = - \frac{u}{n}Applying len’s formula,
\frac{1}{v}-\frac{1}{u} = \frac{1}{f} \Rightarrow  u = –(n + 1)f0