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CBSE Class 12 Physics 2022 Term 2 Outside Delhi Set 2 Solved Paper

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Question : 3 of 4
Marks: +1, -0
Monochromatic light of wavelength 600nm600 \text{nm} is incident from air on a water surface. The refractive index of water is 1.33. Find the (i) wavelength, (ii) frequency and (iii) speed, of reflected and refracted light.
33
Solution:  
Given : λ=600nm\lambda = 600 \text{nm}
μ=1.33\mu = 1.33
(i) In reflection, the ray will reflect back in the same medium as that of incident ray.
Hence, wavelength (λ)=600nm(\lambda) = 600 \text{nm}
frequency =v=cλ=3×108600×10−9= v = \frac{c}{\lambda} = \frac{3 \times 10^8}{600 \times 10^{-9}}
=0.5×1015Hz= 0.5 \times 10^{15} \text{Hz}
speed =3×108ms−1= 3 \times 10^8 \text{ms}^{-1}
(ii) In refraction, the speed and wavelength change while frequency remains same.
Hence, speed =v=cμ=3×1081.33= v = \frac{c}{\mu} = \frac{3 \times 10^8}{1.33}
=2.26×108ms−1= 2.26 \times 10^8 \text{ms}^{-1}
wavelength =λ=νv=2.26×1080.5×1015= \lambda = \frac{\nu}{v} = \frac{2.26 \times 10^8}{0.5 \times 10^{15}}
=4.52×10−7m= 4.52 \times 10^{-7} \text{m}
=452nm= 452 \text{nm}
frequency =v=0.5×1015Hz= v = 0.5 \times 10^{15} \text{Hz}
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