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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 9 of 35
Marks: +1, -0
A beam of light travels from air into a medium. Its speed and wavelength in the medium are 1.5×1081.5 \times 10^{8} ms1\mathrm{ms}^{-1} and 230 nm230\ \mathrm{nm} respectively. The wavelength of light in air will be
In the medium
c=nλc = n \lambda
or, 1.5×108=n×230×1091.5 \times 10^{8} = n \times 230 \times 10^{-9}
or, n=1.5×108230×109n = \frac{1.5 \times 10^{8}}{230 \times 10^{-9}}
n=1.5230×1017 Hz\therefore n = \frac{1.5}{230} \times 10^{17}\ \mathrm{Hz}
In air,
Frequency remains unchanged. So,
n=1.5230×1017 Hzn = \frac{1.5}{230} \times 10^{17}\ \mathrm{Hz}
v=nλv' = n \lambda'
 or, 3×108=1.5230×1017×λ\text{ or, } 3 \times 10^{8} = \frac{1.5}{230} \times 10^{17} \times \lambda'
λ=3×2301.5×109\therefore \lambda' = 3 \times \frac{230}{1.5} \times 10^{-9}
=460 nm= 460\ \mathrm{nm}
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