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Question: A monochromatic light of frequency $5 \times 10^{14}$ Hz travelling through air, is incident on a me...

A monochromatic light of frequency 5×10145 \times 10^{14} Hz travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

A

500 nm

B

400 nm

C

600 nm

D

300 nm

Answer

300 nm

Explanation

Solution

The frequency ff remains constant in all media. The speed of light in the medium is:

v=cn=3×108 m/s2=1.5×108 m/s.v = \frac{c}{n} = \frac{3 \times 10^8 \, \text{m/s}}{2} = 1.5 \times 10^8 \, \text{m/s}.

Now, the wavelength in the medium is:

λ=vf=1.5×1085×1014=3×10−7 m=300 nm.\lambda = \frac{v}{f} = \frac{1.5 \times 10^8}{5 \times 10^{14}} = 3 \times 10^{-7} \, \text{m} = 300 \, \text{nm}.