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Question: If the wavelength of red light in air is \(7500\,\,{A^o},\) then the frequency of light in air is ……...

If the wavelength of red light in air is 7500Ao,7500\,\,{A^o}, then the frequency of light in air is …………
A. 7.5×1010Hz7.5 \times {10^{10}}Hz
B. 3×1014Hz3 \times {10^{14}}Hz
C. 4×1014Hz4 \times {10^{14}}Hz
D. 5×1014Hz5 \times {10^{14}}Hz

Explanation

Solution

A wave travels λ\lambda distance in one time period T. so the average velocity of the wave is given by v=λTv = \dfrac{\lambda }{T}. Also the frequency f=1Tf = \dfrac{1}{T}so v=fλv = f\lambda .

Complete step by step solution:
Velocity of wave ==wavelength ×\timesfrequency
v=fλ....(i)v = f\lambda ....\left( i \right)
The velocity red light is v=3×108v = 3 \times {10^8}m/s and the wavelength of the red light is given λ=7500A0\lambda = 7500\,{A^0}
λ=7500×1010m\lambda = 7500 \times {10^{ - 10}}m
λ=7.5×107m\lambda = 7.5 \times {10^{ - 7}}m
Now putting the values of v and x in equation (i) we get,
v=fλv = f\lambda
Or, 3×108=f×7.5×1073 \times {10^8} = f \times 7.5 \times {10^{ - 7}}
Or, f=3×1087.5×107f = \dfrac{{3 \times {{10}^8}}}{{7.5 \times {{10}^{ - 7}}}}
f=4×1014Hzf = 4 \times {10^{14}}Hz

Thus, option C is correct.

Additional Information: The refractive index can be seen as the factor by which the speed and the wavelength of the light wave are reduced with respect to their vacuum/ air values. The speed of light wave in a medium is vm=cμ{v_m} = \dfrac{c}{\mu } and similarly the wavelength in that medium is λm=λ0μ{\lambda _m} = \dfrac{{{\lambda _0}}}{\mu } where λ0{\lambda _0} is the wavelength of that light in vacuum/air.

Note: The frequency of light waves means the number of crests that pass through a fixed point in the medium in unit time. So the frequency depends on the source but not medium.