Question

If the wavelength of red light in air is $7500\,\,{A^o},$ then the frequency of light in air is …………
A. $7.5 \times {10^{10}}Hz$
B. $3 \times {10^{14}}Hz$
C. $4 \times {10^{14}}Hz$
D. $5 \times {10^{14}}Hz$

Answer 1

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

Complete step by step solution:
Velocity of wave $ = $wavelength $ \times $frequency
$v = f\lambda ....\left( i \right)$
The velocity red light is \[v = 3 \times {10^8}\]m/s and the wavelength of the red light is given $\lambda = 7500\,{A^0}$
$\lambda = 7500 \times {10^{ - 10}}m$
$\lambda = 7.5 \times {10^{ - 7}}m$
Now putting the values of v and x in equation (i) we get,
$v = f\lambda $
Or, $3 \times {10^8} = f \times 7.5 \times {10^{ - 7}}$
Or, $f = \dfrac{{3 \times {{10}^8}}}{{7.5 \times {{10}^{ - 7}}}}$
$f = 4 \times {10^{14}}Hz$

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 ${v_m} = \dfrac{c}{\mu }$ and similarly the wavelength in that medium is ${\lambda _m} = \dfrac{{{\lambda _0}}}{\mu }$ where ${\lambda _0}$ 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.