Question

The frequency of light in air is $5 \times {10^{14}}Hz$. What will be the frequency of light, when it enters the water?
$
  A.2.5 \times {10^{14}}Hz \\
  B.5 \times {10^{14}}Hz \\
  C{.10^{15}}Hz \\
  D.2.5 \times {10^{12}}Hz \\
 $

Answer 1

Hint: When light ray enters a denser medium, the speed of light decreases. The refractive index of the body is equal to the ratio of velocity of light in air to the velocity of light in the medium.
$\mu = \dfrac{c}{V}$
c = speed of light in air = $3 \times {10^8}m{s^{ - 1}}$

Complete step-by-step answer:
The velocity of light is equal to the product of frequency and wavelength of the light.
$c = \upsilon \times \lambda $
Whenever light enters another medium, the speed of light changes. This change in speed is due to the change in wavelength and not due to the change in the frequency. This is because, when it reaches a new denser medium, the light has to travel a larger distance to make up for the constant time.
Hence, refractive index $\mu = \dfrac{c}{V} = \dfrac{{\upsilon {\lambda _1}}}{{\upsilon {\lambda _2}}}$
Here, we can see that even in air and water, the frequency of the light remains the same and only the wavelength changes.
So therefore, if the frequency of the light in air is $5 \times {10^{14}}Hz$, the frequency of light in water shall remain the same i.e. $5 \times {10^{14}}Hz$

Hence, the correct option is Option B.

Note: Sometimes, we could be confused by the formula for the refractive index in terms of velocity. Whether it is $\dfrac{c}{v}$ or $\dfrac{v}{c}$. Instead of getting confused by the formula and thus, ending up byhearting the formula, we can use a small validation check as explained:
Solve both of the fractions $\dfrac{c}{v}\& \dfrac{v}{c}$.
In one of the fractions, the answer will be lesser than 1 and the other answer will be more than 1.
Refractive index of air is 1. Hence, the minimum value that the refractive index can take, is 1. Hence, the fraction with the answer lesser than 1 is not an option.