Question
Question: A light ray of frequency \(\upsilon \)and wavelength \(\lambda \)enter a liquid of refractive index ...
A light ray of frequency υand wavelength λenter a liquid of refractive index 3/2. The ray travels in the liquid with:
A. Frequency υ and wavelength (32)λ
B. Frequency υ and wavelength (23)λ
C. Frequency υ and wavelength λ
D. Frequency (23)υ and wavelength λ
Solution
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.
μ=Vc
C – speed of light in air = 3×108ms−1
Complete step-by-step answer:
The velocity of light is equal to the product of frequency and wavelength of the light.
c=υ×λ
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, μ=Vc=υλ2υλ1
∴μ=λ2λ1 λ1λ2=μ1
Given, μ=23
Substituting, we get:
λ1λ2=32 λ2=32λ1
Hence, the new wavelength is 32λ
The correct option is Option A.
Note: Sometimes, we could be confused by the formula for the refractive index in terms of velocity.
Whether it is vc or cv. 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 vc&cv.
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.