Question

7) We know that the frequency of the light remains constant. How do we account for the shift in wavelength of the refracted ray? 3) What happens when the index of the top layer is larger than the index of the bottom layer? Does Snell's Law still apply?

Answer 1

Answers :

Question (7) Answer :

As light travels from one medium to another medium of different refractive indices. The frequency of the light remains constant.

But, The speed of the light changes according to the refractive index of the material. If the refractive index of the material is n, Then the speed of the light in that medium is given by : v = c / n

Where, c is the speed of light in vacuum (c = 3 x 10⁸ m/s)

Which means the light will travel slower in denser medium.

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Now, Speed of an wave is given by : v = λ f

Where, λ is wavelength and f is the frequency of wave.

Thus, When the light travels from one medium to another medium - The shift in wavelength of the light occurs according to the shift in the speed of the light and frequency of the light remains constant.

Here, If frequency of light is constant : Then, Wavelength of the light will be directly proportional to the speed of the light. Thus, If the speed of light in given medium after refraction increases (When light travels from denser to rarer medium) - Then, the wavelength of the light also increases.

Similarly, If the speed of light in given medium after refraction decreases (When light travels from rarer to denser medium) - Then, the wavelength of the light also decrease

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Question (8) Answer :

When the index of refraction of the top layer is larger than the index of refraction of the bottom layer the ray of light traveling from bottom layer to top layer will bend towards the normal. And, The speed and the wavelength of the light in top layer will decreases.

And, Snell's law will still work for this scenario.

Here, n₂ > n₁

According to Snell's law : n₁ sinθ₁ = n₂ sinθ₂

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