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A monochromatic light wave with wavelength $λ_{1}$ and frequency $v_{1}$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength $λ_{2}$ and frequency $v_{2}$ of the refracted wave are :

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$λ_{2} = \sqrt{2} λ_{1}, v_{2} = v_{1}$

$λ_{2} = λ_{1}, v_{2} = \sqrt{2} v_{1}$

$λ_{2} = \frac{1}{\sqrt{2}} λ_{1}, v_{2} = v_{1}$

$λ_{2} = λ_{1}, v_{2} = \frac{1}{\sqrt{2}} v_{1}$

answer is C.

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Detailed Solution

We know that frequencies do not change on changing medium.

$\Rightarrow v_{1} = v_{2}$

$v = frequency$

By Snell’s law

$μ_{1} \sin i = μ_{2} \sin r$

$μ_{1} \times \frac{1}{\sqrt{2}} = μ_{2} \times \frac{1}{2}$

$But μ = \frac{C}{V}$

$V = velocity$

$\frac{C}{V_{1}} \times \frac{1}{\sqrt{2}} = \frac{C}{V_{2}} \frac{1}{2}$

$V_{1} = \sqrt{2} V_{2}$

$v_{1} λ_{1} = \sqrt{2} v_{2} λ_{2}$

$λ = wavelength$

$λ_{1} = \sqrt{2} λ_{2}$

$λ_{2} = \frac{1}{\sqrt{2}} λ_{1}$

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