Question

when a Ray is refracted from one medium to another the wavelength changes from 6000 Armstrong to 4000 Armstrong the critical angle of the interface will be​

Answer 1

Answer:Refractive Index n = sin i/sin r => n = wavelength(air)/wavelength(medium) = 6000/4000 = 1.5. Critical angle c occurs when i = 90°. n = sin 90°/ sin c => 1.5 = 1/sin c => c arcsin (1/1.5) = 41.81°.

Explanation:

Answer 2

The critical angle of the interface is $\sin^{-1}(\dfrac{2}{3})$

Explanation:

Given that,

Wavelength $\lambda_{1}=6000\ \AA$

Wavelength  $\lambda_{1}=4000\ \AA$

We need to calculate the refractive index

Using formula of refractive index

$_{1}\mu^{2}=\dfrac{\mu_{2}}{\mu_{1}}$

$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{v_{1}}{v_{2}}=\dfrac{\lambda_{1}}{\lambda_{2}}$

$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{\lambda_{1}}{\lambda_{2}}$

Put the value into the formula

$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{6000}{4000}$

$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{3}{2}$

We need to calculate the critical angle of the interface

Using formula of critical angle

$\mu_{2}\sin C=\mu_{1}\sin90$

$\sin C=\dfrac{\mu_{1}}{\mu_{2}}$

Put the value into the formula

$\sin C=\dfrac{2}{3}$

$C=\sin^{-1}(\dfrac{2}{3})$

Hence, The critical angle of the interface is $\sin^{-1}(\dfrac{2}{3})$

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Topic : critical angle

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