Question

A ray of light travelling in air falls on the surface of a glass slab at an angle of incidence 45°. Find the angle made by the refracted ray with the normal within the slab where refractive index for glass is 3/2.​

Answer 1

$by \: using \: snells \: law = > \\ \\ mew1 \: \times sin \: i = mew2 \times sin \: r \\ \\ 1 \times \frac{1}{ \sqrt{2} } = \frac{3}{2} \times sin \: r \\ sin \: r = \frac{ \sqrt{2} }{3} \\ r = {sin}^{ - 1} ( \frac{ \sqrt{2} }{3} ) \\ \\ \\ \\ hope \: it \: helps \: uh....$

Answer 2

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf \theta_2=sin^{-1}\frac{2\sqrt{2}}{3} }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Angle of incidence ($\sf{\theta_1}$)=45°

• Refractive index of air ($\sf{\mu_1)=1}$

• Refractive index of glass ($\sf{\mu_{2})=\frac{3}{2}}$

$\large\underline\pink{\sf To\:Find: }$

• Angle made by refracted ray ($\sf{\theta_2})$=?

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According to Snell's Law

$\LARGE{♡}$$\Large{\boxed{\sf \mu_1sin\theta_1=\mu_2sin\theta_2}}$

On Putting value

$\large\implies{\sf 1×sin45°=\frac{3}{2}×sin\theta_2 }$

Here ,

$\sf{\theta_1=angle\:of\: incidence\:ray}$

$\sf{\theta_2=angle\:of \: refracted\:ray}$

$\large\implies{\sf 1×\frac{1}{\sqrt{2}}=\frac{3}{2}×sin\theta_2 }$

$\large\implies{\sf sin\theta_2=\frac{2\sqrt{2}}{3} }$

$\large\implies{\sf \theta_2=sin^{-1}\frac{2\sqrt{2}}{3} }$

$\Huge\red{♡}$$\large\red{\boxed{\sf \theta_2=sin^{-1}\frac{2\sqrt{2}}{3} }}$

Hence ,

Angle of Refracted ray is $\underline{\sf \theta_2=\frac{2\sqrt{2}}{3}}$