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Let the $x-z$ plane be the boundary between two transparent media. Medium 1 in $z\geq 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z< 0$ has a refractive index of $\sqrt{3}$ . . A ray of light in medium 1 given by the vector $\vec{A}= 6\sqrt{3}\: \hat{i}+8\sqrt{3}\: \hat{j}-10\: \hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is

Option 1)
$30^{\circ}$

Option 2)
$45^{\circ}$

Option 3)
$60^{\circ}$

Option 4)
$75^{\circ}$

Answers (1)

best_answer

As we learnt in

Relation between angle of incidence and angle of refaction -

$\mu _{1}\sin i= \mu _{2}\sin r$

- wherein

$\mu _{1}=$ refractive index of medium of incidence.

$\mu _{2}=$ refractive index of medium where rays is refracted.

$i=$ angle of incidence.

$r=$ angle of refraction.

$\vec{A}=6\sqrt{3}\ \widehat{i}+8\sqrt{3}\ \widehat{j}-10\widehat{k}$

$cos \widehat{i}=\frac{10}{(6\sqrt{3})^{2}+(8\sqrt{3})^{2}+(-10)^{2}}=\frac{10}{20}$

$cos i=\frac{1}{2}=i=cos^{-1}(1/2)$

$i=60^{o}$

From Snell's law

$\mu_{1}sini=\mu_{2}sinr$

$\sqrt{2}sin60^{o}=\sqrt{3}sinr$

$\Rightarrow\ \; r = 45^{o}$

Correct option is 2.

Option 1)

$30^{\circ}$

This is an incorrect option.

Option 2)

$45^{\circ}$

This is the correct option.

Option 3)

$60^{\circ}$

This is an incorrect option.

Option 4)

$75^{\circ}$

This is an incorrect option.

Posted by

prateek

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