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A transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$ It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.

The incident angle $\Theta$ for which the light ray grazes along the wall of the rod is

Option 1)
$\sin ^{-1}\left ( \frac{1}{2} \right )$

Option 2)
$\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$

Option 3)
$\sin ^{-1}\left ( \frac{2}{\sqrt{3}} \right )$

Option 4)
$\sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$

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.

Applying Snell's law at Q

$\eta=\frac{sin90^{o}}{sin(90-\alpha)}=\frac{1}{cos\alpha}$

$cos\alpha=\frac{1}{n}$

$sin\alpha=\sqrt{1-cos^{2}\alpha}=\sqrt{1-\frac{1}{n^{2}}}=\sqrt{\frac{n^{2}-1}{n^{2}}}$ (i)

Snell's law at P

$\eta=\frac{sin\theta}{sin\alpha}=sin\theta=nsin\alpha=\sqrt{n^{2}-1}$

$sin\theta=\sqrt{\left ( \frac{2}{\sqrt{3}} \right )^{2}-1}=\sqrt{\frac{4}{3}-1}=\frac{1}{\sqrt{3}}$

$\theta=sin^{-1}\left(\frac{1}{\sqrt{3}} \right )$

Correct option is 4.

Option 1)

$\sin ^{-1}\left ( \frac{1}{2} \right )$

This is an incorrect option.

Option 2)

$\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$

This is an incorrect option.

Option 3)

$\sin ^{-1}\left ( \frac{2}{\sqrt{3}} \right )$

This is an incorrect option.

Option 4)

$\sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$

This is the correct option.

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