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Answer please! - Optics - JEE Main-2

A transparent cube of side d, made of a material of refractive index $\mu _{2}$ , is immersed in a liquid of refractive index $\mu _{1}(\mu _{1}<\mu _{2})$ . A ray is incident on the face AB at an angle $\theta$ (shown in the figure). Total internal reflection takes place at point E on the face BC.

Then $\theta$ must satisfy:

Option 1)
$\Theta <sin^{-1}\frac{\mu _{1}}{\mu _{2}}$
Option 2)
$\Theta >sin^{-1}\sqrt{\frac{\mu ^{2}_{2}}{\mu ^{2}_{1}}-1}$
Option 3)
$\Theta <sin^{-1}\sqrt{\frac{\mu ^{2}_{2}}{\mu ^{2}_{1}}-1}$
Option 4)
$\Theta >sin^{-1}\frac{\mu _{1}}{\mu _{2}}$

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S solutionqc

$\mu_{1}\sin \Theta =\mu_{2}\sin r_{1}$

$90-r_{1}>c$

$sin(90-r_{1})>sin c$

$cosr_1>\frac{\mu_1}{\mu_2}$

$\sqrt{1-\left ( \frac{\mu _{1}\sin \Theta }{\mu _{2}} \right )^{2}}> \frac{\mu _{1}}{\mu _{2}}$

$1-\frac{\mu _{1}^{2}}{\mu ^{2}_{2}}\sin ^{2}\Theta > \frac{\mu _{1}^{2}}{\mu ^{2}_{2}}$

$\frac{\mu ^{2}_{2}-\mu _{1}^{2}}{\mu ^{2}_{1}}> \sin ^{2}\Theta$

$\Theta < \sin ^{-1}\sqrt{\frac{\mu_2 ^{^{2}}}{\mu ^{2}_{1}}-1}$

Option 1)

$\Theta <sin^{-1}\frac{\mu _{1}}{\mu _{2}}$

Option 2)

$\Theta >sin^{-1}\sqrt{\frac{\mu ^{2}_{2}}{\mu ^{2}_{1}}-1}$

Option 3)

$\Theta <sin^{-1}\sqrt{\frac{\mu ^{2}_{2}}{\mu ^{2}_{1}}-1}$

Option 4)

$\Theta >sin^{-1}\frac{\mu _{1}}{\mu _{2}}$

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