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A rectangular glass slab ABCD, of refractive index \mathrm{n}_{1}, is immersed in water of refractive index \mathrm{n_{2}\left(n_{1}>\right.\mathrm{n}_{2} )}.A ray of light is incident on the surface AB of the slab as shown. The maximum value of the angle of incidence \mathrm{\alpha_{\max }}, such that the ray emerges only from the surface CD is given by

Option: 1

\sin ^{-1}\left[\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} \cos \left(\sin ^{-1} \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)\right]


Option: 2

\sin ^{-1}\left[n_{1} \cos \left(\sin ^{-1} \frac{1}{n_{2}}\right)\right]


Option: 3

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


Option: 4

\sin ^{-1}\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)


Answers (1)

best_answer

Here, total internal reflection must take place at AD Applying Snell’s law at P we obtain,


\frac{\sin \alpha_{m}}{\sin r}=\frac{n_{1}}{n_{2}}
\text{Geometrically} \: r+\theta_{c}=\pi / 2
where critical angle \theta_{c}=\sin ^{-1}\left(n_{2} / n_{1}\right)

\Rightarrow \alpha_{m}=\sin ^{-1}\left[(\sin r)\left(n_{1} / n_{2}\right)\right]
\text{Putting} \sin r=\sin \left(\pi / 2-\theta_{c}\right)=\cos \theta_{c}=\cos \left[\sin ^{-1}\left(n_{2} / n_{1}\right)\right] in the above equation we obtain
\alpha_{m}=\sin ^{-1}\left[\frac{n_{1}}{n_{2}} \cos \left(\sin ^{-1} \frac{n_{2}}{n_{1}}\right)\right].

 

Posted by

Gautam harsolia

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