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Light travels from an optically denser medium 'A' into the optically rarer medium ${ }^{\prime} \mathrm{B}^{\prime}$ with speeds $1 \cdot 8 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and $2-7 \times 10^{8} \mathrm{~m} / \mathrm{s}$ respectively. The critical angle
between them is $\left(\mu_{1}\right.$ and $\mu_{2}$ are the refractive indices of media $\mathrm{A}$ and $\mathrm{B}$
respectively.)
Options:
between them is $\left(\mu_{1}\right.$ and $\mu_{2}$ are the refractive indices of media $\mathrm{A}$ and $\mathrm{B}$
respectively.)
Solution:
1560 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1}\left(\frac{2}{3}\right)$
$\frac{\mu_{\mathrm{A}}}{\mu_{\mathrm{B}}}={ }_{\mathrm{B}} \mu_{\mathrm{A}}=\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{V}_{\mathrm{A}}}=\frac{2.7 \times 10^{8}}{1.8 \times 10^{8}}=\frac{3}{2}$
${ }_{\mathrm{B}} \mu_{\mathrm{A}}=\frac{1}{\sin \mathrm{C}}$
$\therefore \frac{3}{2}=\frac{1}{\sin \mathrm{C}}$
$\therefore \sin \mathrm{C}=\frac{2}{3}$
or $\mathrm{C}=\sin ^{-1} \frac{2}{3}$
${ }_{\mathrm{B}} \mu_{\mathrm{A}}=\frac{1}{\sin \mathrm{C}}$
$\therefore \frac{3}{2}=\frac{1}{\sin \mathrm{C}}$
$\therefore \sin \mathrm{C}=\frac{2}{3}$
or $\mathrm{C}=\sin ^{-1} \frac{2}{3}$
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