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A material is embedded between two glass plates. Refractive index $\mathrm{n}$ of the material varies with thickness as shown below. The maximum incident angle (in degrees) on the material for which beam will pass through the material is-
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Verified Answer
The correct answer is:
$53.1$
$\begin{array}{l}
1.5 \times \operatorname{sini}=1.2 \sin r \\
\sin r=\frac{1.5}{1.2} \operatorname{sini}
\end{array}$
$\mathrm{T} / \mathrm{R}$ should not take place
$\begin{array}{l}
\therefore \operatorname{sinr} < 1 \\
\frac{1.5}{1.2} \operatorname{sini} < 1 \\
\operatorname{sini} < \frac{12}{15} \\
\sin i < 0.8 \\
\sin 45=\frac{1}{\sqrt{2}}=0.707 \\
\mathrm{i}_{\max }>45
\end{array}$
1.5 \times \operatorname{sini}=1.2 \sin r \\
\sin r=\frac{1.5}{1.2} \operatorname{sini}
\end{array}$
$\mathrm{T} / \mathrm{R}$ should not take place
$\begin{array}{l}
\therefore \operatorname{sinr} < 1 \\
\frac{1.5}{1.2} \operatorname{sini} < 1 \\
\operatorname{sini} < \frac{12}{15} \\
\sin i < 0.8 \\
\sin 45=\frac{1}{\sqrt{2}}=0.707 \\
\mathrm{i}_{\max }>45
\end{array}$
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