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Q. Question
As shown in figure light $P$ enters to slab at an angle $60^\circ $ with normal and inside slab $Q$ makes total internal reflection. Find minimum refractive index of slab.

NTA AbhyasNTA Abhyas 2022

Solution:


At point $A$
$1sin60^\circ =\mu sinr\Rightarrow sinr=\frac{\sqrt{3}}{2 \mu }$
At point $B$
$sin\left(90 ^\circ - r\right)>sinC$
$\Rightarrow cosr>\frac{1}{\mu }\Rightarrow \sqrt{1 - \frac{3}{4 \mu ^{2}}}>\frac{1}{\mu }$
$\Rightarrow 1-\frac{3}{4 \mu ^{2}}>\frac{1}{\mu ^{2}}\Rightarrow 1>\frac{7}{4 \mu ^{2}}$
$\Rightarrow \mu ^{2}>\frac{7}{4}\Rightarrow \mu >1.32$