For a situation shown in figure, find the ref | vedclass

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Question

(c)

Clearly $r+i_c=90^{\circ}$

$r=90-i_c$

$\mu=\frac{\sin i}{\operatorname{sinr}}=\frac{\sin 60^{\circ}}{\sin \left(90-i_c\right)}$

$\mu=\

Vedclass · Accepted Answer

$1.732$

Vedclass · Answer

(c)

Clearly $r+i_c=90^{\circ}$

$r=90-i_c$

$\mu=\frac{\sin i}{\operatorname{sinr}}=\frac{\sin 60^{\circ}}{\sin \left(90-i_cight)}$

$\mu=\frac{\sqrt{3}}{2 \cos i_c}$

$\because \quad \mu=\frac{1}{\operatorname{sini} i_c}$

From eqns $(i)$ and $(ii)$

$\sin i_c=\frac{2 \operatorname{cosi} i_c}{\sqrt{2}} \Rightarrow 	an i_c=\frac{2}{\sqrt{3}}$

or $i_c \approx 49^{\circ} 11^{\prime} \approx 50^{\circ}$

$\Rightarrow \quad \mu=\frac{1}{\operatorname{sini}_c}=\frac{1}{\sin 50^{\circ}}=1.31$

For a situation shown in figure, find the refrective index of glass so that it will suffer total internal reflection at the vertical surface.

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