Question 87: Ray Optics

Question

Consider light travelling from a medium $A$ to medium $B$ separated by a plane interface. If the light undergoes total internal reflection during its travel from medium $A$ to $B$ and the speed of light in media $A$ and $B$ are $2.4 	imes 10^8 \mathrm{~m} / \mathrm{s}$ and $2.7 	imes 10^8 \mathrm{~m} / \mathrm{s}$, respectively, then the value of critical angle is :

Accepted Answer

For total internal reflection to occur, light must travel from a denser medium (lower speed) to a rarer medium (higher speed).
The speed of light in medium A (Denser) is $\mathrm{v}_{\mathrm{A}}=2.4 	imes 10^8 \mathrm{~m} / \mathrm{s}$
The speed of light in medium $B$ (rarer) is $v_B=2.7 	imes 10^8 \mathrm{~m} / \mathrm{s}$
Since $\mathrm{v}_{\mathrm{A}}<\mathrm{v}_{\mathrm{B}}$, medium A is indeed optically denser than medium B , satisfying the condition for TIR.
The critical angle is determined by Snell's Law at the boundary where the angle of refraction is $90^{\circ}$ :
$$ n_A \sin \left(	heta_cight)=n_B \sin \left(90^{\circ}ight) \Rightarrow \sin \left(	heta_cight)=\frac{n_B}{n_A} $$
Since the refractive index n is inversely proportional to the speed of light $\left(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}ight)$,
$$ \frac{n_B}{n_A}=\frac{c / v_B}{c / v_A}=\frac{v_A}{v_B} $$
So,
$$ \sin \left(	heta_cight)=\frac{v_A}{v_B} $$
$$\Rightarrow $$ $\sin \left(	heta_Cight)=\frac{2.4 	imes 10^8}{2.7 	imes 10^8}=\frac{8}{9}$
$$ 	an \left(	heta_cight)=\frac{8}{\sqrt{9^2-8^2}}=\frac{8}{17} \Rightarrow 	heta_c=	an ^{-1}\left(\frac{8}{\sqrt{17}}ight) $$
Therefore, the critical angle is $	an ^{-1}\left(\frac{8}{\sqrt{17}}ight)$.
Hence, the correct option is (D).

Answer

$\cos ^{-1}\left(\frac{8}{9}\right)$

Answer

$\sin ^{-1}\left(\frac{9}{8}\right)$

Answer

$\cot ^{-1}\left(\frac{3}{\sqrt{13}}\right)$

Answer

$	an ^{-1}\left(\frac{8}{\sqrt{17}}ight)$

Ray Optics

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