Question 83: Waves Optics

Question

At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $	heta_{1C}$. The $\mathrm{n}_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_3$ materials is $	heta_{2 C}$. If $n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\sin 	heta_{2 C}-\sin 	heta_{1 C}=\frac{1}{2}$, then $	heta_{1 C}$ is :

Accepted Answer

Let the wave be incident from the denser medium onto the rarer one in each case.  Since $n_3>n_2>n_1$ the criticalÃ¢ÂÂangle sines are
Ã¢ÂÂ at the $n_2\kern1pt|\kern1ptn_1$ interface:
$ \sin	heta_{1C}=\frac{n_1}{n_2}, $
Ã¢ÂÂ at the $n_3\kern1pt|\kern1ptn_1$ interface:
$ \sin	heta_{2C}=\frac{n_1}{n_3}. $
WeÃ¢ÂÂre given
$ \sin	heta_{1C}-\sin	heta_{2C}=\frac12 \quad	ext{and}\quad \frac{n_2}{n_3}=\frac25. $
Hence
$ \frac{n_1}{n_2}-\frac{n_1}{n_3} =\;n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =\frac12, $
and with $n_2=	frac25\,n_3$ this becomes
$ n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =n_1\Bigl(\frac{5}{2n_3}-\frac1{n_3}\Bigr) =n_1\frac{3}{2n_3} =\frac12 \;\Longrightarrow\; n_3=3\,n_1,\quad n_2=\frac{2}{5}n_3=\frac{6}{5}n_1. $
Therefore
$ \sin	heta_{1C} =\frac{n_1}{n_2} =\frac{n_1}{	frac{6}{5}n_1} =\frac56, $
so
$ 	heta_{1C}=\sin^{-1}\!\Bigl(\frac56\Bigr). $
That corresponds to Option C.

Answer

$ \sin^{-1}\left( \frac{1}{6n_1} \right) $

Answer

$ \sin^{-1}\left( \frac{1}{3n_1} \right) $

Answer

$ \sin^{-1}\left( \frac{5}{6n_1} \right) $

Answer

$ \sin^{-1}\left( \frac{2}{3n_1} \right) $

Waves Optics

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