written 2.0 years ago by
prajapatijaimin
• 3.2k
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Solution:
We know that the numerical aperture is,
Since,
$ \begin{aligned} & & N A & =\sqrt{n_1^2-n_2^2} \\\\ & & n_1 & =1.5, n_2=1.45 \\\\ \therefore & & N A & =\sqrt{\left.(1.5)^2-1.45\right)^2}=0.38\\ \end{aligned} $
Also, the acceptance angle $\alpha=\sin ^{-1}$ NA
$\therefore \quad \alpha=\sin ^{-1}(0.38)=22.8^{\circ} \quad$
The critical angle,
$ \phi_c=\sin ^{-1}\left(\frac{n_2}{n_1}\right)=\sin ^{-1}\left(\frac{1.45}{1.5}\right)=75.2^{\circ}\\ $
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