Question

Question no. 19 plese solve

Q19. A ray of light passes through a plane boundary separating two media, 'I' and '2' of refractive indices n₁ = $\frac{3}{2}\mathrm{and}{\mathrm{n}}_2=\frac{4}{3}$.

(i) If the ray passes from medium I to medium 2 at an angle of incidence 30°. What is the angle of refraction at same angle of incidence.

(ii) If the ray passes from medium 2 and medium 1, what is the angle of refraction?
[Ans . (i) sin^–1$\frac{9}{16}$ (ii) sin^–1$\frac{4}{9}$]

Answer 1

Dear Student

$$\begin{gathered} U\sin gSnell\text{'}slaw \\ \frac{\sin i}{\sin r}=\frac{n_2}{n_1}(lightgoingfrommedium1tomedium2) \\ \left(i\right)\frac{\sin {30}^0}{\sin r}=\frac{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \sin r=0.5\times \frac{9}{8}=\frac{9}{16} \\ r={\sin }^{-1}\frac{9}{16} \\ \left(ii\right)\frac{\sin {30}^0}{\sin r}=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}(ightgoingfrommedium2tomedium1) \\ \sin r=0.5\times \frac{8}{9}=\frac{4}{9} \\ r={\sin }^{-1}\frac{4}{9} \end{gathered}$$

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