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In a double-slit experiment the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \mathrm{~m}$ away. The wavelength of light used is $600 \mathrm{~nm}$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be $4 / 3$.
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Angular width, $\beta_1=0.2^{\circ}, \lambda_1=600 \times 10^{-9} \mathrm{~m}, \mathrm{n}=\frac{4}{3}, \lambda_2=?$
As, $\beta_1=\frac{\lambda_1}{\mathrm{~d}} ; \beta_2=\frac{\lambda_2}{\mathrm{~d}}$
$$
\begin{aligned}
&\because \mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{n}=\frac{\mathrm{v} \lambda_1}{\mathrm{v} \lambda_2} \Rightarrow \mathrm{n}=\frac{\lambda_1}{\lambda_2} \\
&\therefore \quad \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda 1}=\frac{1}{\mathrm{n}}=\frac{3}{4} \\
&\therefore \quad \beta_2=\frac{3}{4} \times \beta_1=\frac{3 \times 0.2^{\circ}}{4}=0.15^{\circ}
\end{aligned}
$$
As, $\beta_1=\frac{\lambda_1}{\mathrm{~d}} ; \beta_2=\frac{\lambda_2}{\mathrm{~d}}$
$$
\begin{aligned}
&\because \mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{n}=\frac{\mathrm{v} \lambda_1}{\mathrm{v} \lambda_2} \Rightarrow \mathrm{n}=\frac{\lambda_1}{\lambda_2} \\
&\therefore \quad \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda 1}=\frac{1}{\mathrm{n}}=\frac{3}{4} \\
&\therefore \quad \beta_2=\frac{3}{4} \times \beta_1=\frac{3 \times 0.2^{\circ}}{4}=0.15^{\circ}
\end{aligned}
$$
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