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In Young's double slit experiment the two slits are illuminated by light of wavelenght $5890 Å$ and the distance between the fringes obtained on the screen is $0.2^{\circ} .$ If the whole apparatus is immersed in water then the angular fringe width will be, if the refractive index of water is $4 / 3$.
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The correct answer is:
$0.15^{\circ}$
$$
\begin{array}{l}
\mathrm{w}_{\mathrm{a}}=\lambda / \mathrm{d} \Rightarrow \mathrm{w}_{\mathrm{a}} \alpha \lambda \\
\frac{\left(\mathrm{w}_{\mathrm{a}}\right)_{\text {water }}}{\mathrm{w}_{\mathrm{a}}}=\frac{\lambda_{\text {water }}}{\lambda}=\frac{\lambda}{\mu_{\mathrm{water}} \lambda} \\
\left(\mathrm{w}_{\mathrm{a}}\right)_{\text {water }}=\frac{2 \times 3}{4}=0.15^{\circ}
\end{array}
$$
\begin{array}{l}
\mathrm{w}_{\mathrm{a}}=\lambda / \mathrm{d} \Rightarrow \mathrm{w}_{\mathrm{a}} \alpha \lambda \\
\frac{\left(\mathrm{w}_{\mathrm{a}}\right)_{\text {water }}}{\mathrm{w}_{\mathrm{a}}}=\frac{\lambda_{\text {water }}}{\lambda}=\frac{\lambda}{\mu_{\mathrm{water}} \lambda} \\
\left(\mathrm{w}_{\mathrm{a}}\right)_{\text {water }}=\frac{2 \times 3}{4}=0.15^{\circ}
\end{array}
$$
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