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A double slit experiment is immersed in water of refractive index 1.33. The slit separation is $1 \mathrm{~mm}$, distance between slit and screen is $1.33 \mathrm{~m}$ The slits are illuminated by a light of wavelength $6300 Å$. The fringe width is
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Verified Answer
The correct answer is:
$6.3 \times 10^{-4} \mathrm{~m}$
$$
\begin{aligned}
& \lambda_{\text {liquid }}=\frac{\lambda_{\text {sir }}}{\mu} \\
& \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33}
\end{aligned}
$$
Fringe width,
$$
\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}
$$
\begin{aligned}
& \lambda_{\text {liquid }}=\frac{\lambda_{\text {sir }}}{\mu} \\
& \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33}
\end{aligned}
$$
Fringe width,
$$
\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}
$$
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