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A double slit experiment is immersed in water of refractive index 1.33. The slit separations $1 \mathrm{~mm}$ and the 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:
$6.3 \times 10^{-4} \mathrm{~m}$
$$
\begin{aligned}
& \mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{D}=1.33 \mathrm{~m} \\
& \lambda=6300 Å=6.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
$\lambda_{\mathrm{w}}=$ wavelength in water $=\frac{6.3 \times 10^{-7}}{1.33} \mathrm{~m}$
Fringe width,
$$
\mathrm{X}=\frac{\lambda_{\mathrm{w}} \mathrm{D}}{\mathrm{d}}=\frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}=6.3 \times 10^{-4} \mathrm{~m}
$$
\begin{aligned}
& \mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{D}=1.33 \mathrm{~m} \\
& \lambda=6300 Å=6.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
$\lambda_{\mathrm{w}}=$ wavelength in water $=\frac{6.3 \times 10^{-7}}{1.33} \mathrm{~m}$
Fringe width,
$$
\mathrm{X}=\frac{\lambda_{\mathrm{w}} \mathrm{D}}{\mathrm{d}}=\frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}=6.3 \times 10^{-4} \mathrm{~m}
$$
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