A double slit experiment is immersed in water of refractive index 1.33. The slit separations $1…

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
  1. $4.9 \times 10^{-4} \mathrm{~m}$
  2. $6.3 \times 10^{-4} \mathrm{~m}$
  3. $8.6 \times 10^{-4} \mathrm{~m}$
  4. $5.8 \times 10^{-4} \mathrm{~m}$

Solution

$\begin{aligned} & \mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{D}=1.33 \mathrm{~m} \\ & \lambda=6300 Å=6.3 \times 10^{-7} \mathrm{~m} \end{aligned}$ $\lambda_{\mathrm{w}}=$ wavelength in water $=\frac{6.3 \times 10^{-7}}{1.33} \mathrm{~m}$ Fringe width, $\mathrm{X}=\frac{\lambda_{\mathrm{w}} \mathrm{D}}{\mathrm{d}}=\frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}}=6.3 \times 10^{-4} \mathrm{~m}$ /

Asked in: MHT CET 2021 (21 Sep Shift 1)

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