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

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

Solution

$\begin{aligned} & \lambda_{\text {liquid }}=\frac{\lambda_{\text {air }}}{\mu} \\ & \lambda_{\text {liquid }}=\frac{6300 \times 10^{-10}}{1.33} \end{aligned}$ Fringe width, $\mathrm{W}=\frac{\lambda_{\text {liquid }} \times \mathrm{D}}{\mathrm{d}}=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 0.001}=6.3 \times 10^{-4} \mathrm{~m}$ *

Asked in: MHT CET 2023 (11 May Shift 1)

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