A double slit experiment is immersed in water of refractive index $1 \cdot 33$. The slit separation is $1…

A double slit experiment is immersed in water of refractive index $1 \cdot 33$. The slit separation is $1 \mathrm{~mm}$, distance between slit and screen is $1 \cdot 33 \mathrm{~m}$. The slits are illuminated by a light of wavelength $6300 Å$. The fringewidth is
  1. $6.9 \times 10^{-4} \mathrm{~m}$
  2. $6.3 \times 10^{-4} \mathrm{~m}$
  3. $5 \cdot 8 \times 10^{-4} m$
  4. 8.6 x $10^{-4} \mathrm{~m}$

Solution

fringe width $=X=\frac{\lambda_{W} D}{d}$ $\lambda_{w}=\frac{6.3 \times 10^{-7}}{1.33}, \quad D=1 \mathrm{~m}, \quad \mathrm{~d}=10^{-3} \mathrm{~m} \quad$ [on substituting and solving ] $X=6.3 \times 10^{-4} \mathrm{~m}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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