In a diffraction pattern, light of wavelength $580 \mathrm{~nm}$ is incident normally on a slit of width '…

In a diffraction pattern, light of wavelength $580 \mathrm{~nm}$ is incident normally on a slit of width ' $a$ ' The distance between slit and the screen is $2.5 \mathrm{~m}$ and the distance of the second order maximum from the center of the screen is $14.5 \mathrm{~mm}$. The value of ' $a$ ' is
  1. $0.26 \times 10^{-3} \mathrm{~m}$
  2. $0.36 \times 10^{-3} \mathrm{~m}$
  3. $0.50 \times 10^{-3} \mathrm{~m}$
  4. $0.12 \times 10^{-3} \mathrm{~m}$

Solution

Distance of $2^{\text {nd }}$ order maximum from the centre of the screen $x=\frac{5 D \lambda}{2 d}$ Here, $\mathrm{D}=2.5 \mathrm{~m}, \mathrm{x}=14 \mathrm{~mm}=14 \times 10^{-3} \mathrm{~m}$ $\begin{aligned} & \lambda=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m} \\ & \therefore \mathrm{d}=\frac{5 \mathrm{D} \lambda}{2 \mathrm{x}}=\frac{\left(5 \times 2.5 \times 580 \times 10^{-9} \mathrm{~m}\right)}{2 \times 14 \times 10^{-3} \mathrm{~m}}=259 \mu \mathrm{m}=0.26 \times 10^{-3} \mathrm{~m}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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