A Fraunhofer diffraction pattern due to a single slit of width $0 \cdot 3 \mathrm{~mm}$ is obtained on a…

A Fraunhofer diffraction pattern due to a single slit of width $0 \cdot 3 \mathrm{~mm}$ is obtained on a screen placed at a distance of $3 \mathrm{~m}$ from the slit. The first minima lie at $5 \cdot 5 \mathrm{~mm}$ on either side of the central maximum on the screen. The wavelength of light used is
  1. $6000 Å$
  2. $5500 Å$
  3. $4500 Å$
  4. $5000 Å$

Solution

$a=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}, \quad D=3 \mathrm{~m}$, $x=5.5 \mathrm{~mm}=5.5 \times 10^{-3} \mathrm{~m}$ $x=\frac{\lambda D}{a}$ $\therefore \lambda=\frac{\mathrm{xa}}{\mathrm{D}}$ $\therefore \frac{\lambda=5.5 \times 10^{-3} \times 0.3 \times 10^{-3}}{3}=5.5 \times 10^{-7} \mathrm{~m}=5500 Å$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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