MCQ
A screen is placed $50\,cm$ from a single slit, which is illuminated with $6000$ \mathring A light. If distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is.....$mm$
  • A
    $0.1$
  • $0.2$
  • C
    $0.3$
  • D
    $0.4$

Answer

Correct option: B.
$0.2$
b
(b) Position of nth minima ${y_n} = \frac{{n\lambda D}}{d}$
==> $({y_3} - {y_1}) = \frac{{\lambda D}}{d}(3 - 1) = \frac{{2\lambda D}}{d}$
==> $3 \times {10^{ - 3}} = \frac{{2 \times 6000 \times {{10}^{ - 10}} \times 0.5}}{d}$
==> $d = 0.2 \times {10^{ - 3}}m = 0.2\,mm$

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