MCQ
In Young's double slit experiment, the distance between sources is $1 mm$ and distance between the screen and source is $1 m$. If the fringe width on the screen is $0.06cm$, then $\lambda$ =........$\mathop A\limits^o $
  • $6000 $
  • B
    $4000 $
  • C
    $1200 $
  • D
    $2400 $

Answer

Correct option: A.
$6000 $
a
(a)$\beta = \frac{{\lambda \,D}}{d} \Rightarrow (0.06 \times {10^{ - 2}}) = \frac{{\lambda \times 1}}{{1 \times {{10}^{ - 3}}}}$ $ \Rightarrow \lambda = 6000\,Å$

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