MCQ
In Young's double slit experiment, the intensity on the screen at a point where path difference is $\lambda$ is $K$. What will be the intensity at the point where path difference is $\lambda / 4$
  • A
    (a) $\frac{K}{4}$
  • B
    (b) $\frac{K}{2}$
  • C
    (c) $K$
  • (d) Zero

Answer

Correct option: D.
(d) Zero
(d) If shift is equivalent to $n$ fringes then
$n=\frac{(\mu-1) t}{\lambda} \Rightarrow n \propto t \Rightarrow \frac{t_2}{t_1}=\frac{n_2}{n_1}$
$ \Rightarrow t_2=\frac{n_2}{n_1} \times t $
$t_2=\frac{20}{30} \times 4.8=3.2 mm .$

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