Optics Ques 9
What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $\left(S_1\right)$ ?
(2019 Main, 10 Jan II)
(a) $\frac{\lambda}{2(5-\sqrt{2})}$
(b) $\frac{\lambda}{(5-\sqrt{2})}$
(c) $\frac{\lambda}{2(\sqrt{5}-2)}$
(d) $\frac{\lambda}{(\sqrt{5}-2)}$
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Answer:
Correct Answer: 9.( c )
Solution:
Formula:
We know that condition for minima in Young’s double slit experiment is path difference,
$ \Delta x=(2 n-1) \lambda / 2 $
For first minima, $n=1$
$ \Rightarrow \quad \Delta x=\lambda / 2 $
Path difference between the rays coming from virtual sources $S_1$ and $S_2$ at point ’ $P$ ’ will be
$ \Delta x=S_2 P-S_1 P $
From triangle $S_1 S_2 P$,
$ S_1 P=2 d $
$ \begin{array}{ll} \text { and } & \left(S_2 P^2\right)=\left(S_1 S_2\right)^2+\left(S_1 P\right)^2=d^2+(2 d)^2 \\ \Rightarrow & \left(S_2 P^2\right)=5 d^2 \text { or } S_2 P=\sqrt{5} d \end{array} $
Substituting the values from Eqs. (iii) and (iv) in Eq. (ii), we get
$ \Delta x=\sqrt{5} d-2 d $
From Eqs. (i) and (v), we get
$ \sqrt{5} d-2 d=\lambda / 2 \Rightarrow d=\frac{\lambda}{2(\sqrt{5}-2)} $
BETA
