In a double slit experiment shown in figure, when light of wavelength $400 nm$ is used, dark fringe is observed at $P$ . If $D = 0.2 m$ , the minimum distance between the slits $S_{1}$ and $S_{2}$ is $α mm$ . Write the value of $10 α$ to the nearest integer.

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TestHub · Accepted Answer

According to the given diagram, the path difference for minima at P is given by

$∆ x = 2 \sqrt{D^{2} + d^{2}} - 2 D . . . (1)$

And, from the condition of minima, it follows that

$∆ x = \frac{λ}{2} . . . (2)$

Equations (1) and (2) imply that

$2 \sqrt{D^{2} + d^{2}} - 2 D = \frac{λ}{2}$

$\Rightarrow \sqrt{D^{2} + d^{2}} - D = \frac{λ}{4}$

$\Rightarrow \sqrt{D^{2} + d^{2}} = \frac{λ}{4} + D$

$\Rightarrow D^{2} + d^{2} = D^{2} + \frac{λ^{2}}{16} + \frac{D λ}{2}$

$\Rightarrow d^{2} = \frac{D λ}{2} + \frac{λ^{2}}{16}$

$\Rightarrow d^{2} = \frac{0.2 \times 400 \times 10^{- 9}}{2} + \frac{16 \times 10^{- 14}}{16}$

$\Rightarrow d^{2} \approx 400 \times 10^{- 10}$

$\begin{array}{rcl} ∴ d & = & 20 \times 10^{- 5} m \\ = & 0.20 mm \end{array}$

Therefore, $10 α = 2$ .

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