In YDSE first minima is found in front of one of the slits. If distance between plane of slits and screen is 5 cm and wavelength of light is 800 nm, the separation between slits is

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$4 \times 10^{- 4} m$

$6 \times 10^{- 4} m$

$2 \times 10^{- 4} m$

$1 \times 10^{- 4} m$

answer is A.

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Detailed Solution

$S_{1} P = D$

$S_{2} P = \sqrt{D^{2} - d^{2}} = D {(1 + \frac{d^{2}}{D^{2}})}^{\frac{1}{2}}$

$= D {(1 + \frac{d^{2}}{2 D^{2}})}^{\frac{1}{2}} = D + \frac{d^{2}}{2 D}$

$Δ x = \frac{d^{2}}{2 D}$

To have 1st minimum, $Δ x = \frac{λ}{2}$

$∴ \frac{d^{2}}{2 D} = \frac{λ}{2}$

$\Rightarrow d^{2} = λ D$

$\Rightarrow d = \sqrt{800 \times 10^{- 9} \times 5 \times 10^{- 2}} = \sqrt{4 \times 10^{- 8}}$

$= 2 \times 10^{- 4} m$

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