In Young's double slit experiment, two slits are illuminated with a light of wavelength $λ$ . The line joining $A_{1} P$ is perpendicular to $A_{1} A_{2}$ as shown in figure. If the first minimum is detected at P, the value of slits separation 'a' will be
(D = distance between source and screen)

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From the given figure, the path difference at point P is given by:

$Δ x = A_{2} P - A_{1} P$

Using Pythagoras theorem in right-angled triangle $A_{1} A_{2} P$ , we have $A_{1} P = D$ and $A_{2} P = D^{2} + a^{2}$ .

So, $Δ x = D^{2} + a^{2} - D$

For the first minimum to be detected at P, the path difference must be $\frac{λ}{2}$ .

$D^{2} + a^{2} - D = \frac{λ}{2}$

$D^{2} + a^{2} = D + \frac{λ}{2}$

Squaring both sides, we get:

$D^{2} + a^{2} = D^{2} + D λ + \frac{λ ^{2}}{4}$

$a^{2} = D λ + \frac{λ ^{2}}{4}$

Since the wavelength of light $λ$ is very small compared to the distance $D$ ( $λ ≪ D$ ), the term $\frac{λ ^{2}}{4}$ is negligibly small and can be ignored.

$a^{2} \approx D λ$

$a = λ D$

Answer: $λ D$

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