In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm. [2023]

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Solution

Q.
In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining $A_{1} P$ is perpendicular to $A_{1} A_{2}$ as shown in the figure. If the first minimum is detected at $P$ , the value of slit separation ' $a$ ' will be

The distance of screen from slits D = 5 cm [2023]

1 0.4 mm

2 0.5 mm

3 0.2 mm

4 0.1 mm

Ans.
(3)

$A_{2} P - A_{1} P = \frac{λ}{2} (Condition of minima)$

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

$D {(1 + \frac{a^{2}}{D^{2}})}^{1 / 2} - D = \frac{λ}{2}$

$D (1 + \frac{1}{2} \times \frac{a^{2}}{D^{2}}) - D = \frac{λ}{2}$

$\frac{a^{2}}{2 D} = \frac{λ}{2} \Rightarrow a = \sqrt{λ D} = \sqrt{800 \times 10^{- 6} \times 50}$

$\Rightarrow a = 0.2 mm$

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