In a Young's double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining A₁ P is perpendicular to A₁ A₂ as shown in the figure. If the first minimum is detected at P, the value of slits separation 'a' will be: The distance of screen from slits D = 5 cm (a) 0.2 mm (b) 0.5 mm (c) 0.4 mm (d) 0.1 mm

Answer: (a). Path difference at P: √(D² + a²) - D ≈ a²/2D. First minimum: a²/2D = (λ)/(2) ⇒ a = √(λ D) a = √800×10⁻⁹ × 0.05 = 2×10⁻⁴ m So the answer is (a) 0.2 mm.

In a Young's double slit experiment, two slits are illuminated with a… | JEE Main 2023 (29 Jan, Shift 1)

Q47JEE Main 2023 · 29 JAN · SHIFT 1Medium

YDSE: Film, Medium & Two Wavelengths

In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800\text{ nm}$ . 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 slits separation 'a' will be:

The distance of screen from slits D = 5 cm

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Answer: (a)

Path difference at P: $\sqrt{D^2 + a^2} - D \approx \dfrac{a^2}{2D}$ .
First minimum: $\dfrac{a^2}{2D} = \dfrac{\lambda}{2} \Rightarrow a = \sqrt{\lambda D}$
$a = \sqrt{800\times10^{-9} \times 0.05} = 2\times10^{-4}\text{ m}$
So the answer is (a) 0.2 mm.

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