NEET 2026, Physics Q18: Young’s Double Slit Experiment

NEET2026Physics

In Young’s double slit experiment, using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen where the path difference is $λ$ , is K units. The intensity of light at a point where the path difference is $\frac{λ}{3}$ will be :

A
K
B
2 K
C
$\frac{K}{2}$
D
$\frac{K}{4}$

Solution

At path difference $λ$ , phase difference is $2 π$ , giving maximum intensity. So $K = 4 I$ , where $I$ is the intensity from a single slit.

For path difference $λ /3$ , phase difference $Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{3} = \frac{2 π}{3} = 12 0^{\circ}$ .

Resultant intensity $I_{R} = K cos^{2} (Δ ϕ /2) = K cos^{2} (6 0^{\circ}) = K \cdot \frac{1}{4} = \frac{K}{4}$ .

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NEET 2026, Physics Q18: Young’s Double Slit Experiment

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