JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Wave Optics

In the Young's double slit experiment the intensity produced by each one of the individual slits is

I_{0}

. The distance between two slits is 2 mm. The distance of screen from slits is 10 m. The wavelength of light is

6000 \overset{˚}{A}

. The intensity of light on the screen in front of one of the slits is ____.

Answer: (B)

Solution

Directly in front of one slit the point is at

y = \frac{d}{2} = 1

mm from the central axis, so the path difference is

Δ = \frac{y d}{D} = \frac{( 1 \times 1 0 ^{- 3} ) ( 2 \times 1 0 ^{- 3} )}{10} = 2 \times 1 0^{- 7}

m, which is

2000 \overset{˚}{A}

. In wavelengths

\frac{Δ}{λ} = \frac{2000}{6000} = \frac{1}{3}

, so the phase difference is

δ = \frac{2 π}{3}

. With equal slit intensities the resultant is

I = 4 I_{0} cos^{2} \frac{δ}{2} = 4 I_{0} cos^{2} \frac{π}{3} = 4 I_{0} (\frac{1}{2})^{2} = I_{0}

. The intensity in front of one slit is

I_{0}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option