JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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 _____.

A
$2 I_{0}$
B
$I_{0}$
C
$\frac{I _{0}}{2}$
D
$4 I_{0}$

Solution

In front of one slit, $y = d /2$ . The path difference at the screen is $Δ = \frac{y d}{D} = \frac{d ^{2}}{2 D}$ .

Substitute $d = 2 \times 1 0^{- 3}$ m, $D = 10$ m, $λ = 6 \times 1 0^{- 7}$ m:

$Δ = \frac{( 2 \times 1 0 ^{- 3} ) ^{2}}{2 \times 10} = 2 \times 1 0^{- 7}$ m. Phase difference $ϕ = \frac{2 π}{λ} Δ = \frac{2 π}{6 \times 1 0 ^{- 7}} \cdot 2 \times 1 0^{- 7} = \frac{2 π}{3}$ .

Resultant intensity: $I = 4 I_{0} cos^{2} (\frac{ϕ}{2}) = 4 I_{0} cos^{2} (π /3) = 4 I_{0} \cdot \frac{1}{4} = I_{0}$ .

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