JEE Main 2026 · 5 Apr · Shift 2 · Physics Q42

← JEE Main 2026 · 5 Apr · Shift 2

PhysicsSingle correct+4 / -1Question 42 of 75

The maximum intensity in a Young's double slit experiment is $I_o$ . Distance between the slits ( $d$ ) is $5\lambda$ , where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D=10d$ is ________.

\dfrac{I_o}4

\dfrac{I_o}2

I_o

\dfrac{3I_o}4

Answer and worked solution

Official answer: B

Path difference at $y=\tfrac d2$ :
$\frac{yd}D=\frac{d^2}{2D}=\frac d{20}=\frac\lambda4,$
which is a phase difference of $\tfrac\pi2$ .

Intensity. $I=I_o\cos^2\tfrac\pi4=\tfrac{I_o}2$ .

(B)

Topics:Interference and Young's double slit

← Previous question Next question →

JEE Main 2026 · 5 Apr · Shift 2, Physics question 42