In YDSE of equal width slits, if intensity at the centre of screen is $I_{0}$ , then intensity at a point on the screen which is at a distance of $\frac{β}{4}$ from central maxima is( $β = f r i n g e w i d t h)$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{I_{0}}{2}$

$\frac{I_{0}}{8}$

Zero

$\frac{I_{0}}{4}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$I = I_{0} \cos^{2} \frac{ϕ}{2}$

Path difference $= dsinθ ≃ dtanθ = d . \frac{β / 4}{D} = d . \frac{λD}{4 . D . d} = \frac{λ}{4}$

$\begin{array}{l} ∴ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} \\ ∴ intensity= \frac{I_{o}}{2} \end{array}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE