In Young's double slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen where path difference is \( \lambda \), is 'T' units. What is the intensity of light at a point where path difference is \( \displaystyle\frac{\lambda}{4} \)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{T}{2} \)
In Young's double slit experiment with two slits of equal intensity, the intensity at a point where the two waves differ in phase by \( \phi \) is \( I = I_{\max}\cos^{2}(\phi/2) \), and the phase difference is \( \phi = \frac{2\pi}{\lambda}\times(\text{path difference}) \).
For a path difference \( \lambda \), the phase difference is \( 2\pi \), a bright fringe, so T is the maximum intensity:
\( I_{\max} = T\cos^{2}\pi = T \)
For a path difference \( \lambda/4 \):
\( \displaystyle \phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{4} = \frac{\pi}{2} \)
\( \displaystyle I = T\cos^{2}\frac{\pi}{4} = T\times\frac{1}{2} = \frac{T}{2} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
