In Young's double slit experiment using monochromatic light of wavelength \( \lambda \) the intensity of light at a point on the screen is 20 units, where the path difference is \( \lambda \). The intensity of light at a point having the path difference \( \lambda/4 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 units
For two coherent sources of equal intensity the intensity at a point is \( I = I_{\max}\cos^{2}(\phi/2) \), where \( \phi \) is the phase difference, related to the path difference \( \Delta \) by \( \phi = 2\pi\Delta/\lambda \).
For a path difference \( \lambda \): \( \phi = 2\pi \), so \( \cos^{2}\pi = 1 \) and the point is a maximum. Hence \( I_{\max} = 20 \) units.
For a path difference \( \lambda/4 \):
\( \displaystyle \phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{4} = \frac{\pi}{2} \)
\( \displaystyle I = 20\cos^{2}\left(\frac{\pi}{4}\right) = 20 \times \frac{1}{2} = 10\ \text{units} \)
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 – …
