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 \( \lambda \) is K units. What is the intensity of light at a point where path difference is \( \displaystyle\frac{\lambda}{3} \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{K}{4} \)
For two identical coherent waves, the resultant intensity at a point with phase difference \( \phi \) is
\( \displaystyle I = I_{\max}\cos^{2}\frac{\phi}{2}, \qquad \phi = \frac{2\pi}{\lambda}\,\Delta x \)
For path difference \( \lambda \): \( \phi = 2\pi \), \( \cos^{2}\pi = 1 \), so this point is a bright fringe and \( I_{\max} = K \).
For path difference \( \lambda/3 \):
\( \displaystyle \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \)
\( \displaystyle I = K\cos^{2}\frac{\pi}{3} = K \times \left(\frac{1}{2}\right)^{2} = \frac{K}{4} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
