In a Young’s double slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on a screen, where the path difference is \( \lambda \), is \( K\ \mathrm{W/m^{2}} \).
What is the intensity \( (\mathrm{W/m^{2}}) \) of light at a point where the path difference is \( \displaystyle\frac{\lambda}{6} \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.75\ K \)
In Young’s experiment the intensity depends on the phase difference \( \phi \) between the two waves, and \( \phi \) is fixed by the path difference \( \Delta x \).
\( \displaystyle \phi = \frac{2\pi}{\lambda}\,\Delta x \)
\( \displaystyle I = I_{\max}\cos^{2}\frac{\phi}{2} \)
Path difference \( \lambda \): \( \phi = 2\pi \), so this point is a bright fringe of maximum intensity.
\( I = I_{\max}\cos^{2}\pi = I_{\max} = K \)
Path difference \( \lambda/6 \):
\( \displaystyle \phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} = \frac{\pi}{3} \)
\( \displaystyle I = K\cos^{2}\frac{\pi}{6} = K\left(\frac{\sqrt{3}}{2}\right)^{2} = 0.75\,K \)
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 – …
