In a Young's double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen is K units where path difference is \( \displaystyle\frac{\lambda}{3} \). The intensity of light at a point where path difference is \( \displaystyle\frac{\lambda}{4} \), will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2K units
When two coherent waves of equal intensity superpose, the resultant intensity depends on their phase difference \( \phi \), which is fixed by the path difference \( \Delta x \).
\( \displaystyle I = I_0\cos^{2}\frac{\phi}{2} \)
\( \displaystyle \phi = \frac{2\pi}{\lambda}\,\Delta x \)
Here \( I_0 \) is the intensity at the central bright fringe.
For a path difference of \( \lambda/3 \):
\( \displaystyle \phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \)
\( \displaystyle \mathrm{K} = I_0\cos^{2}60^\circ = \frac{I_0}{4} \)
\( I_0 = 4\mathrm{K} \)
For a path difference of \( \lambda/4 \):
\( \displaystyle \phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \)
\( \displaystyle I = I_0\cos^{2}45^\circ = \frac{I_0}{2} \)
\( \displaystyle I = \frac{4\mathrm{K}}{2} = 2\mathrm{K} \)
Shortcut: Take the ratio directly: \( \displaystyle\frac{I}{\mathrm{K}} = \frac{\cos^{2}45^\circ}{\cos^{2}60^\circ} = \frac{1/2}{1/4} = 2 \), so I = 2K without finding \( I_0 \).
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 – …
