In Young’s double-slit experiment, one of the slits, say S1, is covered with a thin filter which absorbs half the intensity of the light passing through it. What happens?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The bright fringes become less bright and the dark fringes have a finite light intensity.
The filter changes only the amplitude of the light from S1. The light through both slits still comes from the same source, so the two waves stay coherent.
The intensity is proportional to the square of the amplitude. So halving the intensity from S1 makes its amplitude \( a/\sqrt{2}\approx0.71a \), while S2 still gives \( a \).
Before the filter, the maxima have intensity proportional to \( (2a)^{2}=4a^{2} \) and the minima have zero intensity.
With the filter, at a maximum the amplitudes add:
\( \begin{aligned} I_{\max} &\propto \left(a+\frac{a}{\sqrt{2}}\right)^{2} \\ &\approx 2.9a^{2} \end{aligned} \)
At a minimum the amplitudes subtract:
\( \begin{aligned} I_{\min} &\propto \left(a-\frac{a}{\sqrt{2}}\right)^{2} \\ &\approx 0.09a^{2} \end{aligned} \)
Option (1) The bright fringes become less bright and the dark fringes have a finite light intensity. — Correct. The maxima fall from \( 4a^{2} \) to about \( 2.9a^{2} \), so the bright fringes are less bright. The minima rise from zero to about \( 0.09a^{2} \), so the dark fringes are no longer fully dark.
Option (2) The bright fringes become brighter and the dark fringes become darker. — Incorrect. The amplitude from S1 is reduced. So the sum of the amplitudes at a maximum is smaller, and their difference at a minimum is no longer zero.
Option (3) The fringe width decreases. — Incorrect. The fringe width \( \lambda D/d \) depends only on the wavelength, the screen distance and the slit separation, none of which the filter changes.
Option (4) No fringes are observed. — Incorrect. The waves from the two slits keep a fixed phase difference, so the fringes remain. Only the contrast between bright and dark fringes is reduced.