Question

For better visibility of fringe pattern

• A. amplitudes of the sources are equal
• B. the width of the slits should not be equal
• C. dark should be the darkest and bright should be the brightest
• D. the widths should be the same

Answer 1

Answer: (A), (C), (D)

The correct options are
A amplitudes of the sources are equal
C dark should be the darkest and bright should be the brightest
D the widths should be the same

The light from the sources interfere in YDSE to form interference pattern.

The intensity at a point on the screen is given by-

$I=I_1+I_2+2\sqrt{I_1{I}_2}cos\delta$ where $\delta$ is the phase difference.

Thus if the amplitude, thus the intensity for the sources is same, the destructive interference fringe would have exactly zero intensity($\delta =\pi$). Thus the dark fringe would be darkest.

Similarly for the bright fringe, intensity would be highest, giving the better visibility of the fringe pattern.