YDSE Fringe Pattern – Rankers Physics
Topic: Wave Optics
Subtopic: Young's Double Slit Experiment

YDSE Fringe Pattern

Assertion (A): In a YDSE, the two slits are at distance 'a' apart. Interference pattern is observed on a screen at a distance D from the slits. At a point on the screen which is directly opposite to the slit, a dark fringe is observed. Then the wavelength of wave is proportional to square of distance between slits.
Reason (R): The light ray coming from two slits do not interfere at the screen.
 
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution:

Assertion (A) is true. If a dark fringe occurs at \(y = a/2\) (point opposite one slit), the path difference is \(a^2/(2D)\). For a dark fringe, \(a^2/(2D) = (n + 1/2)\lambda\), implying \(lambda \propto a^2\).
Reason (R) is false. The core principle of YDSE is the interference of light waves from two coherent slits, which produces the observed pattern on the screen.

Leave a Reply

Your email address will not be published. Required fields are marked *