YDSE with white light – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): In case of YDSE, if monochromatic light is replaced by white light then closest on either side of central white fringe will be blue and farthest will appear red. Reason (R): Fringe width for blue will be greater than that for red for same bright fringe.  
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

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Young's Double Slit Experiment. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A) is true: Fringe width is \(\beta = \frac{\lambda D}{d}\). Since \(\lambda_{\text{red}} > \lambda_{\text{blue}}\), it follows that \(\beta_{\text{red}} > \beta_{\text{blue}}\). Thus, red fringes are wider and appear farther from the center, while blue fringes are closer. Reason (R) is false: Fringe width for blue light is smaller than that for red light. Hence, (A) is true, (R) is false.

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