Question:medium

In Young’s double slit experiment, if the distance between the slits is increased while keeping all other parameters constant, the fringe width will:

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Remember the inverse relationship: $\beta \propto \frac{1}{d}$. If you push the two slits further apart ($d \uparrow$), the interference pattern on the screen shrinks and compresses, meaning the fringe width must decrease ($\beta \downarrow$).
Updated On: Jun 17, 2026
  • Increase
  • Decrease
  • Remain unchanged
  • First increase then decrease
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The Correct Option is B

Solution and Explanation

Step 1 : Understanding the Question:
The topic of this question is Wave Optics, specifically focusing on the phenomenon of Interference as demonstrated in Young's Double Slit Experiment (YDSE). This experiment is a fundamental proof of the wave nature of light, where two coherent sources produce a pattern of bright and dark bands known as fringes on a distant screen. The question asks for the mathematical and physical consequence of increasing the separation between the two slits on the width of these fringes, provided that the wavelength of light and the distance to the screen remain fixed.
Step 2 : Key Formulas and approach:
The approach involves using the standard expression for fringe width in a vacuum or air medium. The fundamental formula is:
\[ \beta = \frac{\lambda D}{d} \]
Where the variables are defined as:
1. $\beta$ (Beta) represents the Fringe Width (distance between two consecutive bright or dark fringes).
2. $\lambda$ (Lambda) represents the Wavelength of the monochromatic light source used.
3. $D$ represents the distance between the plane of the slits and the observation screen.
4. $d$ represents the distance between the two slits.
Step 3 : Detailed Explanation:

To understand the relationship, we look at the proportionality derived from the fringe width formula: $\beta \propto \frac{1}{d}$.

This equation tells us that the fringe width ($\beta$) is inversely proportional to the separation between the slits ($d$).

When we "increase" the distance between the slits ($d$), the denominator in our primary equation becomes larger.

Mathematically, if the denominator of a fraction increases while the numerator ($\lambda D$) stays the same, the overall value of the fraction decreases.

Physically, this means that as the slits move further apart, the interference maxima and minima on the screen crowd closer together, making each individual fringe narrower.

Conversely, if the slits were brought closer together ($d$ decreased), the interference pattern would expand, and the fringe width would increase.

Since the question specifically mentions that $d$ is increased, the only logical result for the fringe width is a decrease.

Step 4 : Final Answer:
Because of the inverse proportionality between slit separation and fringe width, increasing the distance between the slits results in a decrease in the fringe width. Therefore, the correct option is (B).
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