Question:medium

In a classic Young’s Double Slit Experiment, how does the fringe width change if the entire apparatus is immersed in water?

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In YDSE, if the experiment is performed in a medium with refractive index \(n\), the new fringe width becomes \[ \beta' = \frac{\beta}{n} \] Hence fringe width decreases.
Updated On: Mar 25, 2026
  • Fringe width increases
  • Fringe width decreases
  • Fringe width remains unchanged
  • Fringe width becomes infinite
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
In Young's Double Slit Experiment, the interference pattern consists of bright and dark fringes. The separation between consecutive bright or dark fringes is the fringe width.
Step 2: Key Formula or Approach:
The formula for fringe width $\beta$ is:
\[ \beta = \frac{\lambda D}{d} \]
where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation.
When light enters a medium of refractive index $\mu$, its new wavelength $\lambda'$ is $\frac{\lambda}{\mu}$.
Step 3: Detailed Explanation:
When the apparatus is immersed in water, the refractive index of water is $\mu \approx 1.33$, which is greater than $1$.
The new wavelength of the light in water becomes:
\[ \lambda' = \frac{\lambda}{\mu} \]
Because $\mu>1$, the wavelength decreases ($\lambda'<\lambda$).
Since the fringe width $\beta$ is directly proportional to the wavelength ($\beta \propto \lambda$), a decrease in wavelength results in a proportional decrease in the fringe width:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu} \]
Thus, the fringe width decreases.
Step 4: Final Answer:
The correct option is (B).
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