Question:medium

If Young's double slit experiment is done in air first and then if the experiment is conducted by immersing the apparatus in water, the fringe width:

Show Hint

Fringe width is directly proportional to refractive index of the medium (indirectly via $\lambda$).
Updated On: Jun 6, 2026
  • remains same
  • decreases
  • increases
  • Becomes zero
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: What sets the fringe width.
In Young's double slit experiment the fringe width is \[ \beta = \frac{\lambda D}{d}, \] where $\lambda$ is the wavelength of light used, $D$ is the screen distance, and $d$ is the slit separation.
Step 2: What changes inside water.
When the whole setup is dipped in water, $D$ and $d$ stay the same, but light slows down and its wavelength shrinks. The new wavelength is $\lambda' = \frac{\lambda}{\mu}$, where $\mu$ is the refractive index of water and $\mu > 1$.
Step 3: New fringe width.
Putting the smaller wavelength back in, \[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}. \]
Step 4: Compare the two.
Since $\mu > 1$ for water (about $1.33$), dividing by $\mu$ makes $\beta'$ smaller than $\beta$.
Step 5: Conclusion.
The fringes move closer together, so the fringe width decreases. \[ \boxed{\text{decreases}} \]
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