If Young.s double slit apparatus is shifted from air to water, then

Options

A. fringe width decreases
B. fringe width increases
C. fringe width remains same
D. fringe system disappers

Answer

To solve the problem of how the fringe width in Young's double slit experiment changes when the apparatus is shifted from air to water, we can follow these steps:

### Step 1: Understand the Formula for Fringe Width
The fringe width (β) in Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \(\beta\) = fringe width
- \(\lambda\) = wavelength of the light used
- \(D\) = distance from the slits to the screen
- \(d\) = distance between the slits

### Step 2: Effect of Medium on Wavelength
When the apparatus is shifted from air to water, the wavelength of light changes. The wavelength of light in a medium is given by:
\[ \lambda' = \frac{\lambda}{n} \]
where:
- \(\lambda'\) = wavelength in the medium (water)
- \(\lambda\) = wavelength in air
- \(n\) = refractive index of the medium (for water, \(n \approx 1.33\))

### Step 3: Substitute the New Wavelength into the Fringe Width Formula
When the apparatus is in water, the new fringe width (\(\beta'\)) can be expressed as:
\[ \beta' = \frac{\lambda' D}{d} = \frac{\left(\frac{\lambda}{n}\right) D}{d} = \frac{\lambda D}{nd} \]

### Step 4: Compare Fringe Widths
Now, we can compare the fringe widths in air and water:
- Fringe width in air: \(\beta = \frac{\lambda D}{d}\)
- Fringe width in water: \(\beta' = \frac{\lambda D}{nd}\)

Since \(n > 1\) (for water), it follows that:
\[ \beta' < \beta \]
This means that the fringe width decreases when the apparatus is shifted from air to water.

### Conclusion
Thus, the correct answer is that the fringe width decreases when Young's double slit apparatus is shifted from air to water.

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