Question

In Young’s double slit experiment, if the distance between the two slits is decreased by 20% and the distance between the slits and the screen is increased by 20%, then the percentage increase in the fringe width is

• A. 50
• B. 20
• C. 40
• D. 100

Hint:

Fringe width \( \beta \propto \frac{D}{d} \). A 20% increase in \(D\) and 20% decrease in \(d\) results in a net 50% increase in \(\beta\).

Answer 1

The Correct Option is A

Fringe width in Young's double slit experiment is given by: \[ \beta = \frac{\lambda D}{d} \] Where: - \(\lambda\) = wavelength of light (constant)
- \(D\) = distance between the screen and the slits
- \(d\) = distance between the two slits
Given:
- \(d\) is decreased by 20% → \(d' = 0.8d\)
- \(D\) is increased by 20% → \(D' = 1.2D\)
New fringe width: \[ \beta' = \frac{\lambda \cdot 1.2D}{0.8d} = \frac{1.2}{0.8} \cdot \frac{\lambda D}{d} = 1.5 \cdot \beta \] So, fringe width increases by 50%.

Answer 2

Step 1: Recall the formula for fringe width
The fringe width \(\beta\) in Young’s double slit experiment is given by:
\[ \beta = \frac{\lambda D}{d} \]
where
\(\lambda\) = wavelength of light (constant),
\(D\) = distance between slits and screen,
\(d\) = distance between the two slits.

Step 2: Apply the given changes
- Distance between slits \(d\) is decreased by 20%, so new slit distance:
\[ d' = 0.8 d \]
- Distance between slits and screen \(D\) is increased by 20%, so new screen distance:
\[ D' = 1.2 D \]

Step 3: Calculate new fringe width \(\beta'\)
\[ \beta' = \frac{\lambda D'}{d'} = \frac{\lambda \times 1.2 D}{0.8 d} = \frac{1.2}{0.8} \times \frac{\lambda D}{d} = 1.5 \beta \]

Step 4: Find percentage increase in fringe width
\[ \text{Percentage increase} = \frac{\beta' - \beta}{\beta} \times 100 = (1.5 - 1) \times 100 = 50\% \]

Step 5: Final answer
The fringe width increases by 50%.