Question

In Young's double-slit experiment, if the distance between the slits is increased to 3 times its initial distance, then the ratio of initial and final fringe widths is.

• A. $ 1 : 9 $
• B. $ 9 : 1 $
• C. $ 3 : 1 $
• D. $ 1 : 3 $

Hint:

In Young's double-slit experiment, the fringe width is inversely proportional to the slit separation (\( d \)). Doubling or tripling \( d \) reduces the fringe width proportionally.

Answer 1

The Correct Option is C

Step 1: Known Information. The fringe width (\( \beta \)) in Young's double-slit experiment is given by: $$ \beta = \frac{\lambda D}{d} $$ where:
\( \lambda \) is the wavelength of light,
\( D \) is the distance between the slits and the screen,
\( d \) is the distance between the two slits.
Step 2: Initial Fringe Width.
Let the initial distance between the slits be \( d_1 \). The initial fringe width (\( \beta_1 \)) is: $$ \beta_1 = \frac{\lambda D}{d_1} $$ Step 3: Final Fringe Width.
If the distance between the slits is increased to 3 times its initial value, the new distance between the slits is \( d_2 = 3d_1 \). The final fringe width (\( \beta_2 \)) is: $$ \beta_2 = \frac{\lambda D}{d_2} = \frac{\lambda D}{3d_1} $$ Step 4: Ratio of Initial and Final Fringe Widths.
The ratio of the initial fringe width to the final fringe width is: $$ \frac{\beta_1}{\beta_2} = \frac{\frac{\lambda D}{d_1}}{\frac{\lambda D}{3d_1}} $$ Simplify: $$ \frac{\beta_1}{\beta_2} = \frac{\lambda D}{d_1} \cdot \frac{3d_1}{\lambda D} = 3 $$ Thus, the ratio of the initial and final fringe widths is: $$ \frac{\beta_1}{\beta_2} = 3 : 1 $$ Final Answer: $ \boxed{3 : 1} $