Question

In a Young's Double Slit Experiment (YDSE), the separation between the two coherent slits is halved, and the perpendicular distance between the plane of the slits and the viewing screen is doubled. If the initial fringe width was \( \beta \), what will the new modified fringe width \( \beta' \) be?

• A. \( \beta \)
• B. \( 2\beta \)
• C. \( 4\beta \)
• D. \( \frac{\beta}{4} \)

Hint:

Fringe width is directly proportional to screen distance (\( D \)) and inversely proportional to slit separation (\( d \)). Halving a value in the denominator always effectively doubles the overall result.

Answer 1

The Correct Option is C

Concept: The fringe width (\( \beta \)) in a double-slit interference pattern represents the physical distance between any two consecutive bright or dark fringes on the screen. It is governed by the formula: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the source wavelength, \( D \) is the distance to the screen, and \( d \) is the slit separation.

Step 1: Express the modified experimental variables. According to the changes described in the problem statement:
• The new slit separation is halved: \( d' = \frac{d}{2} \)
• The new screen distance is doubled: \( D' = 2D \)
• The wavelength remains unchanged: \( \lambda' = \lambda \)

Step 2: Substitute the new variables into the fringe width equation. Set up the equation for the modified fringe width \( \beta' \): \[ \beta' = \frac{\lambda' D'}{d'} = \frac{\lambda \cdot (2D)}{\left(\frac{d}{2}\right)} \] Rearranging the fraction by moving the denominator's divisor to the numerator: \[ \beta' = 2 \cdot 2 \cdot \left(\frac{\lambda D}{d}\right) = 4 \left(\frac{\lambda D}{d}\right) \] Since the original fringe width expression is \( \beta = \frac{\lambda D}{d} \), we substitute it back in: \[ \beta' = 4\beta \]