Question

In Young's double slit experiment, if the distance between 5th bright and 7th dark fringes is 3 mm, then the distance between 5th dark and 7th bright fringes is

• A. 6 mm
• B. 3 mm
• C. 5 mm
• D. 4 mm

Hint:

It's helpful to visualize the fringes. The nth dark fringe is always halfway between the (n-1)th and nth bright fringes. The key to solving these problems quickly is to express all positions in terms of the fringe width, \(\beta\).

Answer 1

The Correct Option is C

Step 1: Formulae for Fringe Positions: Position of \( n^{\text{th}} \) bright fringe: \( y_{nB} = \frac{n \lambda D}{d} \). Position of \( m^{\text{th}} \) dark fringe: \( y_{mD} = \frac{(2m-1) \lambda D}{2d} \). Let fringe width \( \beta = \frac{\lambda D}{d} \).
Step 2: Analyze Given Data: Distance between \( 5^{\text{th}} \) bright (\( n=5 \)) and \( 7^{\text{th}} \) dark (\( m=7 \)) is 3 mm. Assuming fringes are on the same side: \( y_{7D} = \frac{(14-1)}{2} \beta = 6.5 \beta \). \( y_{5B} = 5 \beta \). Difference \( \Delta y_1 = y_{7D} - y_{5B} = 6.5 \beta - 5 \beta = 1.5 \beta \). Given \( 1.5 \beta = 3 \) mm \( \implies \beta = 2 \) mm.
Step 3: Calculate Required Distance: Distance between \( 5^{\text{th}} \) dark (\( m=5 \)) and \( 7^{\text{th}} \) bright (\( n=7 \)). \( y_{7B} = 7 \beta \). \( y_{5D} = \frac{(10-1)}{2} \beta = 4.5 \beta \). Required Distance \( \Delta y_2 = y_{7B} - y_{5D} = 7 \beta - 4.5 \beta = 2.5 \beta \).
Step 4: Final Calculation: \( \Delta y_2 = 2.5 \times 2 \) mm \( = 5 \) mm.