Question

In Young's double slit experiment, the fringe width is 0.4 mm. What is the distance between \(4^{\text{th}}\) dark band and \(6^{\text{th}}\) bright band on the same side of the interference pattern?

• A. \(0.5\) mm
• B. \(0.75\) mm
• C. \(1.0\) mm
• D. \(1.5\) mm

Hint:

Dark fringes are at (n - 1/2) beta and bright fringes at n beta.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
On the same side of the central fringe, the \(n\)-th bright fringe is at \(n\beta\) and the \(n\)-th dark fringe is at \(\left(n - \tfrac12\right)\beta\).

Step 2: Positions:
4th dark: \(3.5\beta = 3.5\times0.4 = 1.4\) mm.
6th bright: \(6\beta = 6\times0.4 = 2.4\) mm.

Step 3: Separation:
\[ 2.4 - 1.4 = 1.0\ \text{mm} \]

Step 4: Check:
Option (C). In terms of fringe width the separation is \(6 - 3.5 = 2.5\beta = 1.0\) mm.

Final Answer:
Positions are 1.4 mm and 2.4 mm, a gap of 1.0 mm. \[ \boxed{\text{(C) }1.0\ \text{mm}} \]