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: Understanding the Concept:
In a YDSE setup, the positions of bright and dark fringes on a screen are determined by their order number (n), the wavelength of light (\(\lambda\)), the distance to the screen (D), and the slit separation (d). The distance between any two fringes can be found by taking the difference of their position coordinates.
Step 2: Key Formula or Approach:
Let the position of a fringe from the central maximum be \(y\).
- Position of the n-th bright fringe: \(y_{n, bright} = n \frac{\lambda D}{d} = n\beta\), where n = 0, 1, 2, ...
- Position of the n-th dark fringe: \(y_{n, dark} = (n - \frac{1}{2}) \frac{\lambda D}{d} = (n - 0.5)\beta\), where n = 1, 2, 3, ...
Here, \(\beta = \frac{\lambda D}{d}\) is the fringe width.
Step 3: Detailed Explanation:
Part 1: Use the given information to find the fringe width (\(\beta\)).
The distance between the 5th bright fringe and the 7th dark fringe is 3 mm.
Position of 5th bright fringe: \(y_{5, bright} = 5\beta\).
Position of 7th dark fringe: \(y_{7, dark} = (7 - 0.5)\beta = 6.5\beta\).
The distance between them is:
\[ \Delta y_1 = y_{7, dark} - y_{5, bright} = 6.5\beta - 5\beta = 1.5\beta \] We are given \(\Delta y_1 = 3\) mm.
\[ 1.5\beta = 3 \text{ mm} \implies \beta = \frac{3}{1.5} = 2 \text{ mm} \] Part 2: Calculate the required distance.
We need to find the distance between the 5th dark fringe and the 7th bright fringe.
Position of 5th dark fringe: \(y_{5, dark} = (5 - 0.5)\beta = 4.5\beta\).
Position of 7th bright fringe: \(y_{7, bright} = 7\beta\).
The distance between them is:
\[ \Delta y_2 = y_{7, bright} - y_{5, dark} = 7\beta - 4.5\beta = 2.5\beta \] Now substitute the value of \(\beta\) we found:
\[ \Delta y_2 = 2.5 \times (2 \text{ mm}) = 5 \text{ mm} \] Step 4: Final Answer:
The distance between the 5th dark and 7th bright fringes is 5 mm. Therefore, option (C) is correct.