Question

In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is

• A. 6
• B. 8
• C. 9
• D. 12

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), the fringes are equally spaced.
The distance between two consecutive bright (or dark) fringes is called the fringe width (\(\beta\)).
The total width of a specific segment or region of the screen containing \(n\) fringes is given by the product of the number of fringes and the fringe width.
Key Formula or Approach:
The formula for fringe width is:
\[ \beta = \frac{\lambda D}{d} \]
Where \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the slit separation.
The total width of the segment is \(W = n \times \beta = n \left( \frac{\lambda D}{d} \right)\).
Since the region on the screen (\(W\)), the distance (\(D\)), and the slit width (\(d\)) are kept constant, we have:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
Step 2: Detailed Explanation:
1. Identify initial values:
Number of fringes \(n_1 = 8\).
Initial wavelength \(\lambda_1 = 600 \text{ nm}\).
2. Identify final values:
Final wavelength \(\lambda_2 = 400 \text{ nm}\).
We need to find the final number of fringes \(n_2\).
3. Set up the equation:
Because the segment of the screen is the same:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
\[ 8 \times 600 \text{ nm} = n_2 \times 400 \text{ nm} \]
4. Solve for \(n_2\):
\[ 4800 = 400 \times n_2 \]
\[ n_2 = \frac{4800}{400} \]
\[ n_2 = 12 \]
Thus, when the wavelength decreases, the fringes become narrower, and more of them fit into the same space.
Step 3: Final Answer:
The number of fringes observed in the same region will be 12.