Question

In biprism experiment, 21 fringes are observed in a given region using light of wavelength 4800 Å. If light of wavelength 5600 Å is used, the number of fringes in the same region will be

• A. 18
• B. 24
• C. 14
• D. 21

Hint:

Remember: For fixed slit separation and screen distance, fringe width \(\beta \propto \lambda\). Hence number of fringes in a fixed length \(N \propto 1/\lambda\). This is useful for quick ratio calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
In a biprism (or double‑slit) interference experiment, for a fixed region on the screen, the number of fringes is inversely proportional to the fringe width, and fringe width \(\beta \propto \lambda\). Thus \(N \propto \frac{1}{\lambda}\).

Step 2: Key Formula or Approach:
\[ \frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1} \] where \(N\) is number of fringes in the same region.

Step 3: Detailed Explanation:
Given \(N_1 = 21\), \(\lambda_1 = 4800\ \text{Å}\), \(\lambda_2 = 5600\ \text{Å}\). Then: \[ N_2 = N_1 \times \frac{\lambda_1}{\lambda_2} = 21 \times \frac{4800}{5600} = 21 \times \frac{48}{56} = 21 \times \frac{6}{7} = 18. \]

Step 4: Final Answer:
The number of fringes is 18, option (A).