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: Understand the situation.
In a biprism experiment $21$ fringes fill a certain region using light of wavelength $4800\ \text{\AA}$. If we switch to light of $5600\ \text{\AA}$, how many fringes fill the same region?

Step 2: Recall fringe width.
The width of one fringe grows with wavelength, $\beta \propto \lambda$. So longer wavelength means fatter fringes.

Step 3: Link fringe count to width.
In a fixed region, fatter fringes mean fewer of them. So the number of fringes $N \propto \frac{1}{\lambda}$.

Step 4: Write the ratio.
This gives $\frac{N_2}{N_1} = \frac{\lambda_1}{\lambda_2}$.

Step 5: Put in the numbers.
$N_2 = 21 \times \frac{4800}{5600} = 21 \times \frac{6}{7}$.

Step 6: Finish the arithmetic.
$21 \times \frac{6}{7} = 3 \times 6 = 18$ fringes.
\[ \boxed{N_2 = 18} \]