Question

In an interference pattern, $10$ fringes are obtained over a certain distance from its central maximum with light of wavelength $480\,\text{nm}$. The number of fringes obtained over the same distance with light of wavelength $600\,\text{nm}$ would be:

• A. $6$
• B. $7$
• C. $8$
• D. $10$

Hint:

Fringe width is proportional to wavelength, so \(N\lambda\) stays constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In Young's double slit experiment, the fringe width is \(\beta = \dfrac{\lambda D}{d}\). The distance of the \(n\)th bright fringe from the centre is \(x_n = n\beta\).

Step 2: Key Formula or Approach:
The distance x is fixed, so \(x = N_1\beta_1 = N_2\beta_2\).
With D and d fixed, \(\beta \propto \lambda\). So \(N_1\lambda_1 = N_2\lambda_2\).

Step 3: Calculate:
\[ N_2 = N_1\,\frac{\lambda_1}{\lambda_2} = 10 \times \frac{480}{600} = 8 \]
A longer wavelength gives wider fringes, so fewer fringes fit in the same distance.

Step 4: Check each option:
(A) 6: this would need a wavelength of 800 nm. So (A) is wrong.
(B) 7: this would need a wavelength of about 686 nm. So (B) is wrong.
(C) 8: this matches the calculation. So (C) is correct.
(D) 10: the number stays the same only if the wavelength is unchanged. So (D) is wrong.

Final Answer:
The number of fringes is 8, which is option (C). \[ \boxed{N_2 = 8} \]