Question

The Young's double slit interference experiment is performed using light consisting of \(480\ \text{nm}\) and \(600\ \text{nm}\) wavelengths to form interference patterns. The least number of the bright fringes of \(480\ \text{nm}\) light that are required for the first coincidence with the bright fringes formed by \(600\ \text{nm}\) light is:

• A. \(4\)
• B. \(8\)
• C. \(6\)
• D. \(5\)

Hint:

For first coincidence of bright fringes, use \(n_1\lambda_1=n_2\lambda_2\) and take the smallest whole number ratio.

Answer 1

The Correct Option is D

Concept:
In Young's double slit experiment, bright fringes coincide when their path differences are equal. For two wavelengths \(\lambda_1\) and \(\lambda_2\), coincidence occurs when: \[ n_1\lambda_1=n_2\lambda_2 \] where \(n_1\) and \(n_2\) are fringe orders.

Step 1: Write the wavelengths.
\[ \lambda_1=480\ \text{nm} \] \[ \lambda_2=600\ \text{nm} \]

Step 2: Apply condition for coincidence.
Let \(n_1\) be the order for \(480\ \text{nm}\) light and \(n_2\) be the order for \(600\ \text{nm}\) light. \[ n_1(480)=n_2(600) \] \[ \frac{n_1}{n_2}=\frac{600}{480} \] \[ \frac{n_1}{n_2}=\frac{5}{4} \]

Step 3: Find least integral values.
The least values satisfying the ratio are: \[ n_1=5,\quad n_2=4 \] So, the \(5^{th}\) bright fringe of \(480\ \text{nm}\) coincides with the \(4^{th}\) bright fringe of \(600\ \text{nm}\).

Step 4: Final conclusion.
Hence, the least number of bright fringes of \(480\ \text{nm}\) light is: \[ \boxed{5} \]