Q66 / 75 · JEE Main 2025 (24 Jan Shift 1) · Physics

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

  • Correct answer
    A
    $5$
  • B
    $4$
  • C
    $6$
  • D
    $8$

Solution

$\begin{aligned} & \frac{n_1 \lambda_1 D}{d}=\frac{n_2 \lambda_2 D}{d} \\ & n 480=m 600 \\ & n_{\min }=5 \end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions