In a Young's double slit experiment, the n^th bright fringe for a light of wavelength 5000 Å coincides with (n + 1)^th bright fringe for light of wavelength 4000 A. The value of $n$ is

4

The correct answer is Option (4) → 4

Condition for bright fringe coincidence:

$n \lambda_{1} = (n+1)\lambda_{2}$

Given $\lambda_{1} = 5000 \, \text{Å}, \, \lambda_{2} = 4000 \, \text{Å}$

$n(5000) = (n+1)(4000)$

$5000n = 4000n + 4000$

$1000n = 4000$

$n = 4$

Answer: $n = 4$