In a Young's double slit experiment, the nth 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
Answer & explanation
Correct answer: option 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: $n = 4$