If the \( 8^{\mathrm{th}} \) bright fringe of wavelength \( \lambda_{1} \) coincides with \( 9^{\mathrm{th}} \) bright fringe of wavelength \( \lambda_{2} \) in Young’s double slit experiment, then the possible values of wavelengths can be respectively:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 450 \) nm and \( 400 \) nm
In Young’s experiment the n-th bright fringe is at \( y_n = n\lambda D/d \). The two fringes coincide when they are at the same distance from the centre:
\( \displaystyle \frac{8\lambda_1 D}{d} = \frac{9\lambda_2 D}{d} \)
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{9}{8} = 1.125 \)
Check Option (1):
\( \displaystyle \frac{450\,\mathrm{nm}}{400\,\mathrm{nm}} = \frac{9}{8} \)
So \( \lambda_1 = 450 \) nm and \( \lambda_2 = 400 \) nm. The longer wavelength is \( \lambda_1 \), because its 8th fringe reaches as far as the 9th fringe of \( \lambda_2 \).