Case: Read the passage and answer the question.
YDSE is the abbreviation used for Young’s double-slit experiment, which shows the interference of light. Interference of light is the phenomenon in which two light waves meet each other while travelling in the same medium. A light source is kept behind a wall in which two slits, S1 and S2, are made. On the screen there appears a pattern of alternate bright and dark fringes. The central fringe is always bright.
The wavelength of red light is \( 7.5\times10^{-5}\,\mathrm{cm} \) and that of blue light is \( 5.0\times10^{-5}\,\mathrm{cm} \). In the same double-slit set-up, for what value of n does the (n + 1)th bright fringe of blue light coincide with the nth bright fringe of red light?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
In Young’s double-slit experiment the nth bright fringe lies at \( x_n=n\lambda D/d \) from the centre. In the same set-up \( D \) and \( d \) are the same for both colours. So two bright fringes coincide when the product \( n\lambda \) is the same for both.
For the (n + 1)th blue fringe to lie on the nth red fringe:
\( (n+1)\,\lambda_{\mathrm{blue}}=n\,\lambda_{\mathrm{red}} \)
\( (n+1)\times5.0\times10^{-5}\,\mathrm{cm}=n\times7.5\times10^{-5}\,\mathrm{cm} \)
\( 5n+5=7.5n \)
\( 2.5n=5 \)
\( n=2 \)
Therefore, the 3rd bright fringe of blue light coincides with the 2nd bright fringe of red light.