In Young's double-slit experiment, the two slits are illuminated by a mixture of two wavelengths, 1200 nm and 1000 nm. At 6.0 mm from the common central bright fringe on a screen 2 m away from the slits, a bright fringe of one interference pattern coincides with a bright fringe of the other. The distance between the slits is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.0 mm
In Young’s double-slit experiment the n-th bright fringe lies at \( x_n=n\lambda D/d \). Two bright fringes of different wavelengths coincide where \( n_1\lambda_1=n_2\lambda_2 \), and the first such point uses the smallest whole numbers that satisfy it.
\( \begin{aligned} \frac{n_1}{n_2} &= \frac{\lambda_2}{\lambda_1} \\ &= \frac{1000\,\mathrm{nm}}{1200\,\mathrm{nm}} \\ &= \frac{5}{6} \end{aligned} \)
So the 5th bright fringe of the 1200 nm light falls on the 6th bright fringe of the 1000 nm light, at \( x=6.0\,\mathrm{mm} \).
With the data in SI units:
\( \lambda_1=1200\,\mathrm{nm}=1.2\times10^{-6}\,\mathrm{m} \)
\( x=6.0\,\mathrm{mm}=6.0\times10^{-3}\,\mathrm{m} \)
\( D=2\,\mathrm{m} \)
The slit separation is:
\( \begin{aligned} d &= \frac{n_1\lambda_1D}{x} \\ &= \frac{5\times1.2\times10^{-6}\,\mathrm{m}\times2\,\mathrm{m}}{6.0\times10^{-3}\,\mathrm{m}} \\ &= 2.0\times10^{-3}\,\mathrm{m} \\ &= 2.0\,\mathrm{mm} \end{aligned} \)