In a Young's experiment, the two slits are separated by 0.5 mm, and they are 0.5 m from the screen. The wavelength of light used is 5000 Å. The distance between 7th maxima and 11th minima on the screen is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.75 mm
In Young’s double slit experiment the fringe width is \( \beta = \lambda D/d \). Measured from the central maximum, the nth bright fringe is at \( n\beta \) and the nth dark fringe at \( (2n - 1)\beta/2 \).
\( \lambda = 5000\,\mathring{\mathrm{A}} = 5\times10^{-7}\,\mathrm{m} \)
\( d = 0.5\,\mathrm{mm} = 5\times10^{-4}\,\mathrm{m} \)
\( D = 0.5\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{5\times10^{-7}\,\mathrm{m} \times 0.5\,\mathrm{m}}{5\times10^{-4}\,\mathrm{m}} \)
\( \beta = 5\times10^{-4}\,\mathrm{m} = 0.5\,\mathrm{mm} \)
\( y_{7,\ \mathrm{bright}} = 7\beta = 3.5\,\mathrm{mm} \)
\( \displaystyle y_{11,\ \mathrm{dark}} = \frac{(2 \times 11 - 1)\beta}{2} = 10.5\beta = 5.25\,\mathrm{mm} \)
\( \Delta y = 5.25\,\mathrm{mm} - 3.5\,\mathrm{mm} = 1.75\,\mathrm{mm} \)
Both fringes are taken on the same side of the central maximum, the usual meaning; on opposite sides they would be 8.75 mm apart, which is not among the options.