In a double-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by \( 5\times10^{-2}\,\mathrm{m} \) towards the slits, the change in fringe width is \( 3\times10^{-5}\,\mathrm{m} \). If the distance between the slits is \( 10^{-3}\,\mathrm{m} \), the wavelength of the light used is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6000 Å
The fringe width in a double-slit experiment is \( \beta=\lambda D/d \). With the same light and slits, \( \beta \) is proportional to the screen distance \( D \). So a change \( \Delta D \) in the distance changes the fringe width by \( \Delta\beta=\lambda\,\Delta D/d \).
\( \begin{aligned} \lambda &= \frac{\Delta\beta\,d}{\Delta D} \\ &= \frac{3\times10^{-5}\,\mathrm{m}\times10^{-3}\,\mathrm{m}}{5\times10^{-2}\,\mathrm{m}} \\ &= 6\times10^{-7}\,\mathrm{m} \end{aligned} \)
Since 1 Å = \( 10^{-10}\,\mathrm{m} \), \( \lambda=6\times10^{-7}\,\mathrm{m} \) = 6000 Å.