Question

Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through a Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm?

Answer 1

Given:
$$\begin{gathered} \lambda =10\mathrm{pm}=100\times {10}^{-12}\mathrm{m} \\ \mathrm{D}=40\mathrm{cm}=40\times {10}^{-2}\mathrm{m} \\ \mathrm{\beta }=0.1\mathrm{mm}=0.1\times {10}^{-3}\mathrm{m} \\ \mathrm{\beta }=\frac{\mathrm{\lambda D}}{d} \\ d=\frac{\mathrm{\lambda D}}{\mathrm{\beta }} \\ =\frac{100\times {10}^{-12}\times 40\times {10}^{-2}}{{10}^{-3}\times 0.1} \\ =4\times {10}^{-7}\mathrm{m} \end{gathered}$$