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 $\lambda =10pm=100\times {10}^{-12}$ m

D= 40 cm = 40 $\times {10}^{-2}$ m

$\beta =0.1mm=0.1\times {10}^{-3}$ m

$\beta =\frac{\lambda D}{d}$

d =$\frac{\lambda D}{\beta }$

= $\frac{100\times {10}^{-12\times 40\times {10}^{-2}}}{{10}^{-3}\times 0.1}$

= 4 $\times {10}^{-7}$ m.