The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

Question

Question

Answer 1

In diffraction

d sin 30° = λ

\( \Rightarrow \lambda = \frac{d}{2}\)

Now, let x is the separation between two slits.

Young’s fringe width (ω) = 1 cm = 10^-2 m

\(\omega = \frac{{\lambda D}}{x}\)

\({10^{ - 2}} = \frac{{d \times 50 \times {{10}^{ - 2}}}}{{2x}}\)

⇒ x = 25 d = 25 × 1 μm = 25 μm