Question

A plane wave of wavelength 6250 $\stackrel{o}{A}$ is incident normally on a slit of width 2 $\times$ 10${}^{-2}$ cm. The width of the principal maximum on a screen distant 50 cm will be

• A. 312.5 $\times$ 10${}^{-3}$ cm
• B. 32.5 $\times$ 10${}^{-3}$ m
• C. 312.5 $\times$ 10${}^{-3}$ m
• D. 312 m

Answer 1

Answer: (A)

312.5 $\times$ 10${}^{-3}$ cm

Width of central maximum
= $\frac{2\lambda D}{a}=\frac{2\times 6250\times {10}^{-10}\times 0.5}{2\times {10}^{-4}}$
= 3125 $\times$ 10${}^{-6}$ m = 312.5 $\times$ 10${}^{-3}$ cm.
Screen position of various minima for Fraunhoffer diffraction pattern of a single slit of width a.