Question

Width of the principal maximum on a screen at a distance of $50 \,cm$ from the slit having width $0.02\, cm$ is $312.5\times 10^{- 3} \, cm$ . If waves were incident normally on the slit, then wavelength of the light from the source will be

• A. $6000 \, Å$
• B. $6250 \, Å$
• C. $6400 \, Å$
• D. $6525 \, Å$

Answer 1

Answer: (B)

Width of central max: $= \frac{2 \lambda D}{a}$
$\lambda = \frac{0.2 \times 10^{- 2} \times 312.5 \times 10^{- 3} \times 10^{- 2} \, m}{2 \times \frac{1}{2}}$
$= 6250 \times 10^{- 10} m = 6250 \, Å$