Question

In a Young’s double-slit experiment the slit separation is $0.5\, mm$ and the screen is $0.5\, m$ from the slit. For a monochromatic light of wavelength $500\, nm$ the distance of third maxima from the second minima on the other side is

• A. 2.75 mm
• B. 2.5 mm
• C. 22.5 mm
• D. 2.25 mm

Answer 1

Answer: (D)

Here, $d = 0.5 \,mm = 0.5 × 10^{-3}\, m$
$D = 0.5\, m, \lambda = 500\, nm = 500 \times 10^{-9}\, m$
The distance of third maxima from the second minima on the other side is
$=\frac{9}{2}\beta$
(where $\beta$ is the fringe width)
$=\frac{9}{2} \frac{\lambda D}{d}=\frac{9\times500\times10^{-9}\times0.5}{2\times0.5\times10^{-3}}$
$=2.25\times10^{-3}\,m=2.25\,mm$