Question

In Young's double slit experiment the slit separation is $0.5m$ from the slits. Distance between the screen and slit is $0.454mm$ For a monochromatic light of wavelengths $500nm$, the distance of $3^{rd}$ maxima from $2^{nd}$ maxima on the other side is

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

Answer 1

Answer: (A)

$2.75mm$

The distance of third maxima from the central maxima=$3\beta =\frac{3d\lambda }{D}$

The distance of second maxima from central maxima=$2\beta =\frac{2d\lambda }{D}$

Hence distance between the two maxima=$\frac{5d\lambda }{D}$

$=\frac{5\times 0.5\times 500\times {10}^{-9}}{0.454\times {10}^{-3}}m$

$=2.75mm$