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

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

Tardigrade Admin · Accepted Answer

Here, d = 0.5 mm = 0.5 Ã 10-3 m D = 0.5 m, λ = 500 nm = 500 × 10-9 m The distance of third maxima from the second minima on the other side is =(9/2)β (where β is the fringe width) =(9/2) (λ D/d)=(9×500×10-9×0.5/2×0.5×10-3) =2.25×10-3 m=2.25 mm

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

2.75 mm

2.5 mm

22.5 mm

2.25 mm

Here, d=0.5mm=0.5×10−3m D=0.5m,λ=500nm=500×10−9m The distance of third maxima from the second minima on the other side is =29​β (where β is the fringe width) =29​dλD​=2×0.5×10−39×500×10−9×0.5​ =2.25×10−3m=2.25mm

Q. 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