Question

In double slit experiment illuminated by coherent 600 nm light, the slit
separation is 0.2 mm and slit to screen distance is 2 m. At distance y=6mm, what is
the average intensity on the screen of the maximum?

Answer 1

average intensity on the screen is equal to maximum intensity.

It has given that wavelength of light , λ = 600 nm

the slit seperation, d = 0.2 mm and distance between slits and distance between slits and screen , D = 2m , distance, y = 6mm

we know,

path difference , ∆x = yd/D

= 6mm × 0.2mm/2m

= 0.6 × 10¯⁶ m = 600 nm

phase difference, Φ = 2π/λ ∆x

= 2π/600nm × 600nm

= 2π

let I is the maximum intensity in Young's double slit experiment.

so, average intensity, I' = Icos²(Φ/2)

= Icos²(2π/2)

= I × (-1)²

= I

therefore, average intensity on the screen is equal to maximum intensity