Question

Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

Answer 1

0.276

0.9

0.756

Step-by-step explanation:

Given that

Wavelength of the light, λ = 588 nm

Distance from the slit to the screen, L = 2.7 m

Width of the slit, a = 0.0351 mm

a point on the screen, y = 1.3 cm = 0.013 m

Sinθ = y/L

Sinθ = 0.013/2.7

sinθ = 0.0081

θ = sin^-1 0.00481

θ = 0.276°

α = (π.a.sinθ)/λ

α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9

α = 5.3*10^-7 / 588*10^-9

α = 0.9 rad

I/i(m) = ((sinα)/α)²

I/I(m) = ((sin 0.9) / 0.9)²

I/I(m) = (0.783/0.9)²

I/I(m) = 0.87²

I/I(m) = 0.756

Note, our calculator has to be set in Rad instead of degree for part C, to get the answer