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

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4