Question

Parallel rays of monochromatic light with wavelength 587 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm.

If the intensity at the center of the central maximum is 3.00×10−4 W/m2 , what is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?

Answer 1

Given :-

= 587 nm = 587 x 10^-9 m

R = 75 cm = 0.75 m

d = 0.640 mm

a = 0.434 mm

Io = 3 x 10^-4 w/m^2

y = 0.710 mm

tan(theta) = y/R = 0.710 x 10^-3 / 0.75 = 9.47 x 10^-4

theta is too small so sin(theta) = tan(theta)

= 2pi*d*sin(theta) /

= (2pi x 0.640 x 10^-3 x 9.47 x 10^-4) / 587 x 10^-9

= 6.487 rad

= 2pi*a*sin(theta) /

= (2pi x 0.434 x 10^-3 x 9.47 x 10^-4) / (587 x 10^-9)

= 4.399 rad

I = Io*cos^2(/2)[(sin(/2) / /2)^2]

I = 3 x 10^-4 x cos^2(3.2435)[(sin(2.2)/2.2)^2]

I = 9.1 x 10^-8 w/m^2