Question

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm? Answer: 0

Answer 1

Question 1)

Givens It is given that the wavelength of the incident light is 450 nm, and that the distance between the centers of the two slits is 0.0500 mm, hence we have d = 0.0500 x 10-3 m X= 450 10-9 m

Calculations Substituting the givens in equation (1) we can find the angular position of the first order maximum m = 1, where 01 = sin-1 (**) 1 sin 450 x 10-9 0.0500 x 10-3 -1 = sin (9.00 10-3) ~0.516° Thus, the angular position of the first maximum is 0.516 degree.

Question 2)

2(720 x 10-9 = = d sin(0) the smallest separation between two slits then sin(0) =1 therefore we have 2(720 10-9 d = = 1.44 x 10-6 m 1

Therefore, the smallest separation between two slits that will produce a second-order maximum for 720-nm red light is,

d = 1440 nm