Examples for interference and diffraction

Example #1

Problem:

A screen is placed 3.0 m from a two-slit setup with the slits separated by 15 mm. If the wavelength of the light is 4000 nm, how far apart are the principal and m = 1 fringes?

Solution:

First, solve for the angle q of the maximum using l = d sin q, where d is the slit separation. Then, solve for the position of the fringe, y, with y / L = tan q . By making the approximation tan q = sinq , which is true for small angles, you can simplify the algebra.

y = 8 cm

Example #2

Problem:

A diffraction grating with 12 thousand lines per cm creates a bright line at 24.5 degrees away from the central bright line. What is the wavelength of the light?

Solution:

The separation between slits d on the grating is 1/12,000. Using l = d sin q, one obtains

l = 345.6 nm

You could also compute where the next bright line appears...

You could also compute how many bright lines appear (based on the fact that they can't extend beyond 90 degrees)...

Example #3

Problem:

Which of the formulae (a or b) does one use to find the thickness of a film to give an interference maximum for reflected light?

a.) 2t = (m+1/2)l         b.) 2t = ml

1.) Light comes from the vacuum and reflects off a soap film floating in air.

Use formula a because only one reflection is from a lower-to-higher n surface.

2.) Light comes from the vacuum and reflects off a soap film floating over glass.

Use formula b because both reflections are from lower-to-higher n surfaces.

3.) Light comes from the glass and reflects off a soap film with vacuum on the other side.

Use formula b because neither reflection is from a lower-to-higher n surface.

Example #4

Problem:

Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit, how far is the first minimum from the central maximum?

Solution:

First, solve for the angle q of the minimum using l = a sin q, where a is the slit wdith. Then, solve for the position of the fringe, y, with y / L = tan q .

y = 6.67 cm

Example #5

Problem:

A spy satellite travels at a distance of 50 km above Earth's surface. How large would the lens have to be so that it can resolve objects of 2 mm and thus read a newspaper? Assume the light has a wavelength of 400 nm.

Solution:

Diffraction limits the resolution according to sin q = 1.22 l /D. We also have tan q = y / L. The height of the object one wishes to resolve is y and the distance to the object is L. Solving for D, one gets

D = 12.2 m

(This is too large to build, even with military budgets; but only by about one order of magnitude, so presumably they can at least read the headlines.)