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 = dsinq, where d is the slit separation. Then, solve for the position of the fringe, y, with y/L = tanq. By making the approximation q = tanq = sinq which is true for small angles, the algegra is simplified.

y = 8 cm

Example #2

Problem:

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

Solution:

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

l = 345.6 nm

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 = asinq, where a is the slit wdith. Then, solve for the position of the fringe, y, with y/L = tanq. By making the approximation q = tanq = sinq which is true for small angles, the algegra can be simplified.

y = 6.67 cm

Example #5

Problem:

A spy satellite travels at a distance of 50 km above Earth's surface. How large must the lens 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 q = 1.22 l/D = y/L. Here 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

Interference and diffraction's index