Question

You are given a visible laser of wavelength λ to study interferences of light in the lab.

Consider the 5 situations below. Treat each question independently

1.In air, you place a screen with two slits, separated by 0.86 mm, in front of the laser of wavelength 507.9 nm. You know that you will see an interference pattern if you place an observation screen some distance away. Determine what the distance between the plane of the fringes and the observation screen should be (in meters) for the adjacent bright fringes to be 1.19 mm apart.

2.In air, you place a screen with a single slit of 0.548 μm in front of the laser of wavelength 433.9 nm. The interference pattern you observe for the single slit is different than the one you observed for the double slit. What is the angle (in degrees) of the first dark fringe above the center maximum on the observation screen?

3.When illuminating the observation screen without any slits, you notice that the laser has a spread, meaning that the light on the screen becomes larger as you increase the distance between you and the screen.

Determine the angular spread (in milli-rad) of the 503.9 nm wavelength laser beam if the laser aperture is 0.89 mm in diameter.

4.Now you consider the phenomenon of interferences in thin films. In the lab, you find a thin film of medium 1 (n=1.4) on top of a glass slide (n=1.51). You use the laser of wavelength 588.9 nm to illuminate the thin film. What is the smallest thickness (in nm) at which you will see a destructive interference?

5.Considering the light of wavelength 323.2 nm travels in air. What will be the wavelength of light (in nm) in the medium of the thin film of index 1.12?

Answer #1

1.

d = separation between the two slits = 0.86 mm = 0.86 x 10-3 m

= wavelength = 507.9 nm = 507.9 x 10-9 m

w = distance between adjacent bright fringes = 1.19 mm = 1.19 x 10-3 m

D = ?

Using the equation

w = D /d

1.19 x 10-3 = ( 507.9 x 10-9) D/(0.86 x 10-3)

D = 2.015 m

2.

d = width of the slit = 0.548 μm = 0.548 x 10-6 m

= wavelength = 433.9 nm = 433.9 x 10-9 m

= Angle

n = 1

Using the equation

d Sin = n

(0.548 x 10-6) Sin = (1) (433.9 x 10-9)

Sin = 0.792

= 52.4 deg

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