April 14, 2009

PHY2054Discussion-Spring‘09

Quiz 9 (Chapter23.6-24.5)

Name: UFID:

**1. (2.5pts)An object is placed 40.0 cm to the left of a converging lens with a focal length of 30.0 cm. A spherical mirror with a radius of curvature of -40.0 cm is located 60.0 cm to the right of the lens. Locate the final image.

The light ray from the object passes the lens from left to right, hits the mirror and then passes the lens from right to left before forming the final image. The position of the image due to the lens is

1/p+1/q1 = 1/f ⇒q1 = (1/f-1/p)-1 = (1/30-1/40 -1 = 120 cm

This image serves as an object for the mirror. The object distance and the image distance are

|p2| = q1-60 = 60 cm ⇒p2 = -60 cm (in the back of the mirror)

1/p2+ 1/q2 = 2/R ⇒q2 = (2/R-1/p2)-1 = [2/(-40)-1/(-60)]-1 = -30 cm (behind the mirror)

This image serves as an object for the lens. Now the front side is to the right of the lens and the back is to the left. The final image is located

p3 = |q2|+60 = 90cm ⇒p3 = +90 cm (in front of the lens)

1/p3+1/q = 1/f ⇒q = (1/f-1/p3)-1 = (1/30-1/90)-1= 45.0 cm to the left of the lens.

**2. (2.5pts)A piece of transparent material is cut into the shape of a wedge with length l = 10.0 cm and height h = 30.0 μm. The angle of the wedge is very small. Monochromatic light of wavelength of 600. nm is normally incident from below. As view from above, the 100th bright fringe is observed at the position x = 7.00 cm. What is the index of refraction of the material?

Let the thickness of the wedge at the 100th fringe be t. Then because of the similarity of triangles we have

t/x = h/l ⇒t = h(x/l) = 30×(7/10) = 21 μm

Since phase changes occur at bottom surface and top surface for transmitted light, the number of phase change is two. Applying an appropriate equation for constructive interference, we get

2nt = mλ ⇒n = mλ/(2t) = 100×600×10-9/(2×21×10-6) = 1.43

Constants & Formulas

Lens eqn:1/p+1/q =2/R, M = -q/p / Interference Condition: δ = mλnor (m+1/2)λn
Young’s experiment: dsinθ = mλ or (m+1/2)λ / Thin films: 2tn = mλ or (m+1/2)λ

*3. (2.5pts) In a Young’s double slit experiment, a beam containing light of wavelengths λ1 and λ2 is incident on a set of parallel slits. In the interference pattern, the sixth bright fringe of the λ1occurs at the same position as the seventh bright fringe of the λ2. What is the value of λ2if λ1is known to be 500. nm.

The angular positions of the 6th bright fringe for λ1 and 7th bright fringe for λ1 are respectively

sinθ6 = 6λ1/d & sinθ7 = 7λ2/d

They locate at the same position, therefore

6λ1/d = 7λ2/d ⇒λ2/d = (6/7)λ1 = (6/7)×500 = 429 nm

***4. (2.5pts) A light with an wavelength of 550 nm is incident on a thin film at an angle of 60.0 to the normal. If the index of refraction of the film is 1.50, what is minimum thickness of the film for destructive interference in reflected light?

The phase shift occurs at the bottom surface at the reflected light, thus the number of phase shift is one. The effective path difference is

δ = r1 –r2 = (AP+PO)-BO = 2AP-BO = 2nt/cosθ2-2ttanθ2sinθ1

Applying Snell’s law, this reduces to

δ = 2nt cosθ2-2t(sinθ2 /cosθ2)nsinθ2 = 2nt/cosθ2 (1-sin2θ2) = 2ntcosθ2

where θ2 = sin-1(sinθ1/n) = 35.3º

Choosing a correct equation for destructive interference, the minimum thickness is

δ = mλ ⇒2ntcosθ2 = 1λ ⇒t = λ/(2ncosθ2) = 550/(2×1.5×cos35.3º) = 225 nm