April 14, 2009
PHY2054 Discussion-Spring ‘09
Quiz 9 (Chapter 23.6-24.5)
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 = mλn or (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 λ1 occurs at
the same position as the seventh bright fringe of the λ2. What is the value of λ2 if λ1 is 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