Homework-21-sln - HW21 PDF

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HW21...

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Solution for Open-Response Homework 21 Light Solution to Open-Response Homework Problem 21.1(Transmission by a Plate of Glass) Problem: A pane of glass, nglass = 1.5, is in air. Light of a certain intensity strikes the pane at normal incidence; some of the light is reflected from the front surface, some of the light goes through the glass itself and is reflected from the back surface. Assume no secondary reflections. (a)What percentage of the original light is reflected from the front surface? (b)What percentage of the original light is reflected from the back surface? (c)What percentage of the original light is transmitted all the way through the pane? Definitions nglass = 1.5 ≡ Index of Refraction of Glass

I0

nair = 1 ≡ Index of Refraction of Air I0 ≡ Incident Intensity

I1

I1 ≡ Reflected Intensity from Top Surface I2 ≡ Reflected Intensity from Bottom Surface

I0 − I1

I2

IT ≡ Transmitted Intensity

IT

Strategy: Use formula for intensity of light reflected at normal incidence at each surface of the glass. Solution to Part (a) Compute Reflected Intensity from First Interface: related to the incident intensity, I0 , by I1 = I0

µ

nglass − nair nglass + nair

¶2

The light reflected from the top surface of the glass is

= I0

µ

1.5 − 1 1.5 + 1

¶2

= 0.04I0 = 4%I0 .

Grading Key: Part (a) 2 Points Solution to Part (b) (a) Compute Transmitted Intensity from First Interface: By conservation of energy, the transmitted intensity from the first interface is I0 − I1 = 0.96I0 . (b) Compute the Reflected Intensity from the Second Interface: The reflected intensity from the second interface is related to the intensity, I0 − I1 , in the glass by ¶2 ¶2 µ µ 1 − 1.5 nair − nglass = (0.96I0 ) I2 = (I0 − I1 ) = (0.96I0 )(0.04) = 3.8%I0 . 1.5 + 1 nair + nglass Grading Key: Part (b) 4 Points Solution to Part (c) Compute Total Transmitted Light: Energy is conserved, so I0 − I = I2 + IT and substituting the values from earlier, 0.96I0 = 0.038I0 + IT . Finally, solving for IT , IT = 0.922I0 = 92.2%I0 . 1

Grading Key: Part (c) 2 Points Total Points for Problem: 8 Points

Solution to Open-Response Homework Problem 21.2(Light in a Cable) Problem: A 300nm laser shines down a fiber optic cable with index of refraction 1.4. (a)What is the speed of the laser light, cf , in the cable? (b)What is the wavelength of the laser light, λf , in the cable? (c)What length of cable, d, is required, if the light is to pass through it in 0.1µs? Solution to Part(a) The speed of light in the material, cf , is the speed of light in a vacuum divided by the index of refraction: cf =

3 × 108 m m c s = = 2.14 × 108 . s n 1.4

Grading Key: Part (a) 2 Points Solution to Part(b) The wavelength of light in a material is reduced from the wavelength in a vacuum by a factor of the index of refraction, λf 300nm λf = = = 214nm. n 1.4 Grading Key: Part (a) 2 Points Solution to Part(c) The time, ∆t, to travel a distance d at a velocity, cf , is d = cf ∆t. d = (2.14 × 108

m )(1 × 10−7 s) = 21.4m s

Grading Key: Part (c) 2 Points Total Points for Problem: 6 Points

2...