Physics 111 Homework Solutions Week #8 - Monday

Transcription

1 Physics 111 Homework Solutions Week #8 - Monay Thursay, February 18, 2010 Chapter 19 Questions - None Multiple-Choice 1914 D 1915 B Problems 1913 The istance away is given by 1914 The istance away is given by converting to astronomical units we have 1916 The laser pointer is rate at 3mW which is 3x10-3 J/s an this energy (per secon) is sprea over the area of the beam spot The area of the beam spot is Thus the raiation pressure is Friay, February 19, 2010 Chapter The spee is inversely proportional to the inex of refraction Therefore the material with the highest inex of refraction will have the lowest spee We have from lowest spee to greatest spee: iamon, crown glass, water, air 205 Refer to class notes for the geometric argument 207 From figure 2010 we see that as the wave fronts enter a higher refractive inex material their spees slow own an the wave fronts bunch of, just like the soliers marching through the stream slow own an bunch up This is consistent with equation 205, which is Snell s law Noting that n is the ratio of the spee of light in vacuum to its spee in the material, the inex of refraction scales with 1/n

2 an also with 1/λ as well since v = fλ in each meium an as the ray crosses a bounary its frequency oes not change, only its wavelength 2010 At each air/glass interface, 4% of the light is reflecte On the first pane at the upper air/glass interface, 96% of the light is transmitte into the glass At the lower glass/air interface 4% of the incient light is reflecte an thus 92% is transmitte into the air pocket between the panes At the upper air/glass interface for the secon pane, 4% is reflecte an 88% transmitte, an at the lower glass/air interface 4% reflecte an 84% transmitte into room Thus the total amount of reflecte light is approximately 16% Glass upper pane air Glass lower pane Multiple-Choice 206 D 207 C 208 B 2010 D 2011 B 2012 A 2017 C 2018 C Problems 206 A narrow pencil of light striking a fish tank a Assuming that the inex of refraction is 155, the angle of refraction is given for the light ray going from air into glass as: b As the ray passes through the glass it will eventually strike the interface between the glass an the water at 188 o For water the inex of refraction is 133 an the angle of refraction in the water is given as:

3 c From the rawing we can see that an so that the ifference between where the beam strikes an where it is aime Δ = 119mm 30 o 112 o 188 o Δ x 221 o 5mm 207 Using the iagram below we have at the upper surface, using Snell s law At the lower surface we have the ray striking at 195 o with respect to the normal an from this we can etermine the exit angle, θ 3 Applying Snell s law we have an the ray leaves parallel to itself However the ray is isplace by an amount from its incient irection To etermine the isplacement of the beam we first realize that 30 o = θ 2 + α, so that α = 30 o 195 o = 105 o Therefore from the geometry we have Where again from the geometry the path the light takes is 30 o 2cm θ 2 L α θ 3

4 208 In the meium we have the spee of light given by an this correspons to a wavelength in the meium of In the air we have the wavelength given as time for the light to cover a istance L in the material as Using the results from problem #7, we have the 2015 We fin the critical angle from The raius of the ring of light is given from 2016 We use Snell s Law with Defining x as the istance from the normal to the surface to where the ray originates, we can express θ 1 in terms of an x an θ 2 in terms of x an as follows: an Since the angles involve are small, we can use the small angle approximation to get an Therefore, Thus the apparent epth is given as A ray iagram is shown below n air = 100 x n water = 133 θ 2 θ 1 Box appears here

5 2017 From the iagram below we apply Snell s Law at the upper surface where we want the ray to strike at the critical angle We have therefore From the geometry we see that the angle of refraction of light off of the front surface of the pipe, α is Therefore θ can be foun by applying Snell s Law on the front surface an we have θ α ο 90 θ c