HW2soln - plot(t,d xlabel'entrance angle[rad ylabel'total...

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1. 2-4 Assuming eyes at top of the head, the law of reflection allows you to see all of yourself with a mirror half your height, the same height off the floor. 2. 2-6 The second image is from reflection off the bottom surface. The ratio of distances (3mm/1.87mm) is the same as the ratio of sines of the angles; by Snell’s law, this is the ratio of refraction index (1.6) θ θ
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3. 2-7 n g sin θ c =n air sin π /2~1 θ c =asin(1/n g )=atan(1.9/2.25) n g =1.55 4. 2-10 Reflected image: Use Eq. 2-12 to find s’=-7.5cm. Refracted image: 5cm/1.5=10/3cm from flat surface. See problem 8 below. The reflected image therefore appears 10cm below the flat surface. 5. 2-32 At any interface i , n i sin θ i =n i+1 sin θ i+1 At interface i+1, n i+1 sin θ i+1 =n i+2 sin θ i+2 so n i sin θ i =n i +2sin θ i+2 by induction, n i sin θ i =n f sin θ f 7.6cm 1.9cm θ c θ c 2.25cm
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6. 3-6 clear ng=1.52; ii=1; for theta=0.55:0.0001:1.5 theta_i=asin(sin(theta)/ng); theta_0=asin(ng*sin(pi/3-theta_i)); delta=theta+theta_0-pi/3; t(ii)=theta; d(ii)=delta; ii=ii+1; end
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Unformatted text preview: plot(t,d); xlabel('entrance angle [rad]'); ylabel('total deviation angle [rad]') %check [a,b]=min(d) 2*t(b)-pi/3 a = 0.67943 b = 3134 ans = 0.67940 π /3 θ θ i θ 0 δ θ-θ i π/3-θ π/3-θ i So minimum deviation is ~0.6794 rad. 7. 3-7 Just re-do the above w/ ng=1.525 and 1.535 and take the difference: 0.7026-0.6872=0.0154 rad (~0.9 deg) 8. Prove that to someone looking straight down into a swimming pool, any object in the water will appear to be 3/4 of its true depth. (HINT: n water =4/3) a a w w n n θ sin sin = 2 2 2 2 x a x x d x n w + = + d a θ a x θ w 2 x a + d For infinitesimal x, 4 / 3 1 = = w n d a 9. Light is incident in air perpendicularly on a sheet of crown glass having an index of refraction of 1.552. Determine both the reflectance and the transmittance. 2 +-= t i t i n n n n R =0.043 ( ) 2 4 t i t i n n n n T + = =0.957...
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HW2soln - plot(t,d xlabel'entrance angle[rad ylabel'total...

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