UPDATE: on number 3, you should get about 2.25m and 2.1m (not 2.6m and 2.1m).
Some stray HW2 hints:
(1) Draw a normal to the prism surface at the incident and exit points. Let θ1 be the incident angle (relative to the normal) and θ4 the exit angle. Let θ2,3be the refracted angles relative to the normal on the incident and exit sides, respectively. The deviation angle is then θ1-θ2+θ4-θ3.
Use the triangle formed by the incident and exit rays and the apex (angle φ), all the angles sum to 180 deg. Apply Snell's law, and you're basically done. About 4.6deg.
(2) Use (3.70) in the text. Presume the 2nd term under the radical is small compared to 1 (try to justify this), and note sqrt(1+x)~1+x/2 for x<<1.
(3). The lateral position of the object is unchanged, only the depth is different. To see this, think about rays emanating from the object at the bottom of the well and trace them to the observer's eye. The apparent position is where the ray reaching the observer would extrapolate back to the horizontal position of the object. About 2.25 and 2.1m, respectively.
(4). Check out this figure. The angular rotation rate of the earth is ω=2π/86400s, and the angular deviation is δθ=ωδt. Double that for sunrise and sunset. About 160s.
(5). Use the result of problem 1, noting by symmetry that θ1=θ4 and θ2=θ3.
(6). Carry the sums through (i.e., you needn't bother just keeping the first term), but approximate to first order 1/(1-x)~1+x and sqrt(1+x)~1+x/2 for x<<1.
(7). We'll do this one in class on Tuesday. The time it takes to cross a distance Δy in air is Δy/c. Crossing the same distance of glass means a time of nΔy/c. The difference between the two is the phase lag the observer sees, basically an offset in the observer's clock. Without the plate, the wave would be exp(iωt - y/c) to account for the frequency and propagation delay across a distance y.
(8). We'll do this in class. Re-write n sin(θ) as a cross product on each side of Snell's law ...
Some stray HW2 hints:
(1) Draw a normal to the prism surface at the incident and exit points. Let θ1 be the incident angle (relative to the normal) and θ4 the exit angle. Let θ2,3be the refracted angles relative to the normal on the incident and exit sides, respectively. The deviation angle is then θ1-θ2+θ4-θ3.
Use the triangle formed by the incident and exit rays and the apex (angle φ), all the angles sum to 180 deg. Apply Snell's law, and you're basically done. About 4.6deg.
(2) Use (3.70) in the text. Presume the 2nd term under the radical is small compared to 1 (try to justify this), and note sqrt(1+x)~1+x/2 for x<<1.
(3). The lateral position of the object is unchanged, only the depth is different. To see this, think about rays emanating from the object at the bottom of the well and trace them to the observer's eye. The apparent position is where the ray reaching the observer would extrapolate back to the horizontal position of the object. About 2.25 and 2.1m, respectively.
(4). Check out this figure. The angular rotation rate of the earth is ω=2π/86400s, and the angular deviation is δθ=ωδt. Double that for sunrise and sunset. About 160s.
(5). Use the result of problem 1, noting by symmetry that θ1=θ4 and θ2=θ3.
(6). Carry the sums through (i.e., you needn't bother just keeping the first term), but approximate to first order 1/(1-x)~1+x and sqrt(1+x)~1+x/2 for x<<1.
(7). We'll do this one in class on Tuesday. The time it takes to cross a distance Δy in air is Δy/c. Crossing the same distance of glass means a time of nΔy/c. The difference between the two is the phase lag the observer sees, basically an offset in the observer's clock. Without the plate, the wave would be exp(iωt - y/c) to account for the frequency and propagation delay across a distance y.
(8). We'll do this in class. Re-write n sin(θ) as a cross product on each side of Snell's law ...
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