PHYS 3410/6750 Problem Solutions #8. Fall 2012, Prof. Bolton. Hecht 8.48: The
density of sucrose in solution in the tube is (10g)/(1000 cm. 3. )=0.01 g cm. −3.
PHYS 3410/6750 Problem Solutions #8 Fall 2012, Prof. Bolton Hecht 8.48: The density of sucrose in solution in the tube is (10 g)/(1000 cm3 ) = 0.01 g cm−3 . Thus its specific rotatory power for sodium light will be 0.01 g cm−3 × (+66.45◦ per 10 cm) = +0.6645◦ per 10 cm . 1 g cm−3 The total rotation of the polarization after passing through the 1-meter tube is 100 cm ×
+0.6645◦ = +6.645◦ . 10 cm
Hecht 9.3: The complex-form electric field for the wave propagating parallel to the z axis is E0 exp[i(kz − ωt)] , where k = 2π/λ. For the wave propagating at angle θ, the wavevector is k = xˆ k cos θ + yˆ k sin θ, and the complex-form electric field is E0 exp[i(k · r − ωt)] = E0 exp[i(kz cos θ + ky sin θ − ωt)] . In the xy plane, z = 0. Adding these two components together, the total complex-form electric field is E(y, z = 0) = E0 [exp(−iωt) + exp(iky sin θ − iωt)] = E0 [1 + exp(iky sin θ)] exp(−iωt) .
The irradiance is then I(y) = = = = = =
E2
T
1 ∗ EE 2 1 2 E {[1 + exp(iky sin θ)] exp(−iωt)} {[1 + exp(−iky sin θ)] exp(iωt)} 2 0 1 2 E [2 + exp(iky sin θ) + exp(−iky sin θ)] 2 0 1 2 E [2 + 2 cos(ky sin θ)] 2 0 E02 [1 + cos(ky sin θ)]
Then, using cos 2α = 2 cos2 α − 1, we get I(y) = 2E02 cos2 (ky sin θ/2) ,
PHYS 3410/6750 or
Problem Solutions #8 2
I(y) = 2E02 cos2 [(πy/λ) sin θ] .
(I don’t know where Hecht gets the additional factor of 2.) Constructive fringes will occur when the argument of the cosine is an integer multiple of π: (πy/λ) sin θ = mπ ⇒ y =
mλ . sin θ
Destructive fringes (zeros of irradiance, part a of Quiz 8) will occur when the argument of the cosine is an “integer-plus-half” multiple of π: (m + 12 )λ 1 . (πy/λ) sin θ = (m + )π ⇒ y = 2 sin θ The fringe separation (part b of Quiz 8) is ∆y = λ/ sin θ, and it decreases as θ increases (part c of Quiz 8). This analysis is similar to the analysis leading up to Equation 9.17, in the limit that the two sources get infinitely far away, but maintain the same angular separation as seen from the region of space where we are considering the irradiance distribution.
