A laser beam enters a prism from air. For a symmetrical prism (one in which
the apex angle lies at the top of an isosceles triangle), the total deviation
angle δ of the beam is minimized when the light ray inside the prism travels
parallel to the prism's base.
(a) If the index of refraction of the glass is n and the apex angle is A, find total deviation angle δmin.
(b) Give a numerical value for δmin if A = 60o and n = 1.5.
A beam of monochromatic light of wavelength λ in vacuum is incident normally on a nonmagnetic dielectric film of refractive index n and thickness d. Calculate the fraction of the incident energy that is reflected.
ki = kr = kt = k, k1 = k2
For z < 0 we have
E(r,t) = i [Ei exp(i(kz - ωt)) - Er exp(i(-kz - ωt))],
B(r,t) = j [Ei exp(i(kz - ωt)) + Er exp(i(-kz - ωt))]/c,
H(r,t) = B(r,t)/μ0.
In the film for 0 < z < d we have
E(r,t) = i [E1 exp(i(nkz - ωt)) - E2 exp(i(-nkz - ωt))],
B(r,t) = j [E1 exp(i(nkz - ωt)) + E1 exp(i(-nkz - ωt))](n/c), H(r,t) = B(r,t)/μ0.
For z > d we have
E(r,t) = i Et exp(i(kz - ωt)),
B(r,t) = j Et exp(i(kz - ωt))/c. H(r,t) = B(r,t)/μ0.
The boundary conditions at z = 0 for the tangential component of E yield
Ei - Er = E1 - E2.
The boundary conditions at z = 0 for the tangential component of H yield
Hi + Hr = H1 + H2. H = B/μ0 in all regions, B = En/c in the film, B = E/c outside the film.
Ei + Er = n(E1 + E2).
Measuring all field strength in units of Ei we have express E1 and E2 in terns of Er.
Er = -E1 + E2 + 1, Er = nE1 + nE2 - 1. E2 = ((n + 1)Er - n + 1)/(2n), E1 = (n + 1 - (n - 1)Er)/2n.
The boundary conditions at z = d for the tangential component of E yield
E1exp(inkd) - E2exp(-inkd) = Etexp(ikd).
The boundary conditions at z = d for the tangential component of H yield
H1exp(inkd) + H2exp(-inkd) = Htexp(ikd).
nE1exp(inkd) + nE2exp(-inkd) = Etexp(ikd).
-(n + 1)E2exp(-inkd) = (n - 1)E1exp(inkd).
Substituting E1 and E2 in terms of Er from above we get
-(n + 1)((n + 1)Er - n + 1)exp(-inkd) = (n - 1)(n + 1 - (n - 1)Er)exp(inkd).
Er[(n + 1)2exp(-inkd) - (n - 1)2exp(inkd)] = (n2 - 1)exp(-inkd) - (n2 - 1)exp(inkd).
Er[4n cos(nkd) + i2(n2 + 1) sin(kd)] = -i(n2 - 1)2sin(kd).
R = |Er|2 = 4(n2 - 1)2sin2(kd)/[16n2 cos2(nkd) + 4(n2 + 1)2 sin2(kd)].
R = (n2 - 1)2/[4n2 cot2(nkd) + (n2 + 1)2]
is the reflectance, i.e. the fraction of the incident energy that is reflected.
(a) Write down Maxwell's equations for a conducting medium with conductivity σ,
permittivity ε0 and permeability μ0,
(b) A plane wave of low frequency ω << σ/ε0 is propagating in the z-direction inside the conducting medium.
Let E = E0exp(i(kz - ωt)), B = B0exp(i(kz - ωt)), where E0, B0, and k are complex.
Use Maxwell's equations to show that k = k1 + ik2, k1 ≈ k2 ≈ (μ0ωσ/2)½.
Calculate the ratio of the complex amplitude of the two fields, E0/B0 (magnitude and phase).
(c) Calculate the energy flux (time averaged Poynting vector) in the conducting medium.
In a double-slit experiment with 500 nm light, the slits each have a width of
(a) If the interference fringes are 5 mm apart on a screen which is 4 m from the slits, determine the separation of the slits.
(b) What is the distance from the center of the pattern to the first diffraction minimum on one side of the pattern?
(c) How many interference fringes will be seen within the central maximum in the diffraction pattern? Draw a sketch of the pattern.
Light is incident along the normal on face AB of a glass prism of index of refraction 1.52, as shown in the figure.
Find the largest value the angle α can have such that there is no light
refracted out of the prism at face AC if
(a) the prism is immersed in air.
(b) the prism is immersed in water.
At the AC face, under the condition for total internal
reflection we have that α + θcrit = 90o.
(a) For a glass-air interface, na = 1.52 and nb = 1.00.
We have that sinθcrit = 1/1.52. Thus, θcrit = 41.1o. We conclude that α = 90o - θcrit = 48.9o.
(b) In this case the refraction problem corresponds to glass -->water, for which nb = 1.33. Therefore sinθcrit = 1.33/1.52 , or θcrit = 61.3o. We thus have that α = 90o - θcrit = 28.7o.