Photonics Midterm 1 – Worked Solutions
Q1. Polarized sunglasses
Chapter: Ch. 6 Electromagnetic Optics → 6.6 Reflection and refraction (Brewster angle)
Specular (mirror-like) reflection of sunlight on a wet road surface can obstruct your view when driving a car.
The specular reflected light will be mostly horizontally polarized when the sun is low in the sky (large angle of
incidence). That is why polarized sunglasses block this polarization to eliminate glare from the road surface.
Calculate the Brewster angle (angle with the normal) for a puddle of water on the road using nwater = 1.3.
Reasoning:
At the Brewster angle the reflected light is purely TE (s-polarized) and
tan θB =
1.3
n2
=
n1
1.0
incident
θB = arctan(1.3) ≈ 52.4◦ .
=⇒
reflected (TE)
θB
n1 = 1
n2 = 1.3
Answer: (b) 52◦
Q2. Transmission at a boundary (TE)
Chapter: Ch. 6 Electromagnetic Optics → 6.6 Reflection and refraction (Fresnel TE)
TE polarized light arrives at a boundary between two media. The incoming P1 light has a power of 1.00 mW.
The refractive indices of the two media are n1 = 3 and n2 = 2. The light is incident on the boundary at an
angle of 30 degrees with the normal. What is the power of the light transmitted through the boundary?
Reasoning:
Snell’s law:
sin θ2 =
n1
3
sin θ1 = sin 30◦ = 0.75 ⇒ θ2 ≈ 48.6◦ .
n2
2
Fresnel TE reflection coefficient:
rT E =
n1 cos θ1 − n2 cos θ2
3(0.866) − 2(0.661)
=
≈ 0.325.
n1 cos θ1 + n2 cos θ2
3(0.866) + 2(0.661)
R = |rT E |2 ≈ 0.106,
T = 1 − R ≈ 0.89.
P2 = T · P1 ≈ 0.89 mW.
n1 = 3
n2 = 2
θ1 =30◦
θ2
Answer: P2 ≈ 0.89 mW
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Q3. Snell’s law on a stack of materials
Chapter: Ch. 3 Geometric Optics → 3.2.5 Interface between homogeneous media (Snell’s law)
Light from a material with index n0 is incident at angle θ0 on a stack of three different materials with different
indices: n1 , n2 , and n3 . What is the angle of the light in the third material (θ3 ) expressed in the refractive
indices of the other materials?
Reasoning:
Snell’s law applies at every interface:
n0 sin θ0 = n1 sin θ1 = n2 sin θ2 = n3 sin θ3 .
The intermediate indices cancel, leaving
θ3 = sin−1
$n0 $
Answer: (a) θ3 = sin−1
n0
sin θ0
n3
n0
sin θ0 .
n3
$n1 $
$n2 $
$n3 $
Q4. TE polarization direction
Chapter: Ch. 6 Electromagnetic Optics → 6.5 Polarization of electromagnetic waves
TE-polarized light hits an interface between two materials with different refractive indices (n0 and n1 ). Part of
the light is reflected. With the coordinate system shown (x vertical, y out of the page, z horizontal along the
interface normal), which of the following statements is true?
Reasoning:
TE (transverse electric, s-polarized) means the electric field is perpendicular to the plane of incidence. With
the plane of incidence being xz, the E-field must lie along y.
Answer: (c) The electric field oscillates in the y-direction.
Q5. Focal length of a lens system
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems (system matrix)
Consider a lens system in air. The system is described by this matrix equation:
′ x
1
5 cm
x
=
.
−0.2 cm−1 −1
nα
n′ α ′
Calculate the focal distance f of this lens system.
Reasoning:
A
For a system matrix M =
C
B
, the power is P = −C, and in air f = f ′ = 1/P .
D
P = −(−0.2 cm−1 ) = 0.2 cm−1 ⇒ f = f ′ =
Answer: (b) f = f ′ = +5 cm
2
1
= +5 cm.
0.2
Q6. Power in dBm
Chapter: Ch. 2 Quantities and Units of Light → 2.4 Energetic Quantities (logarithmic units)
We detect an optical signal of 25 µW. How much is that expressed in dBm?
Reasoning:
PdBm = 10 log10
P
1 mW
= 10 log10 (0.025) ≈ −16 dBm.
Answer: (b) −16 dBm
Q7. Camera obscura
Chapter: Ch. 3 Geometric Optics → 3.6 Applications (imaging systems)
We are comparing two types of projecting imaging systems. The camera obscura with a pin-hole and the camera
obscura with a bigger opening in which a lens is inserted. Which of the following statements is true?
Reasoning:
A pin-hole admits very little light (poor light efficiency), but every point on the object maps to (essentially)
a point on the screen regardless of distance, so the depth of focus is essentially infinite. A lens collects much
more light (better light efficiency), but only objects at one conjugate distance are sharply focused, so depth of
focus is limited.
Answer: (b) The pin-hole has better depth of focus; the lens has better light efficiency.
Q8. Refraction at a single spherical surface
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems (refractive power)
A light ray passes from a medium with refractive index n1 = 1.51 (glass) into another medium with refractive
index n2 = 1.33 (water). The boundary between the two media is a convex spherical surface with a radius of
curvature R = 15 cm. What is the refractive power P of the surface?
Reasoning:
P =
1.33 − 1.51
n2 − n1
=
= −1.2 diopter.
R
0.15 m
n1 = 1.51
n2 = 1.33
Answer: (d) P = −1.2 diopter
Q9. Change of refractive index
Chapter: Ch. 4 Scalar Wave Optics → 4.4 Reflection and refraction
A wave travels from a medium with a constant refractive index to a different medium with a different constant
refractive index. What happens to the phase velocity, wavelength and frequency of the wave?
Reasoning:
Frequency is set by the source and is conserved across the boundary (continuity of the field oscillation in
time). The phase velocity is v = c/n, so it changes. Since λ = v/f , the wavelength also changes.
Answer: (c) The frequency remains the same, but the phase velocity and wavelength change.
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Q10. Imaging (1:1 upright)
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems
We want to create an upright image of an object with a lens system in air. The image should have the same
size as the object (1 to 1 imaging). The lens system has a focal distance f . Which choice of object distance S
will achieve this?
Reasoning:
Requiring magnification m = +1 (upright, same size) is a special degenerate case satisfied trivially when
S = 0: the object is placed at the lens itself, and the image coincides with the object, same size and upright.
Answer: (b) S = 0
Q11. Transmission through a transparent plate
Chapter: Ch. 6 Electromagnetic Optics → 6.8 Layered structures (Fabry–Pérot fringes)
We have two plates in air, plate A and plate B. We measure the transmission of visible light through each of
them sequentially. The light hits the plates perpendicular to the surface. The transmitted light is measured with
a spectrometer. The two curves have the same fringe contrast (amplitude) but different fringe spacing. What
can you conclude about the plates?
Reasoning:
The Fabry–Pérot fringes in transmission through a parallel plate have:
ˆ Fringe contrast (amplitude) determined by surface reflectivity, hence by the refractive index n.
ˆ Fringe spacing (free spectral range) determined by the optical thickness nd.
Same amplitude but different periodicities ⇒ same n, different thickness.
Answer: The two plates have the same refractive index but different thicknesses.
Q12. Principal planes of a lens system
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems (principal planes)
Light is traveling from left to right through two lenses in air with focal length f and f /2, respectively, that are
spaced a distance f apart. Calculate the distances D and D′ of the principal planes to the respective vertex
planes of this lens system.
Reasoning:
Combined power:
2
1 2
3
2
1
1
+ −f · · = − = .
f
f
f f
f
f
f
So fsys = f . Distances of principal planes from the respective vertex planes:
P = P1 + P2 − dP1 P2 =
D=−
d P2
· fsys = −2f,
P
D′ = −
d P1
· fsys = −f.
P
Neither D = D′ = ±f nor D = −D′ matches; therefore none of the listed pairs is correct.
Answer: (a) None of the answers is correct.
Q13. Light polarization
Chapter: Ch. 6 Electromagnetic Optics → 6.5 Polarization of electromagnetic waves
Consider a monochromatic plane wave travelling in the z-direction. The E-field components in the x and y
direction are the following:
Ey = π cos 2πf (t − z/c) − π2 .
Ex = −π cos 2πf (t − z/c) + π2 ,
What is the polarization state of this field?
Reasoning:
Use cos(α + π/2) = − sin α and cos(α − π/2) = sin α with α = 2πf (t − z/c):
Ex = −π(− sin α) = π sin α,
Ey = π sin α.
Ex = Ey at all times: a real linear oscillation along the diagonal x̂ + ŷ, i.e. linear at 45◦ .
Answer: (b) Linear polarization at 45◦ .
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Q14. Focusing a Gaussian beam
Chapter: Ch. 5 Gaussian Beam Optics → 5.2 Gaussian beams in lens systems
We focus a red laser beam (λ0 = 633 nm) with a flat phase-front and a beam diameter of 5 mm with a lens that
has a refractive power of P = 9.5 diopter. What is the approximate beam diameter (beam waist) in the focal
point?
Reasoning:
For a flat-phase (collimated) Gaussian focused by a thin lens, with f = 1/P ≈ 0.1053 m and win = 2.5 mm:
wf =
λ0 f
633 × 10−9 · 0.1053
≈ 8.3 µm.
=
π win
π · 2.5 × 10−3
Beam diameter at the waist: 2wf ≈ 16.6 µm.
Answer: (c) 16.6 µm
Q15. Snell’s law on a slab of unknown refractive index
Chapter: Ch. 3 Geometric Optics → 3.2.5 Interface between homogeneous media (Snell’s law)
A light ray is incident from air on a slab of material with a thickness of 5 (arbitrary units) and with a refractive
index of nslab . The light hits the slab at a height of 4 with an angle of 50 degrees with the normal. What value
should nslab be in order for the light ray to exit the slab on the opposite side at height 5?
Reasoning:
Inside the slab the ray travels along a straight line of horizontal length 5 and vertical length 5 − 4 = 1.
Therefore:
1
tan θ2 = = 0.2 ⇒ θ2 ≈ 11.31◦ .
5
Snell at the entry surface:
nair sin 50◦ = nslab sin θ2 ⇒ nslab =
sin 50◦
≈ 3.9.
sin 11.31◦
nslab ≈ 3.9
θ2
50◦
n=1
n=1
Answer: nslab ≈ 3.9
Q16. Biconvex lens in water
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems (lensmaker)
What is the focal length f of a biconvex lens with radii 20 cm and 30 cm and refractive index n = 1.5 when it
is immersed in water (n = 4/3)?
Reasoning:
Sign convention: biconvex means R1 = +20 cm, R2 = −30 cm. Lensmaker’s equation in a surrounding
medium:
1
1
1
1
P = (nl − n)
−
= (1.5 − 1.33)
+
≈ 1.417 D.
R1
R2
0.20 0.30
Focal length in the surrounding medium:
f=
1.33
n
=
≈ 0.96 m.
P
1.417
Answer: (c) f ≈ 0.96 m
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Q17. Positive lens
Chapter: Ch. 3 Geometric Optics → 3.3 Paraxial theory of imaging systems (imaging equation)
We have a positive lens in air. The focal distance of the lens is 20 cm. We place an object at a distance of
15 cm in front of the lens. Which of the following statements is true?
Reasoning:
With object distance |S| < f , the imaging equation gives an image on the same side as the object: a virtual,
upright, magnified image in front of the lens.
Answer: (d) A virtual image is formed in front of the lens.
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