SIMG-733 Optics Final Exam Closed Book

THREE HOURS (though targeted for two hours)UNPROGRAMMED CALCULATORS ALLOWED, SHOW YOUR WORKSelect ANY FIVE (5) problems (equally weighted even if not equally difficult)Submit problems stapled together IN NUMERICAL ORDERSTANDARD HINT: MAKE SKETCHES BEFORE WRITING EQUATIONS

1. Consider the apparatus shown below. Light with wavelength λ0 = 600nm encountersan opaque screen with two point apertures separated by the distance d0. Light throughone hole then encounters a variable neutral-density filter whose transmittance cyclesfrom 1 to 0 and back to 1 in a sinusoidal fashion at the rate at the rate ν0 = 1

cyclesec,

where 1 second is much longer than the measurement time T0. At all transmittances,the neutral-density filter induces a constant phase of π

4radians. Derive the expression

for visibility V of the irradiance pattern as a function of time that is observed on theobservation screen located a distance z1 “downstream” from the aperture plane in theFraunhofer diffraction region.

SOLUTION: Name the transmittance function of the “upper” aperture to be:

τ [t] =1

2(1 + cos [ω0t])

At times when the transmittance τ [t] = 1, this is the two-aperture problem with aphase difference; when the transmittance τ [t] = 0, this is single-aperture diffraction.From the drawing, I will write down the aperture function as (but you could use thecomplementary form):

f [x, y; z = 0] =

µδ

∙x− d0

2

¸+ δ

∙x+

d02

¸· τ [t] · exp

h+i

π

4

i¶· δ [y]

=

µδ

∙x− d0

2

¸· exp [i · 0] + δ

∙x+

d02

¸· τ [t] · exp

h+i

π

4

i¶· δ [y]

1

=

µδ

∙x− d0

2, y

¸· exp

h+i

π

8

i· exp

h−iπ8

i¶+

µδ

∙x+

d02, y

¸· τ [t] · exp

h+i

π

8

i· exp

h+i

π

8

i¶= exp

h+i

π

8

i·µδ

∙x− d0

2, y

¸· exp

h−iπ8

i+ δ

∙x+

d02, y

¸· τ [t] · exp

h+i

π

8

i¶Now split the contribution from the “unattenuated” aperture into the equal part and itscomplement:

f [x, y; z = 0] = exph+i

π

8

i· δ∙x− d0

2, y

¸· (τ [t] + (1− τ [t])) · exp

h−iπ8

i+exp

h+i

π

8

i· δ∙x+

d02, y

¸· τ [t] · exp

h+i

π

8

i=

µexp

h+i

π

8

i· τ [t] ·

µδ

∙x− d0

2

¸· exp

h−iπ8

i+ δ

∙x+

d02

¸· exp

h+i

π

8

i¶¶· δ [y]

+ exph+i

π

8

i· (1− τ [t]) ·

µδ

∙x− d0

2

¸· exp

h−iπ8

i¶· δ [y]

The 2-D Fourier transform is:

F [ξ, η] = exph+i

π

8

i·τ [t]·

µexp

∙−2πiξ · d0

2

¸exp

h−iπ8

i+ exp

∙−2πiξ · −d0

2

¸exp

h+i

π

8

i¶·1 [η]

+ exph+i

π

8

i· (1− τ [t]) · exp

∙−2πiξ · d0

2

¸exp

h−iπ8

i· 1 [η]

= exph+i

π

8

i· τ [t] · 2 · cos

∙2π

ξd02+

π

8

¸· 1 [η]

+ exph+i

π

8

i· (1− τ [t]) · exp

∙−2πiξ · d0

2

¸exp

h−iπ8

i· 1 [η]

Now propagate into the Fraunhofer diffraction region:

g [x, y; z = z1] ∼=µ

1

iλ0z1· exp

∙+2πi

z1λ0

¸¶· F∙

x

λ0z1,

y

λ0z1

¸

=

µ1

iλ0z1· exp

∙+2πi

z1λ0

¸· exp

h+i

π

8

i· 1∙

y

λ0z1

¸¶

·

⎛⎝⎛⎝2 · τ [t] · cos⎡⎣2π x³

2λ0z1d0

´ + π

8

⎤⎦⎞⎠+⎛⎝(1− τ [t]) · exp

⎡⎣−2πi x³2λ0z1d0

´⎤⎦ exp h−iπ

8

i⎞⎠⎞⎠This is the superposition of the translated cosine fringe pattern and a linear-phase term.Now calculate the squared magnitude:

|g [x, y; z = z1]|2 =µ1

λ0z1

¶2·¯̄̄̄F

∙x

λ0z1,

y

λ0z1

¸¯̄̄̄22

=

µ1

λ0z1

¶2 ¯̄̄̄exp

∙+2πi

z1λ0

¸· exp

h+i

π

8

i· 1∙

y

λ0z1

¸¯̄̄̄2·

⎛⎝⎛⎝4 · (τ [t])2 · cos2⎡⎣2π x³

2λ0z1d0

´ + π

8

⎤⎦⎞⎠+ ¡(1− τ [t])2 · 1¢⎞⎠

+

⎛⎝2 · (1− τ [t]) · τ [t] · cos

⎡⎣2π x³2λ0z1d0

´ + π

8

⎤⎦ 2 cos⎡⎣2π x³

2λ0z1d0

´ + π

8

⎤⎦⎞⎠

=

µ1

λ0z1

¶2·

⎛⎝¡(1− τ [t])2¢+ 4 · τ [t] · cos2

⎡⎣2π x³2λ0z1d0

´ + π

8

⎤⎦⎞⎠=

µ1

λ0z1

¶2·

⎛⎝¡(1− τ [t])2¢+ 2 · τ [t]

⎛⎝1 + cos⎡⎣2π x³

λ0z1d0

´ + π

4

⎤⎦⎞⎠⎞⎠=

µ1

λ0z1

¶2·

⎛⎝¡(1− τ [t])2¢+ 2 · τ [t] + 2 · τ [t] · cos

⎡⎣2π x³λ0z1d0

´ + π

4

⎤⎦⎞⎠=

µ1

λ0z1

¶2·

⎛⎝¡1 + (τ [t])2¢+ 2 · τ [t] · cos⎡⎣2π x³

λ0z1d0

´ + π

4

⎤⎦⎞⎠which is a biased cosine with amplitude 2 · τ [t] and bias 1 + (τ [t])2.Since the oscillation of the amplitude is slow compared to the observation time, thevisibility is:

V=2 · τ [t]

1 + (τ [t])2=

2 · 12(1 + cos [ω0t])

1 +¡12(1 + cos [ω0t])

¢2 = 1 + cos [ω0t]54+ 1

2cos [ω0t] +

14cos2 [ω0t]

V = 1 if τ = 1

V = 0 if τ = 0

Just for fun, I’ll graph the rsult as a function of angle θ = ω0t in degrees:

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 3600.0

0.2

0.4

0.6

0.8

1.0

theta

V

3

2. An optical system consists of two lenses with f1 = −100mm and f2 = +50mm. Thediameters of the lenses are d1 = d2 = +50mm. The two lenses are separated byt0 = +100mm

(a) Determine the focal length of the system.

(b) Locate the principal planes of the system.

(c) Determine which element is the aperture stop of the system.

(d) Locate the entrance and exit pupils of the system

(e) Determine the f-number (also called the focal ratio) of the system.

(f) Determine the distances from the object to the first lens and from the second lensto the image in the case where the transverse magnification of the image createdby the system is MT = −1.SOLUTION:(a)

feff = ϕ1 + ϕ2 − ϕ1ϕ2t =1

f1+1

f2− t

f1f2

=

µµ1

−100mm

¶+

µ1

+50mm

¶−µ

+100mm

(−100mm) (+50mm)

¶¶−1= +

100

3mm

feff = +100

3mm

(b) bring in a ray from an object at infinity and refract at the first lens:

L1 :1

f1=1

z1+1

z01=1

∞ +1

z01=⇒ z01 = f

01 = −100mm

the distance to the second lens is:

z2 = t− z01 = 100mm− (−100mm) = +200mm

so the image created by the second lens is located at:

z02 ≡ V0F0 =

µ1

f2− 1

z2

¶−1=

µ1

+50mm− 1

+200mm

¶−1= +

200

3mm

this is the location of the system focal point F0 measured from the rear vertex V0.From the system focal length;

feff ≡ H0F0 = +100

3mm

4

we can calculate the distance from the principal point to the rear vertex:

H0F0 = H0V0 +V0F0

+100

3mm = H0V0 +

µ+200

3mm

¶=⇒ H0V0 = +

100

3mm−

µ+200

3mm

¶= −100

3mm

=⇒ V0H0 = −H0V0 = +100

3mm

so the image-space principal point is located “after” the second lens at a distanceof 100

3mm from the rear system vertex. The image-space focal point is located the

same distance of +2003mm “after” the the second lens.

To find the object-space principal and focal points, we must “reverse” the system:the image distance from the positive lens is located at a distance of z02 = f2 =+50mm, so that the object distance for the negative lens is

z1 = t− z02 = 100mm− (50mm) = +50mm

The object-space focal point located at the image from the negative lens:

z01 = FV =

µ1

f1− 1

z1

¶−1=

µ1

−100mm −1

50mm

¶−1= −100

3mm

=⇒ VF = −FV = +100

3mm

so the object-space focal point is located “between” the lenses and closer to the first(negative) lens. Since the system focal length is positive, the object-space principalpoint is located so that FH = +100

3mm, which means that

VH = VF+ FH =

µ+100

3mm

¶+

µ+100

3mm

¶= +

175

3mm

VH = +200

3mm

The system is graphed below:

5

(c) from the diagram, it is obvious that the second lens L2 is the aperture stop,since the ray entering at the edge (margin) of the first lens does not get throughthe system:

L2 is the aperture stop

(d) Since the aperture stop is the second lens, it is also the exit pupil (which isthe image of the stop seen from image space). Since the exit pupil is the aperturestop, the transverse magnification of the exit pupil is +1. The entrance pupil isthe image of the stop seen through the system from the object side, so the use thedistance t0 between the lenses and the focal length of the first lens in the imagingequation:

z0 =

µ1

f1− 1

t0

¶−1=

µ1

−100mm −1

+100mm

¶−1= −50mm

which means that the entrance pupil appears to be midway between the lenses. Thetransverse magnification of the entrance pupil is:

MT = −z0

z= − (−50mm)

(+100mm)= +

1

2

and thus the diameter of the entrance pupil is:

dNP = dstop ·MT = +50mm ·µ+1

2

¶= 25mm

(e) The focal ratio is the ratio of the system focal length to the diameter of the

6

entrance pupil:

F/# ≡ feffdNP

=1003mm

25mm=4

3

F/# =4

3

: (side comment: this is an unreasonably “fast” system — the focal length is justbarely longer than the diameter of the entrance pupil).(f) From the given transverse magnification:

MT = −z0

z= −1 =⇒ z0 = z

where the distances are measured from the principal points

1

feff=

1

z+1

z0=⇒ z0 = z = 2 · feff = 2 ·

µ+100

3mm

¶=200

3mm

OH = H0O0 = +200

3mm

So the object is 2003mm “in front” of the object-space principal point H and the

image is 2003mm “behind” of the image-space principal point H0. Since we know

the distances of the principal points from the system vertices, we can locate theobject and image:

OH = OV +VH

=⇒ +200

3mm = OV +

200

3mm

=⇒ OV =200

3mm− 200

3mm = 0mm

so the object is located AT L1.On the image side, the image distance is:

H0O0 = H0V0 +V0O0

+200

3mm = −100

3mm+V0O0

V0O0 = +200

3mm−

µ−1003mm

¶= +100mm

V0O0 = +100mm

So the image is located 100mm after lens L2Check: brute-force method: Object at lens =⇒ image at lens with MT = +1

z2 = t0 − z01 = +100mm− 0 = +100mm

z02 =

µ1

f2− 1

z2

¶−1=

µ1

+50mm− 1

+100mm

¶−1= 100mm

M2 = −z02

z2= −(+100mm)

(+100mm)= −1

7

3. Assuming normally incident plane-wave illuminationand neglecting finite aperture ex-tent, find the Fresnel diffraction pattern of a screen with the following transmittancefunction:

f [x, y] =1

2(1 +m cos [2πξ0x])

SOLUTION:

F2 {t [x, y]} = F [ξ, η] =1

2δ [ξ, η] +

m

2

µ1

2δ [ξ + ξ0] +

1

2δ [ξ − ξ0]

¶δ [η]

=1

2δ [ξ, η] +

m

4(δ [ξ + ξ0] + δ [ξ − ξ0]) δ [η]

=1

2δ [ξ, η] +

m

4δ [ξ + ξ0, η] +

m

4δ [ξ − ξ0, η]

We know that the “impulse response of light propagation” into the Fresnel diffractionregion is:

h [x, y; z] =1

iλzexp

h2πi

z

λ

iexp

∙iπx2 + y2

λz

¸So the “transfer function of light propagation” is:

H [ξ, η; z] =1

iλzexp

h2πi

z

λ

iF2

(exp

"iπ

µx√λz

¶2#· exp

"iπ

µy√λz

¶2#)= exp

h2πi

z

λ

iexp

£−iπλz

¡ξ2 + η2

¢¤The spectrum of the diffracted signal is the product of the input spectrum and thetransfer function:

G [ξ, η; z] = F [ξ, η; z] ·H [ξ, η; z]

=

µ1

2δ [ξ, η] +

m

4δ [ξ + ξ0, η] +

m

4δ [ξ − ξ0, η]

¶· exp

h2πi

z

λ

iexp

£−iπλz

¡ξ2 + η2

¢¤= exp

h2πi

z

λ

iµ12δ [ξ, η] +

m

4δ [ξ + ξ0, η] +

m

4δ [ξ − ξ0, η]

¶exp

£−iπλz

¡ξ2 + η2

¢¤= exp

h2πi

z

λ

iµ12δ [ξ, η] +

m

4δ [ξ + ξ0, η] e

−iπλz(−ξ0)2 +m

4δ [ξ − ξ0, η] e

−iπλz(+ξ0)2¶

= exph2πi

z

λ

iµ12δ [ξ, η] +

m

4(δ [ξ + ξ0, η] + δ [ξ − ξ0, η]) e−iπλzξ

20

¶where the identity has been used:

f [x] δ [x− x0] = f [x0] δ [x− x0]

(a) The output pattern is the inverse Fourier transform of this function:

g [x, y] = e+2πizλ1

2

³1 [x, y] +m cos [2πξ0x] 1 [y] e

−iπλzξ20´

8

4. Attack or defend the following statement, giving evidence (equations, examples, etc.)to buttress your position: “The optical transfer function of an optical system acting inincoherent light is nonnegative at all spatial frequencies.”

This is a giveaway question — clearly the statement is not true, because it is tantamountto saying that the autocorrelation of all functions is nonnegative. We saw this in thediscussion of defocus, where the transfer function is real-valued and negative.

9

5. Give a qualitative explanation of the physical mechanism for refractive index, includingdescriptions of normal and anomalous dispersion and the reason why glasses havedifferent refractive indices and dispersions. Sketches and graphs are always helpful.

refractive index arises from electron resonances. Electrons in atoms (or molecules) inmedium have natural oscillation frequencies due to atomic forces. Incident electric fieldgenerates driving force on electrons, Electrons “respond” differently when absorbing andre-emitting electric fields with different oscillation frequencies (this is the process of“scattering”). The process is analogous to a “damped harmonic oscillator” in physics.see notes from class of 1/14/2009.

10

6. Consider two ideal linear polarizers that are oriented so that their axes are orthogo-nal. Unpolarized light is incident on the first polarizer with “intensity” (analogous toirradiance) I0.

(a) How much light (amplitude and irradiance) “escapes” out the second polarizeronto an observation screen?Another giveaway question: since the polarizers are ideal and orthogonal, thetransmittance of the system of two polarizers is 0, so the output amplitude is:

g [t] = f [t] · τ [t]τ = 0 =⇒ g = 0 =⇒ |g|2 = 0

(b) A third identical polarizer is placed between these two with its axis orientedmidway between. Determine the amplitude and irradiance on the observationscreen in this case.A polarizer transmits an amplitude proportional to the cosine of the relative angleθ:

Eout = E0 cos [θ]

So in this case, the first polarizer produces linearly polarized light at its specificangle, call it θ = 0. The relative angle for the second polarizer is:

E1 = E0 coshπ4

iand the output amplitude from the third polarizer is:

E2 = E1 coshπ4

i=³E0 cos

hπ4

i´· cos

hπ4

i= E0 cos

2hπ4

iE2E0

= cos2hπ4

i=1

2

The intensity out (or the irradiance on a sensor) is the time average of the squared

11

magnitude:

I2 =­|E2 [t]|2

®=­|E2 cos [ω0t]|2

®= |E2|2

­|cos [ω0t]|2

®=1

2|E2|2

=

¯̄̄̄1

2

¯̄̄̄2· 12|E0|2 =

1

8

compared to the intensity in:­|E0 cos [ω0t]|2

®= |E0|2

­|cos [ω0t]|2

®=1

4|E0|2

12

7. The bore of a cylindrical thermometer (i.e., the hole in the middle for the mercury)has diameter equal to half the outside diameter of the stem. The refractive index ofthe glass is 1.5.

(a) Determine the apparent width of the column of mercury relative to the apparentwidth of the stem, assuming that the viewing distance is large compared to thediameter of the thermometer.

1.5 · sin [θ1] = 1.0 · sin [θ2]M ABC is a right triangle, the right angle is located at B

ifR

r= 2, then angle B is 30 ◦

1.5 · sinhπ6

i= 1.0 · sin [θ2]

θ2 = sin−1∙3

2· 12

¸= sin−1

∙3

4

¸' 0.85 radians

Apparent width = 2 ·R sin [θr] = 2R ·3

4= D · 3

4

Viewing the mercury (red) in the thermometer; the eyeis at the bottom

(b) Determine the apparent width if the diameter of the bore is 2/3 the diameter ofthe stem? (HINT: draw the picture)Same picture:

1.5 · sin [θ1] = 1.0 · sin [θ2]M ABC is a right triangle, the right angle is located at B

R

r=

3

2=⇒ sin [θ1] =

2

33

2· sin [θ1] =

3

2· 23= 1 = 1.0 · sin [θ2]

θ2 = sin−1 [1] =π

2radians = 90◦

Apparent width = 2 ·R = D

The apparent diameter of the mercury column is equal to the diameter of thethermometer; the mercury “fills” the tube!

13