Optics - Index of Refraction

A light source is at the bottom of a pool of water (the index of refraction of water is 1.33). At what minimum angle of incidence will a ray be totally reflected at the surface?

A. 0^o  
B. 25^o
C. 50^o 
D. 75^o
E. 90^o 

(GR9677 #56)

Solution:

Snell's Law: n_w sin θ_i = n_a sin θ_r

For total internal reflection, θ_r = 90^o
n_w sin θ_i  =  n_a
sin θ_i = n_a/n_w = 1/1.33 ≈ 3/4
Convert to degree, θ = ¾ × 180/π ≈ 45^o closer to 50^o 

Answer: C

Optics - Diffraction Grating

Consider a single-slit diffraction pattern for a slit of width d. It is observed that for a light of wavelength 400 nanometers, the angle between the first minimum and the central maximum is 4 × 10⁻³ radians. The value of d is

A. 1 × 10⁻⁵ m
B. 5 × 10⁻⁵  m
C. 1 × 10⁻⁴  m
D. 2 × 10⁻⁴ m
E. 1 × 10⁻³ m

(GR9677 #57)

Solution:

Single slit diffraction: d sin θ = mλ

λ = 400 nm = 4 × 10⁻⁷ m
θ = 4 × 10⁻³ radians
sin θ ≈ θ for small θ

dθ = mλ
d = mλ / θ = 1 × (4 × 10⁻⁷) / (4 × 10⁻³) = 10⁻⁴ m

Answer: C

Optics - Telescope

A collimated laser beam emerging from a commercial HeNe laser has a diameter of about 1 millimeter. In order to convert this beam into a well-collimated beam of diameter 10 millimeter, two convex lenses are to be used. The first lens if of focal length 1.5 centimeters and is to be mounted at the output of the laser. What is the focal length, f of the second lens and how far from the first lens should it be placed?

A. f  = 4.5 cm, Distance = 6.0 cm
B. f  = 10 cm, Distance = 10 cm
C. f  = 10 cm, Distance = 11.5 cm
D. f  = 15 cm, Distance = 15 cm
E. f  = 15 cm, Distance = 16.5 cm

(GR9677 #58)

Solution:

Since two convex lenses are used, we can treat this system as a telescope.

Magnification, M =  f_object /  f_eye = 10
  f_object = 10 f_eye  = 10 × 1.5 cm = 15 cm

D =  f_object +  f_eye  = 10 + 1.5 = 16.5 cm

Answer: E

Optics - Thin Film

Consider two horizontal glass plate with a thin film of air between them. For what values of the thickness of the film of air will the film, as seen by reflected light, appear bright if it is illuminated normally from above by blue light of wavelength 488 nanometers?

A. 0, 122 nm, 244 nm
B. 0, 122 nm, 366 nm
C. 0, 244 nm, 488 nm
D. 122 nm, 244 nm, 366 nm
E. 122 nm, 366 nm, 610 nm

(GR9677 #82)

Solution:

See Thin Film.

Glass to Air: n_g  n_a → Δ_a = 0
Air to Glass: n_a  n_g →  Δ_b = 2t  +½λ

Δ = Δ_b − Δ_a = 2t  + ½λ

Blue Light → Constructive Interference: Δ = mλ
→ mλ =  2t  + ½λ
→ t = ½λ(m − ½)
with λ_a = 488 nm and m = 0,1,2,3...
m = 1 →  t = ½λ(1 − ½) = ¼ λ = 122 nm
m = 2 →  t = ½λ(2 − ½) = ¾ λ = 366 nm
m = 3 →  t = ½λ(3 − ½) = ⁵/4 λ = 610 nm

Answer: E

Optics - Interference

In a double-slit interference experiment, d is the distance between the centers of the slits and w is the width of each slit, as shown in the figure above. For incident plane waves, an interference maximum on a distant screen will be “missing” when

A. d = √2 w
B. d = √3 w
C. 2d = w
D. 2d = 3w
E. 3d = 2w

(GR9277 #20)

Solution:

For double-slit interference, d is always bigger than w
→ (C) and (E) are FALSE

Constructive or destructive patterns of interference (single or double slits) only deals with integer and half integer factors.
→ (A) and (B) are FALSE

Answer: D


Calculation:

“missing” = destructive pattern (minimum intensity).
For double slit: d sin θ = (m + ¹/₂) λ; m = 0, 1, 2, 3, ...
For single slit: w sin θ = mλ; m = 1, 2, 3, ...

^d/_w =  (m + ¹/₂)/m
Take m = 1,
^d/_w = ³/₂
2d = 3w

Optics - Thin Film

A soap film with index of refraction greater than air is formed on a circular wire frame that is held in a vertical plane. The film is viewed by reflected light from a white-light source. Bands of color are observed at the lower parts of the soap film, but the area near the top appears black. A correct explanation for this phenomenon would involve which of the following?

I. The top of the soap film absorbs all the light incident on it; none is transmitted.

II. The thickness of the top part of the soap film has become much less than a wavelength of visible light.

III. There a phase change of 180o for all wave-lengths of light reflected from the front surface of the soap film.

IV. There is no phase change for any wavelength of light reflected from the back surface of the soap film

A. I only
B. II and III only
C. III and IV only
D. I, II and III
E. II, III and IV

(GR9277 #21)

Solution:

I. FALSE. 

Soap film does not absorb light.

II. TRUE.

The gravity effect on water molecules decreases the thickness, t of the top of the soap film, become much less than a wavelength, λ of visible light.

When t ≪ λ, the path difference between the front and the rear surface reflections is small compared to λ. It causes an insignificant phase shift. The result is complete destructive interference of the 2 light waves, creating a black band.

III and IV. TRUE.

Thin film: 

n₁〈  n₂ → there is 180^o phase change
n₁ 〉n₂ → no phase change 

n = refractive index

From air to soap (front surface):

n_air〈  n_soap → 180^o phase change

From soap to air (back surface):
n_soap 〉n_air → no phase change 

Answer: E

Optics - Telescope

A simple telescope consists of two convex lenses, the objective and the eyepiece, which have a common focal point P, as shown in the figure above. If the focal length of the objective is 1.0 meter and the angular magnification of the telescope is 10, what is the optical path length between objective and eyepiece? 

A. 0.1 m
B. 0.9 m
C. 1.0 m
D. 1.1 m
E. 10 m

(GR9277 #22)

Solution:

Magnification of the telescope: M = f₀/ f_e

The optical path length between objective and eyepiece d =  f₀ +  f_e

f₀ = focal length of the objective
f_e = focal length of the eyepiece

Since M = 10,  f₀ = 1

f_e = f₀/M = 0.1

d =  f₀ +  f_e = 1 + 0.1 = 1.1 m

Answer: D

Optics - Diffraction Grating

Light of wavelength 5200 angstroms is incident normally on a transmission diffraction grating with 2000 lines per centimeter. The first-order diffraction maximum is at an angle, with respect to the incident beam, that is most nearly

A. 3^o
B. 6^o
C. 9^o
D. 12^o
E. 15^o

(GR9277 #35)

Solution:

d sin θ = mλ 

λ = 5200 Angstrom = 5200  × 10⁻¹⁰ m = 5.2 × 10⁻⁷ m
d = 1 cm / 2000 = 5 × 10⁻⁴ cm =  5 × 10⁻⁶ m
m = 1

sin θ = mλ / d  =  (5.2 × 10⁻⁷) / (5 × 10⁻⁶) ≈ 0.1

→ arcsin θ = 6^o 

Or, since sin θ ≪ 1 → sin θ ≈ θ 

Convert the angle from radians to degrees: 0.1  × 180/π  ≈ 18/3 =  6^o


Answer: B

Optics - Polarization

A steady of light is normally incident on a piece of Polaroid. As the Polaroid is rotated around the beam axis, the transmitted intensity varies as A + B cos 2θ where θ is the angle of rotation, and A and B are constants with . Which of the following may be correctly concluded about the incident light? 

A. The light is completely unpolarized
B. The light is completely plane polarized
C. The light is partly plane polarized and partly unpolarized
D. The light is partly circularly polarized and partly unpolarized
E. The light is completely circularly polarized

(GR9277 #67)

Solution:

Malus' Law: for plane-polarized light, I = I₀ cos²θ
Trig identity: cos²θ =  ½ (1 + cos 2θ) 

If A = B, the light is completely plane polarized, but A > B > 0

When cos²θ = 0, I = A = constant → unpolarized.

Thus, the light is partly plane polarized and partly unpolarized

Answer: C

Optics - Telescope

The angular separation of the two components of a double star is 8 microradians, and the light from the double star has a wavelength of 5500 angstroms. The smallest diameter of a telescope mirror that will resolve the double star is most nearly 

A. 1 mm
B. 1 cm
C. 10 cm
D. 1 m
E. 100 m

(GR9277 #68)

Solution:

Rayleigh telescope resolution limit: sin θ = 1.22 λ/D

D = 1.22 λ/sin θ = 1.22 (5500 × 10⁻¹⁰ m) / sin (8×10⁻⁶)

Since θ ≪, assume sin θ ~ θ

 D = (1.22 × 5.5 / 8) × 10³ × 10⁻¹⁰ × 10⁶ ≈ 10⁻¹ m = 10 cm

Answer: C

Optics - Index of Refraction

A fast charged particle passes perpendicularly through a thin glass sheet of index of refraction 1.5. The particle emits light in the glass. The minimum speed of particle is 

A.  ¹/₃ c
B.  ⁴/₉ c
C. ⁵/₉ c
D. ²/₃ c
E. c

(GR9277 #69)

Solution:

Index of refraction, n = ^c/_v
n = 1.5 = ³/₂
v = ^c/_n = ²/₃ c

Answer: D

Optics - Dispersion Curve

The dispersion curve shown above relates the angular frequency to the wave number k. For waves with wave numbers lying in the range k₁ < k< k₂ which of the following is true of the phase velocity and the group velocity?

A. They are in opposite directions.
B. They are in the same direction and the phase velocity is larger.
C. They are in the same direction and the group velocity is larger.
D. The phase velocity is infinite and the group velocity is finite.
E. They are the same in direction and magnitude.

(GR9277 #79)

Solution:

Phase velocity: v_p = ω/k
Group velocity: v_g = dω/dk

In the range k₁ < k< k₂:
v_p → positive quantity
v_g → negative quantity (negative slope)
→ v_p and v_g are in opposite directions.

Answer: A

Optics - Interference

Two coherent sources of visible monochromatic light form an interference pattern on a screen. If the relative phase of the source is varied from 0 to 2π at a frequency of 500 hertz, which of the following best describes the effect, if any, on the interference pattern?

1. It is unaffected because the frequency of the phase change is very small compared to the frequency of visible light.
2. It is unaffected because the frequency of the phase change is an integral multiple of π.
3. It is destroyed except when the phase difference is 0 to π.
4. It is destroyed for all phase differences because the monochromaticity of the sources is destroyed.
5. It is not destroyed but simply shifts positions at a rate too rapid to be detected by the eye.

(GR8677 #13)

Solution:

Interference pattern is observed only when the sources are coherent (both sources have identical λ, f, and phase relationship).

The problem states that the light sources are coherent, thus the interference patterns are observed, not destroyed → (C) and (D) are FALSE.

Phase change will affect the constructive and destructive patterns of interference → (A) and (B) are FALSE.

Note: 500 Hz or 500/second is too rapid for human eye which only can perceive 60 Hz to 80 Hz flickering light.

Answer: E

Optics - Hologram

In an ordinary hologram, coherent monochromatic light produces a 3-dimensional picture because wave information is recorded for which of the following?

I. Amplitude
II. Phase
III. Wave-front angular frequency

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

(GR8677 #58)

Solution:



Answer: B

Optics - Thin Film

It is necessary to coat a glass lens with a non-reflecting layer. If the wavelength of the light in the coating is λ, the best choice is a layer of material having an index of refraction between those of glass and air and a thickness of

A. λ/4
B. λ/2
C. λ/√2
D. λ
E. 1.5λ

(GR8677 #73)

Solution:

Index of refraction: 

Air to Layer,
phase difference: Δ_a = λ/2

Layer to Glass,
phase difference: Δ_b = 2t + λ/2

The relative shift: Δ = Δ_b − Δ_a =  2t + λ/2 − λ/2 = 2t

Non-reflecting → destructive interference
Δ = (m +½)λ

m = 0 (the thinnest the better)
2t = (0 +½)λ
t = λ/4

Answer: A


Notes: Click HERE for more info on Thin Film.

Optics - Polarization

Unpolarized light is incident on two ideal polarizers in series. The polarizers are oriented so that no light emerges through the second polarizer. A third polarizer is now inserted between the first two and its orientation direction is continuously rotated through 180^o. The maximum fraction of the incident power transmitted through all three polarizers is

A. Zero
B. 1/8
C. 1/2
D. 1/√2
E. 1

(GR8677 #74)

Solution:

Power ~ Intensity

The maximum fraction of intensity is: I = I₀/8 (see GR0177 #51)

→ P = P₀/8

Answer: B

Optics - Converge Lens

An object is located 40 centimeters from the first of two thin converging lenses of focal lengths 20 cm and 10 cm, respectively, as shown in figure above. The lenses are separated by 30 cm. The final image formed by the two lenses is located 

A. 5.0 cm to the right of the second lens
B. 13.3 cm to the right of the second lens
C. infinitely to the right of the second lens
D. 13.3 cm to the left of the second lens
E. 100 cm to left right of the second lens

(GR0177 #11)

Solution:

¹/_S + ¹/_S' = ¹/_f

For the first lens:

¹/_S1 + ¹/_S1' = ¹/_f1

¹/₄₀ + ¹/_S1'  = ¹/₂₀

S₁' = 40 cm

The image is located 40 cm behind (to the right of) the first lens, which is 10 cm behind the second lens. Therefore, S₂ = −10.

For the second lens:

¹/_S2 + ¹/_S2' = ¹/_f2

¹/ ₍₋₁₀₎ + ¹/_S2' = ¹/₁₀

S₂' = 5 cm

The final image is located 5 cm to the right of the second lens.

Answer: A

Optics - Concave Mirror

A spherical concave mirror is shown in the figure above. The focal point F and the location of the object O are indicated. At what point will the image be located?

A. I
B. II
C. III
D. IV
E. V

(GR0177 #12)

Solution:

The mirror equation:



The picture shows:   is negative.
The image is VIRTUAL (inside the mirror) i.e. at point V

Answer: E

Note:
For concave mirror:

• If an object is located BEFORE the focal point then image is VIRTUAL (inside the mirror).
• If an object is located AFTER the focal point then image is REAL (outside the mirror). 

Optics - Telescope

Two stars are separated by an angle of 3 × 10⁻⁵ radians. What is the diameter of the smallest telescope can resolve the two stars using visible light (λ = 600 nanometers)? (ignore any effect due to Earth’s atmosphere)

A. 1 mm
B. 2.5 cm
C. 10 cm
D. 2.5 m
E. 10 m

(GR0177 #13)

Solution:

Rayleigh telescope resolution limit: sin θ = 1.22 λ/D

Since θ ≪, assume sin θ ~ θ

D = ^1.22λ/_θ 

λ = 600 nanometers = 600 × 10⁻⁹ m

D = ^{1.22(600×10−9)}/_{(3×10⁻⁵)} = 2.44 × 10^-2 m ≈  2.5 cm

Answer: B

Optics - Polarization

Unpolarized light of intensity I₀ is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45^o to that of the first filter, while the axis of the third filter is oriented at 90^o to that of the first filter. What is the intensity of the light transmitted through the third filter?

A. 0
B. I₀/8
C. I₀/4
D. I₀/2
E. I₀/√2

(GR0177 #51)

Solution:

Polarizer:
The first filter always reduces the intensity of the light to half, I₁=½ I₀
The next filter reduces the intensity by I_n = I_(n-1) cos² θ
θ is the angle with respect to the n^th filter.

Thus,

I₁ = ½ I₀

I₂ = I₁ (cos 45)² = I₁ (½ √2)² = ½ I₁ = ¼ I₀

I₃=I₂ (cos 45)² = ½ I₂ = I₀/8

Answer: B