Optics - Michelson Interferometer

A Michelson interferometer is configured as a wave-meter, as shown in the figure above, so that a ratio of fringe counts may be used to compare the wavelength of two lasers with high precision. When the mirror in the right arm of the interferometer is translated through a distance d, 100,000 interference fringes pass across the detector for green light and 85,865 fringes pass across the detector for red (λ = 632.82 nanometers) light. The wavelength of the green laser light is

A. 500.33 nm

B. 543.37 nm

C. 590.19 nm

D. 736.99 nm

E. 858.65 nm

(GR0177 #100)

Solution

λ_red = 632.82 nm

λ_red  λ_green → (D) and (E) are FALSE.

Constructive path difference = Nλ

N_gλ_g = N_rλ_r

λ_g = N_rλ_r / N_g

= (85,865 × 632.82) / 100,000

≈ (8.6 × 10⁴ × 6.3 × 10²) / 10⁵

= 541.8

Answer: B

Optics - Pinhole Camera

The screen of a pinhole camera is at a distance D from the pinhole, which has a diameter d. The light has an effective wavelength λ. (λ ≪ D) For which of the following value of d will the image be sharpest?

A. √(λD)
B. λ
C. λ/10
D. λ²/D
E. D²/λ

(GR8677 #100)

Solution:

See Pinhole Camera

Order of magnitude:
λ of visible light is 400 to 700 nm or in the order of 10⁻⁷ meter.
D is around 10 to 100 cm or in the order of 1 meter.
d is expected to be around 1 mm or in the order of 10⁻³ meter.

A. TRUE
√(λD) = √(10⁻⁷× 1) ≈ 10⁻³ about the right size of d

B. FALSE
λ → d cannot be in the order of 10⁻⁷ meter

C. FALSE
λ/10 → d cannot be in the order of 10⁻⁸ meter

D. FALSE
λ²/D ≈ 10⁻¹⁴→ d cannot be too small

E. FALSE
D²/λ ≈ 1/10⁻⁷ = 10⁷→ d cannot be too big

Answer: A

Classical Mechanics - Rotational Motion

A uniform rod of length 10 meters and mass 20 kilograms is balanced on a fulcrum with a 40 kg mass on one end of the rod and 20 kg mass on the other end, as shown above. How far is the fulcrum located from the center of the rod?

A. 0 m
B. 1 m
C. 1.25 m
D. 1.5 m
E. 2 m

(GR9277 #100)

Solution:

∑τ = 0

200(5+d) + 200d - 400(5-d) = 0

1000 + 200d + 200d - 2000 + 200d = 0

800d = 1000

d = 10/8 = 1.25

Answer: C

Quantum Mechanics – Ladder Operator

The operator,  when operating on a harmonic energy eigenstate ψ_n with energy E_n, produces another energy eigenstate whose energy is E_n − ħω₀. Which of the following is true? 

  I.  commutes with the Hamiltonian. 

 II.  is a Hermitian operator and therefore an observable. 

III. The adjoint operator  

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

(GR9677 #100)

Solution:

I. FALSE
Commutes if [H, a] = 0
But a is a ladder operator, a raises the energy level so that [H, a] = − ħωa

II. FALSE
Hermitian operator if  





III. TRUE
See II.

Answer: C