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 - 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 - 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