12th Class Physics Unit 15 Physical Optics Numericals

NP 15.1.Coherent light of wavelength 500 nm is incident on two very narrow and closely spaced slits. The interference pattern is observed on a very tall screen that is 2.00 m from the slits. Near the center of the screen the separation between two adjacent interference maxima is 3.53 cm. What is the distance on the screen between the m = 49 and m = 50 maxima?

NP 15.2.Young's experiment is performed with light from excited helium atoms (λ = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

Given λ = 502 nm = 502 × 10⁻⁹ m, L = 1.20 m, m = 20, y₂0 = 10.6 mm = 10.6 × 10⁻³ m.

Formula y_m = mλL/d ⟹ d = mλL/y_m

Substitution d = (20)(502 × 10⁻⁹ m)(1.20 m) / (10.6 × 10⁻³ m)
d = (1.2048 × 10⁻⁵ m²) / (10.6 × 10⁻³ m)
d = 1.14 × 10⁻³ m

Result d ≈ 1.14 mm (1.14 × 10⁻³ m).

NP 15.3.Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Given λ = 450 nm = 450 × 10⁻⁹ m, L = 1.80 m, spacing between adjacent dark fringes Δy = 3.90 mm = 3.90 × 10⁻³ m.

The spacing between adjacent dark fringes equals the spacing between adjacent bright fringes, Δy = λL/d, so:
d = λL/Δy

Substitution d = (450 × 10⁻⁹ m)(1.80 m) / (3.90 × 10⁻³ m)
d = (8.10 × 10⁻⁷ m²) / (3.90 × 10⁻³ m)
d = 2.08 × 10⁻⁴ m

Result d ≈ 2.08 × 10⁻⁴ m (0.208 mm).

NP 15.4.Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

Given d = 0.450 mm = 4.50 × 10⁻⁴ m, L = 75.0 cm = 0.750 m, λ = 500 nm = 5.00 × 10⁻⁷ m.

Dark fringes occur at y_dark = (m + 1/2)λL/d, m = 0, 1, 2, ... The second dark line (m = 1) and third dark line (m = 2) are therefore separated by one full fringe spacing:
Δy = λL/d

Substitution Δy = (5.00 × 10⁻⁷ m)(0.750 m) / (4.50 × 10⁻⁴ m)
Δy = (3.75 × 10⁻⁷ m²) / (4.50 × 10⁻⁴ m)
Δy = 8.33 × 10⁻⁴ m

Result Δy ≈ 8.33 × 10⁻⁴ m (0.833 mm).

NP 15.5.Two very narrow slits are spaced 1.80 mm apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light of wavelength 550 nm?

Given d = 1.80 mm = 1.80 × 10⁻³ m, L = 35.0 cm = 0.350 m, λ = 550 nm = 5.50 × 10⁻⁷ m.

Consecutive dark fringes are separated by one fringe spacing:
Δy = λL/d

Substitution Δy = (5.50 × 10⁻⁷ m)(0.350 m) / (1.80 × 10⁻³ m)
Δy = (1.925 × 10⁻⁷ m²) / (1.80 × 10⁻³ m)
Δy = 1.07 × 10⁻⁴ m

Result Δy ≈ 1.07 × 10⁻⁴ m (0.107 mm).

NP 15.6.Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of 11.3°. What is the angular position of the fourth-order maximum?

Given grating equation d sinθ = nλ, first order n = 1 at θ₁ = 11.3°.

From n = 1 λ = d sin(11.3°)

For n = 4 d sinθ₄ = 4λ = 4d sin(11.3°) ⟹ sinθ₄ = 4 sin(11.3°)

Substitution sin(11.3°) = 0.1959
sinθ₄ = 4 × 0.1959 = 0.7835
θ₄ = sin⁻¹(0.7835)

Result θ₄ ≈ 51.6°.

NP 15.7.If a diffraction grating produces a third-order bright spot for red light of wavelength 700 nm at 65° from the central maximum, at what angle will the second-order bright spot be for violet light of wavelength 400 nm?

Given n₁ = 3, λ₁ = 700 nm, θ₁ = 65°; find θ₂ for n₂ = 2, λ₂ = 400 nm, same grating (same d).

Step 1 — find grating spacing d from the red-light data, using d sinθ = nλ:
d = n₁λ₁ / sinθ₁ = (3)(700 nm) / sin(65°) = 2100 nm / 0.9063 = 2318 nm

Step 2 — use this d to find θ₂ for violet light:
sinθ₂ = n₂λ₂ / d = (2)(400 nm) / 2318 nm = 800 / 2318 = 0.345
θ₂ = sin⁻¹(0.345)

Result θ₂ ≈ 20.2°.

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Unit 15: Physical Optics — Numericals