Wave Optics

1 exercises8 questions solved

Exercise 10.1Interference, Diffraction and Polarisation

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? μ of water = 1.33.

Solution

λ = 589 nm = 589 × 10⁻⁹ m, μ_water = 1.33 (a) Reflected light (stays in air): • Wavelength: λ_reflected = 589 nm (unchanged) • Frequency: f = c/λ = 3 × 10⁸/(589 × 10⁻⁹) = 5.09 × 10¹⁴ Hz (unchanged) • Speed: c = 3 × 10⁸ m/s (unchanged) (b) Refracted light (enters water): • Frequency: f = 5.09 × 10¹⁴ Hz (frequency never changes on refraction) • Speed: v = c/μ = 3 × 10⁸/1.33 = 2.26 × 10⁸ m/s • Wavelength: λ_water = v/f = 2.26 × 10⁸/(5.09 × 10¹⁴) = 4.44 × 10⁻⁷ m = 444 nm

What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Solution

(a) Light diverging from a point source: The wavefront is spherical — concentric spheres centered at the point source. (b) Light emerging from a convex lens (point source at focus): The convex lens converts the diverging spherical waves to parallel rays. The wavefront is planar (plane wavefront). (c) Light from a distant star intercepted by Earth: The star is so far away that the spherical wavefronts have a very large radius. The portion intercepted by Earth (small compared to the radius) appears as a plane wavefront.

In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Solution

d = 0.28 mm = 0.28 × 10⁻³ m, D = 1.4 m Distance between central fringe and 4th bright fringe = 4β = 1.2 cm = 0.012 m Fringe width β = 0.012/4 = 0.003 m Using: β = λD/d λ = βd/D = (0.003 × 0.28 × 10⁻³) / 1.4 = (8.4 × 10⁻⁷) / 1.4 = 6 × 10⁻⁷ m = 600 nm Wavelength of light = 600 nm

In Young's experiment, interference fringes are observed on a screen placed 1.0 m away from the double slit of slit separation 0.5 mm. If a monochromatic source of wavelength 600 nm is used, calculate: (a) fringe width (b) change in fringe width if screen is moved 50 cm closer.

Solution

(a) d = 0.5 mm = 5 × 10⁻⁴ m, D = 1.0 m, λ = 600 nm = 6 × 10⁻⁷ m Fringe width β = λD/d = (6 × 10⁻⁷ × 1.0)/(5 × 10⁻⁴) = 1.2 × 10⁻³ m = 1.2 mm (b) New D = 1.0 – 0.5 = 0.5 m: β' = λD'/d = (6 × 10⁻⁷ × 0.5)/(5 × 10⁻⁴) = 0.6 × 10⁻³ m = 0.6 mm Change = β – β' = 1.2 – 0.6 = 0.6 mm (fringe width decreases)

In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

Solution

Angular fringe width in air: θ = 0.2°, λ_air = 600 nm, μ = 4/3 Angular fringe width θ = λ/d, so d = λ/θ In water: λ_water = λ_air/μ = 600/(4/3) = 600 × 3/4 = 450 nm New angular width: θ_water = λ_water/d = (λ_water/λ_air) × θ = (450/600) × 0.2° = (3/4) × 0.2° = 0.15°

What is the Brewster angle for air-to-glass transition? (Refractive index of glass = 1.5)

Solution

Brewster's law: tan(i_B) = μ tan(i_B) = 1.5 i_B = tan⁻¹(1.5) ≈ 56.3° The Brewster angle for air-to-glass transition is approximately 56.3°. At this angle, the reflected light is completely plane polarised (the electric field vector oscillates only perpendicular to the plane of incidence).

Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Solution

λ = 5000 Å = 5000 × 10⁻¹⁰ m = 5 × 10⁻⁷ m = 500 nm Wavelength of reflected light = 5000 Å (unchanged — reflection doesn't change wavelength) Frequency = c/λ = 3 × 10⁸ / (5 × 10⁻⁷) = 6 × 10¹⁴ Hz (unchanged) For reflected ray to be normal to incident ray: The angle between incident and reflected rays = 90° Since angle of reflection = angle of incidence (i): i + r = 90° and i = r → 2i = 90° → i = 45° The angle of incidence must be 45°.

Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Solution

Aperture a = 4 mm = 4 × 10⁻³ m, λ = 400 nm = 4 × 10⁻⁷ m Ray optics is a good approximation when the Fresnel distance Z_F >> distance from aperture. Fresnel distance: Z_F = a²/λ = (4 × 10⁻³)² / (4 × 10⁻⁷) = 16 × 10⁻⁶ / (4 × 10⁻⁷) = 4 × 10¹ = 40 m Ray optics is a good approximation for distances much less than 40 m. For distances much greater than 40 m, diffraction effects become significant.

More chapters

← All chapters: Class 12 Physics

Wave Optics

Appearing for the May Phase 2 Board Exam? Practice with AI-ranked questions.

SST

SST