Wave Optics: CBSE Class 12 Physics NCERT Guide

Fundamental Concepts of Wave Optics

Wavefront

A wavefront is the locus of all points in a medium that are vibrating in the same phase. For a point source, the wavefronts are spherical; for a distant source, they are planar.

Huygens' Principle

This principle states that every point on a given wavefront acts as a fresh source of new disturbances, called secondary wavelets. The forward envelope of these secondary wavelets at any instant gives the new wavefront.

Coherent Sources

Two sources of light are said to be coherent if they emit light waves of the same frequency (or wavelength) and have a constant phase difference between them. This is a crucial condition for sustained interference.

Interference

The modification in the distribution of light intensity resulting from the superposition of two or more waves. It leads to a pattern of bright (constructive interference) and dark (destructive interference) fringes.

Diffraction

The phenomenon of bending of light waves around the sharp corners of an obstacle or an aperture and their subsequent spreading into the regions of the geometrical shadow.

The Principle of Superposition and Interference

Imagine two waves meeting at a point. What happens? The principle of superposition gives us the answer: the resultant displacement at that point is the vector sum of the individual displacements of each wave. This simple rule leads to the profound phenomenon of interference. When two coherent light waves superimpose, their energies get redistributed. At some points, the crest of one wave meets the crest of another (or trough meets trough). Here, the amplitudes add up, creating a region of maximum intensity called constructive interference. The path difference between the waves arriving at this point is an integral multiple of the wavelength (Δx = nλ). At other points, the crest of one wave meets the trough of another. Here, they cancel each other out, resulting in a region of minimum or zero intensity called destructive interference. The path difference for this is a half-integral multiple of the wavelength (Δx = (2n+1)λ/2). Young's Double Slit Experiment is the classic demonstration of this effect, producing a stable pattern of bright and dark 'fringes' on a screen.

Worked Examples: Young's Double Slit Experiment (YDSE)

• Problem 1: Calculating Fringe Width 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. Step 1: Identify the given quantities. Slit separation, d = 0.28 mm = 0.28 × 10⁻³ m Screen distance, D = 1.4 m Position of the 4th bright fringe from the center, x₄ = 1.2 cm = 1.2 × 10⁻² m Order of the fringe, n = 4 Step 2: Relate the position of the bright fringe to the fringe width. The distance of the nth bright fringe from the central maximum is given by xₙ = nβ, where β is the fringe width. We can also write it as xₙ = n(λD/d). Step 3: Solve for the wavelength (λ). We have 1.2 × 10⁻² = 4 (λ 1.4 / (0.28 × 10⁻³)) Rearranging for λ: λ = (1.2 × 10⁻² × 0.28 × 10⁻³) / (4 × 1.4) λ = (0.336 × 10⁻⁵) / 5.6 λ = 0.06 × 10⁻⁵ m = 6 × 10⁻⁷ m Final Answer: The wavelength of light used is 600 nm.
• Problem 2: Finding the Angular Width Monochromatic light of wavelength 500 nm is incident on a single slit of width 0.1 mm. Find the angular width of the central maximum in the resulting diffraction pattern. Step 1: Identify the given quantities. Wavelength, λ = 500 nm = 500 × 10⁻⁹ m Slit width, a = 0.1 mm = 0.1 × 10⁻³ m Step 2: Recall the formula for the position of the first minimum in diffraction. The condition for the first minimum in a single-slit diffraction pattern is a sinθ = nλ. For the first minimum, n = 1, so a sinθ = λ. This angle θ gives the angular position of the first minimum on one side of the central maximum. Step 3: Calculate the angular width. The total angular width of the central maximum is the separation between the first minima on both sides of the center, which is 2θ. For small angles, sinθ ≈ θ. So, aθ ≈ λ, which gives θ ≈ λ/a. The total angular width = 2θ = 2λ/a. Step 4: Substitute the values and calculate. Angular width = 2 * (500 × 10⁻⁹ m) / (0.1 × 10⁻³ m) Angular width = (1000 × 10⁻⁹) / (1 × 10⁻⁴) Angular width = 1000 × 10⁻⁵ = 10⁻² radians. Final Answer: The angular width of the central maximum is 0.01 radians.

Exam Tip: Interference vs. Diffraction Formulas

A very common mistake students make is mixing up the formulas for interference and diffraction. Remember this key difference:

• Interference (YDSE): The fringe width (β) is given by β = λD/d, where 'd' is the separation between the two slits. All bright fringes have nearly the same intensity.
• Diffraction (Single Slit): The width of the central maximum is given by 2λD/a, where 'a' is the width of the single slit. This is twice the width of the other secondary maxima. The intensity of successive maxima decreases rapidly.

Always double-check whether the question is about interference (usually two slits) or diffraction (usually one slit) before applying a formula.

Practice Questions with Solutions

• Q: Two coherent sources have intensities in the ratio 81:1. Calculate the ratio of the intensities of the maxima and minima in the interference pattern. A: Step 1: Relate intensity to amplitude. The intensity (I) is proportional to the square of the amplitude (A), so I₁/I₂ = (A₁/A₂)². Step 2: Find the ratio of the amplitudes. A₁/A₂ = √(I₁/I₂) = √(81/1) = 9/1. Let A₁ = 9k and A₂ = k. Step 3: Calculate the amplitudes of the maxima and minima. A_max = A₁ + A₂ = 9k + k = 10k. A_min = A₁ - A₂ = 9k - k = 8k. Step 4: Calculate the ratio of maximum to minimum intensity. I_max / I_min = (A_max / A_min)² = (10k / 8k)² = (10/8)² = (5/4)² = 25/16. Final answer: The ratio of intensities of maxima to minima is 25:16.
• Q: In a YDSE, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, what will be the new fringe width? A: Step 1: Write the formula for fringe width in air. β = λD/d, where λ is the wavelength in air. Step 2: Determine the new wavelength in water. When light enters a medium of refractive index μ, its wavelength changes to λ' = λ/μ. Step 3: Write the formula for the new fringe width (β') in water. β' = λ'D/d = (λ/μ)D/d. Step 4: Relate the new fringe width to the old one. From the formulas in Step 1 and 3, we can see that β' = β/μ. Step 5: Calculate the new fringe width. β' = 0.4 mm / (4/3) = (0.4 * 3) / 4 = 1.2 / 4 = 0.3 mm. Final answer: The new fringe width will be 0.3 mm.
• Q: Light of wavelength 6000 Å falls on a single slit of width 0.5 mm. What is the distance between the two first-order dark fringes on either side of the central bright fringe on a screen 2 m away? A: Step 1: Identify the given values. λ = 6000 Å = 6000 × 10⁻¹⁰ m, a = 0.5 mm = 0.5 × 10⁻³ m, D = 2 m. Step 2: Understand the question. The question asks for the width of the central bright fringe, which is the distance between the first dark fringe (n=1) on one side and the first dark fringe on the other side. Step 3: Find the position of the first dark fringe. The position of the nth dark fringe is given by y = nλD/a. For the first dark fringe (n=1), y₁ = λD/a. Step 4: Calculate the total width. The total distance is 2y₁ = 2λD/a. Step 5: Substitute the values. Width = 2 (6000 × 10⁻¹⁰ m) (2 m) / (0.5 × 10⁻³ m) = (24000 × 10⁻¹⁰) / (0.5 × 10⁻³) = 48000 × 10⁻⁷ m = 4.8 × 10⁻³ m = 4.8 mm. Final answer: The distance is 4.8 mm.
• Q: What is Brewster's angle for air to glass transition if the refractive index of glass is 1.5? A: Step 1: State Brewster's Law. Brewster's Law states that when unpolarized light is incident on a transparent medium at a particular angle, known as Brewster's angle (i_p), the reflected light is completely polarized. The law is given by μ = tan(i_p). Step 2: Identify the given quantity. The refractive index of glass, μ = 1.5. Step 3: Solve for Brewster's angle (i_p). We have tan(i_p) = 1.5. Step 4: Calculate the angle. i_p = tan⁻¹(1.5). Using a calculator, this gives i_p ≈ 56.3°. Final answer: Brewster's angle for the air to glass transition is approximately 56.3°.

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