Wave Optics
Tamil Nadu Board · Class 12 · Physics
Most important questions from Wave Optics for Tamil Nadu Board Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
According to Malus' Law, plane polarised light of intensity I₀ passes through an analyser whose transmission axis makes an angle of 60° with the polariser. What is the intensity of the emergent light?
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I₀/4
Step 1: Malus' Law states I = I₀ cos²θ. Step 2: θ = 60°, so cos 60° = 1/2. Step 3: I = I₀ × (1/2)² = I₀ × 1/4 = I₀/4. Option B (I₀/2) is incorrect — students often apply cos θ instead of cos²θ. Option C (3I₀/4) would correspond to θ = 30°. Option D involves the wrong trigonometric function. The law specifically uses the square of cosine.
In a single slit diffraction experiment, the slit width is 0.5 mm and the screen is 1 m away. If the wavelength of light is 500 nm, the width of the central maximum on the screen is:
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2 mm
Step 1: The central maximum extends from the first minimum on one side to the first minimum on the other side. Step 2: Position of first minimum: y₁ = λD/a = (500 × 10⁻⁹ × 1) / (0.5 × 10⁻³) = 1 × 10⁻³ m = 1 mm. Step 3: Width of central maximum = 2y₁ = 2 × 1 mm = 2 mm. Step 4: The factor of 2 is because the central maximum spreads on BOTH sides. Option B (1 mm) is just y₁ — a common mistake of forgetting both sides. Option A is less than y₁. Option D doubles incorrectly.
Brewster's law states that when light is incident at the polarising angle iₚ on a surface with refractive index n, which of the following relations is correct?
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tan iₚ = n
Step 1: At the polarising angle iₚ, the reflected and refracted rays are perpendicular to each other. Step 2: This means rₚ = 90° − iₚ. Step 3: Applying Snell's law: sin iₚ / sin rₚ = n → sin iₚ / sin(90° − iₚ) = n. Step 4: Since sin(90° − iₚ) = cos iₚ, we get sin iₚ / cos iₚ = tan iₚ = n. This is Brewster's Law. Options A, B, D use wrong trigonometric ratios. Note that for glass (n = 1.5), iₚ = tan⁻¹(1.5) ≈ 56.3°.
The wavelength of light from a sodium source in vacuum is 5893 Å. When this light travels in water (refractive index 1.33), its wavelength in water is approximately:
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4431 Å
Step 1: Use the relation λ₁/λ₂ = n₂/n₁. Step 2: λ₁ = 5893 Å (in vacuum, n₁ = 1), n₂ = 1.33. Step 3: λ₂ = λ₁ × (n₁/n₂) = 5893 × (1/1.33). Step 4: λ₂ = 5893 / 1.33 ≈ 4431 Å. The wavelength decreases in a denser medium because light slows down. Option A (5893 Å) — frequency stays the same but wavelength changes. Option B (7837 Å) is wrong — wavelength increases (incorrect direction). Option D is too small. Note: frequency remains unchanged in all media.
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