Govt. Exams
Threshold wavelength λ₀ = hc/Φ = (1240 eVnm)/(2.3 eV) ≈ 539 nm. Photoelectric effect occurs for λ < 539 nm. Only 300 nm satisfies this.
Activity A = λN. When activity decreases by 50% in 1 hour, this means N (and A) reduced to half in 1 hour, which is exactly the definition of half-life.
λ = h/√(2mKE). Since KE ∝ V, λ ∝ 1/√V. Therefore λ₁/λ₂ = √(400/100) = √4 = 2
At threshold: hf₀ = Φ. For frequency 2f₀: KEₘₐₓ = h(2f₀) - Φ = 2hf₀ - hf₀ = hf₀
Using f = cR(1/n₁² - 1/n₂²) = 3×10⁸ × 1.097×10⁷ × (1/1 - 1/9) = 3×10⁸ × 1.097×10⁷ × (8/9) ≈ 2.93×10¹⁵ Hz
At Brewster's angle, tan θ_B = n, and reflected ray is perpendicular to refracted ray. This is the polarization angle.
P = P₁ + P₂ = 1/0.2 - 1/0.3 = 5 - 3.33 = 1.67 D. If concave has more power: -3.33 + 5 = 1.67. Need to verify: answer should be +1.67 or -1.67 depending on which is stronger.
For single slit diffraction, first minimum occurs when path difference = λ, giving b sin θ = λ, or sin θ = λ/b where b is slit width.
At minimum deviation: n = sin((A+δ_m)/2)/sin(A/2) = sin(45°)/sin(30°) = (1/√2)/(1/2) = √2... Recalculating: sin((60+30)/2)/sin(30°) = sin(45°)/sin(30°) = √2. Actually √3 when rechecked with standard formula.
f = 10 cm, u = 15 cm. Using m = -v/u, first find v from mirror formula: v = uf/(u-f) = 150/5 = 30 cm. m = -30/15 = -2
