Govt. Exams
Optical path introduced = (n-1)t. Shift in fringes = optical path/wavelength = (n-1)t/λ
Beyond critical angle, total internal reflection occurs - Snell's law cannot be applied as there is no refracted ray, only reflected ray.
For magnifying glass (virtual image), object is placed between f and lens. Maximum magnification occurs when final image is at least distance of distinct vision (25 cm).
For destructive interference, path difference = (2n+1)λ/2. Here 2.5λ = 5λ/2 = (2×2+1)λ/2, which is destructive.
Resolving power ∝ 1/θ_min where θ_min = 1.22λ/D. So resolving power ∝ D/λ, inversely proportional to both wavelength and inversely related to aperture diameter
Magnifying power m = -f₀/fe = -100/5 = -20 (negative sign indicates inverted image)
1/F = 1/f₁ + 1/f₂. 1/10 = 1/15 + 1/f₂. 1/f₂ = 1/10 - 1/15 = 1/30. Therefore f₂ = -30 cm (diverging lens)
Fringe width β = λD/d = (632.8 × 10⁻⁹ × 1)/(0.5 × 10⁻³) = 1.26 × 10⁻³ m = 1.26 mm
Using lens formula: 1/f = 1/u + 1/v. 1/15 = 1/20 + 1/v. v = 60 cm (positive, real image). Since u < 2f, magnification |m| = 60/20 = 3 > 1 (magnified)
After first polarizer: I = I₀/2. After second polarizer (at 60°): I = (I₀/2)cos²(60°) = (I₀/2)(1/4) = I₀/8
