Govt. Exams
Entrance Exams
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
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
