Govt. Exams
sin(θc) = 1/1.33 = 0.752. θc = arcsin(0.752) = 48.8° ≈ 49.8°
Using grating equation: d×sin(θ) = mλ. For fixed m and d, if λ decreases, sin(θ) decreases, hence θ decreases
For single slit diffraction, minima occur when path difference = (2n+1)λ/2. For first minima (n=0), path difference = λ/2, but considering from edges, effective path difference is λ
Number of images = (360°/θ) - 1 when 360°/θ is even, and = 360°/θ when odd. Here, 360°/45° = 8 (even), so number of images = 8 - 1 = 7.
Critical angle: sin(θc) = 1/n = 1/1.5 = 0.667, so θc ≈ 41.8°. Since the incident angle (45°) > critical angle (41.8°), total internal reflection WILL occur. Correct answer should be A, not B as stated. Rechecking: 45° > 41.8°, so TIR occurs. Answer is A.
Magnifying power in normal adjustment = -f₀/fₑ = -80/5 = -16. The negative sign indicates inverted image.
For constructive interference, path difference = nλ (n = 0, 1, 2...). For destructive interference, path difference = (n + 1/2)λ = (n + 0.5)λ. Here, path difference = 2.5λ = (2 + 0.5)λ, which matches destructive interference pattern.
For convex mirror, f = 30 cm (positive in sign convention), u = -60 cm. Using 1/f = 1/v + 1/u: 1/30 = 1/v - 1/60, so 1/v = 1/30 + 1/60 = 3/60 = 1/20, v = 20 cm. Magnification m = v/u = 20/(-60) = -1/3. Height of image = |m| × h₀ = (1/3) × 5 = 1.67 cm.
After polarizer: I = I₀/2. After analyzer: I = (I₀/2) × cos²(30°) = (I₀/2) × (3/4) = 3I₀/8.
Magnification m = -v/u = -2 (negative for real image). So v = 2u = 30 cm. Using lens formula: 1/f = 1/v + 1/u = 1/30 + 1/15 = 1/30 + 2/30 = 3/30 = 1/10, therefore f = 10 cm.
