Govt. Exams
For concave lens, images are always virtual, erect, and diminished regardless of object position. Using 1/v = 1/f - 1/u = -1/20 - 1/10 = -3/20, v = -20/3 ≈ -6.67 cm (virtual).
At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.
For bright fringes: y = (m·λ·D)/d. For 5th bright fringe: 2.5 × 10⁻³ = (5 × λ × 1)/(1 × 10⁻³). Therefore λ = 500 nm.
Magnification m = -v/u = -2, so v = 2u. Using lens equation: 1/f = 1/u + 1/v = 1/u + 1/(2u) = 3/(2u). Therefore u = 3f/2.
Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²
Only transverse waves can be polarized. Polarization demonstrates the transverse nature of electromagnetic waves.
Using lens formula: 1/(-20) = 1/10 + 1/v. v = -6.67 cm. Magnification m = -v/u = 6.67/10 = 0.67
Higher frequency → higher refractive index → slower speed in medium → greater deviation. All statements are correct.
Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))
sin(θ₂) = sin(45°)/1.6 = 0.441. θ₂ = 26.19°. Lateral displacement = t×sin(θ₁-θ₂)/cos(θ₂) = 5×sin(18.81°)/cos(26.19°) ≈ 2.16 cm
