JEE Physics - MCQ Practice Questions

Diffraction and interference are characteristic wave phenomena. Other options demonstrate particle nature of light.

Using lens formula: 1/f = 1/u + 1/v. $\frac{1}{15}$ = $\frac{1}{30}$ + 1/v. 1/v = $\frac{1}{15}$ - $\frac{1}{30}$ = $\frac{1}{30}$. v = 30 cm (real image)

sin(θc) = n₂/n₁ = $\frac{1}{1}$.5 = $\frac{2}{3}$. θc = arcsin(0.667) = 41.81°

In a plane mirror, image distance equals object distance. Total separation = u + v = 50 + 50 = 100 cm

Fringe width β = λD/d = (500×10⁻⁹ × 1)/(0.5×10⁻³) = 1×10⁻³ m = 1 mm

When object is at center of curvature (u = R = 2f), image forms at the same position (v = R). Magnification = -1

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

Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))

Higher frequency → higher refractive index → slower speed in medium → greater deviation. All statements are correct.

Using lens formula: 1/(-20) = $\frac{1}{10}$ + 1/v. v = -6.67 cm. Magnification m = -v/u = 6.$\frac{67}{10}$ = 0.67

Only transverse waves can be polarized. Polarization demonstrates the transverse nature of electromagnetic waves.

Power P = 1/f. f = $\frac{1}{5}$ = 0.2 m = 20 cm. Positive power indicates convex lens.

Optical path = μt, geometrical path = t. Additional path difference = μt - t = (μ-1)t

Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²

Lens maker's formula: 1/f = (n_lens/n_medium - 1)(1/R₁ - 1/R₂). When medium changes from air to water, (n_lens/n_medium) decreases, so f increases.

All three conditions are equivalent and occur at minimum deviation: symmetric path, equal angles, and ray parallel to base.

tan(θ_B) = n = 1.5. θ_B = arctan(1.5) = 56.31°. At this angle, reflected light is completely polarized.

Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂). So 1.5 × sin(30°) = 1.0 × sin(θ₂). This gives sin(θ₂) = 0.75, therefore θ₂ = 48.6°

For a concave mirror, when object is between F and C, the image is real, inverted, and magnified. This is the magnification region used in shaving mirrors.

For single-slit diffraction, minima occur at positions where a·sin(θ) = mλ, where m = 1, 2, 3... Since sin is approximately proportional to θ for small angles, the second minimum occurs at approximately 2θ.