JEE Physics - MCQ Practice Questions

Fringe width β = λD/d = (632.8 × 10⁻⁹ × 1)/(0.5 × 10⁻³) = 1.26 × 10⁻³ m = 1.26 mm

1/F = 1/f₁ + 1/f₂. $\frac{1}{10}$ = $\frac{1}{15}$ + 1/f₂. 1/f₂ = $\frac{1}{10}$ - $\frac{1}{15}$ = $\frac{1}{30}$. Therefore f₂ = -30 cm (diverging lens)

Magnifying power m = -f₀/fe = -$\frac{100}{5}$ = -20 (negative sign indicates inverted image)

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

For destructive interference, path difference = (2n+1)λ/2. Here 2.5λ = 5λ/2 = (2×2+1)λ/2, which is destructive.

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).

Beyond critical angle, total internal reflection occurs - Snell's law cannot be applied as there is no refracted ray, only reflected ray.

Optical path introduced = (n-1)t. Shift in fringes = optical path/wavelength = (n-1)t/λ

f = 10 cm, u = 15 cm. Using m = -v/u, first find v from mirror formula: v = uf/(u-f) = $\frac{150}{5}$ = 30 cm. m = -$\frac{30}{15}$ = -2

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 Brewster's angle, tan θ_B = n, and reflected ray is perpendicular to refracted ray. This is the polarization angle.

Using f = cR(1/n₁² - 1/n₂²) = 3×10⁸ × 1.097×10⁷ × ($\frac{1}{1}$ - $\frac{1}{9}$) = 3×10⁸ × 1.097×10⁷ × ($\frac{8}{9}$) ≈ 2.93×10¹⁵ Hz

At threshold: hf₀ = Φ. For frequency 2f₀: KEₘₐₓ = h(2f₀) - Φ = 2hf₀ - hf₀ = hf₀

λ = h/√(2mKE). Since KE ∝ V, λ ∝ 1/√V. Therefore λ₁/λ₂ = √($\frac{400}{100}$) = √4 = 2

Activity A = λN. When activity decreases by 50% in 1 hour, this means N (and A) reduced to half in 1 hour, which is exactly the definition of half-life.

Threshold wavelength λ₀ = hc/Φ = (1240 eVnm)/(2.3 eV) ≈ 539 nm. Photoelectric effect occurs for λ < 539 nm. Only 300 nm satisfies this.

For hydrogen-like atoms: E = -13.6Z²/n² eV. For He⁺ (Z=2): E = -13.6×4 = -54.4 eV. For H (Z=1): E = -13.6 eV. Ratio is 4:1

Characteristic X-rays arise from inner-shell electron transitions (e.g., L→K shell). Each element has specific wavelengths, hence 'characteristic'.

ΔxΔp ≥ h/4π. If Δx = 0, then Δp must be ≥ ∞ to satisfy the inequality, making momentum completely uncertain.

Pair production creates an electron-positron pair. Minimum energy = 2mₑc² = 2 × 0.511 MeV = 1.022 MeV