Question

1. A disc shaped photoelectric detector of area 0.5 cm² generates current when light of frequency > that of yellow light falls on it. The photoelectron generation efficiency is 20%. In the arrangement shown, S is an isotropic 100 watt point monochromatic light source with $\lambda$ = 4000 Å. The focal of the lens is 20 cm and its circular area is 4 cm². Find the photocurrent (in mA) in the detector assuming all rays to be paraxial w.r.t. lens.

Answer 1

1.13 mA

Answer 2

Solution:

1. Power intercepted by the lens:
    • Lens area = 4 cm² = 4×10⁻⁴ m²
    • At a distance r = 30 cm = 0.3 m, the spherical area is
     A_sphere = 4π(0.3)² ≈ 1.13 m²
    • Fraction intercepted = (4×10⁻⁴) / 1.13 ≈ 3.54×10⁻⁴
    • Hence, power on lens = 100 W × 3.54×10⁻⁴ ≈ 0.0354 W

2. Illumination of the detector:
    • Lens forms an image of the point source at v given by
     1/f = 1/u + 1/v → 1/20 = 1/30 + 1/v ⇒ v = 60 cm
    • After focus the converging rays diverge as though emanating from the focus. At the detector (placed 90 cm from the lens, i.e. 30 cm beyond the focus), the beam radius is proportional to (30/60) times the lens radius.
     Lens radius, rₗ = √(Area/π) = √(4/π) ≈ 1.13 cm
     Beam radius on detector ≈ 1.13/2 ≈ 0.565 cm
    • Beam area ≈ π (0.565)² ≈ 1.0 cm²
    • The detector has an area of 0.5 cm² so it collects half the beam’s power.
     ⇒ Power on detector ≈ 0.0354 W × 0.5 = 0.0177 W

3. Photon flux and photoelectron current:
    • Wavelength λ = 4000 Å = 4×10⁻⁷ m
    • Energy per photon:
     E = hc/λ = (6.626×10⁻³⁴ J·s × 3×10⁸ m/s) / (4×10⁻⁷ m) ≈ 5×10⁻¹⁹ J
    • Number of photons incident per second:
     N = 0.0177 W / (5×10⁻¹⁹ J) ≈ 3.54×10¹⁶ s⁻¹
    • With 20% efficiency, photoelectrons generated per second = 0.2×3.54×10¹⁶ ≈ 7.08×10¹⁵ s⁻¹
    • Photocurrent:
     I = (number/s) × (electron charge) = 7.08×10¹⁵ × 1.6×10⁻¹⁹ C ≈ 1.13×10⁻³ A = 1.13 mA

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Answer:
The photocurrent is approximately 1.13 mA.

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