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Electronic Devices question

2015 · Shift 0 · Q44
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Electronic Devices question

2015 · Shift 0 · Q44

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
A red LEDLEDLED emits light at 0.10.10.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1m1m1m from the diode is :
  1. A
    5.48V/m5.48V/m5.48V/m
  2. B
    7.75V/m7.75V/m7.75V/m
  3. C
    1.73V/m1.73V/m1.73V/m
  4. D
    2.45V/m2.45V/m2.45V/m
View written solutionFree

Correct answer: D

  1. Given data
  • Power emitted by LED: P=0.1 WP = 0.1\,\text{W}P=0.1W
  • Distance from LED: r=1 mr = 1\,\text{m}r=1m
  • Emission is uniform in all directions, so assume isotropic radiation.
  1. Find the intensity at distance rrr

For isotropic emission,

I=P4πr2I = \frac{P}{4\pi r^2}I=4πr2P​

Substituting values:

I=0.14π(1)2=0.14πI = \frac{0.1}{4\pi (1)^2} = \frac{0.1}{4\pi}I=4π(1)20.1​=4π0.1​ I≈0.112.566≈7.96×10−3 W/m2I \approx \frac{0.1}{12.566} \approx 7.96 \times 10^{-3}\,\text{W/m}^2I≈12.5660.1​≈7.96×10−3W/m2
  1. Relate intensity to electric field amplitude

For an electromagnetic wave in free space, average intensity is

I=12cε0E02I = \frac{1}{2} c \varepsilon_0 E_0^2I=21​cε0​E02​

So,

E0=2Icε0E_0 = \sqrt{\frac{2I}{c\varepsilon_0}}E0​=cε0​2I​​

Using:

  • c=3×108 m/sc = 3 \times 10^8\,\text{m/s}c=3×108m/s
  • ε0=8.854×10−12 F/m\varepsilon_0 = 8.854 \times 10^{-12}\,\text{F/m}ε0​=8.854×10−12F/m
cε0=3×108×8.854×10−12=2.6562×10−3c\varepsilon_0 = 3 \times 10^8 \times 8.854 \times 10^{-12} = 2.6562 \times 10^{-3}cε0​=3×108×8.854×10−12=2.6562×10−3

Then,

E0=2(7.96×10−3)2.6562×10−3E_0 = \sqrt{\frac{2(7.96 \times 10^{-3})}{2.6562 \times 10^{-3}}}E0​=2.6562×10−32(7.96×10−3)​​ E0=5.99≈2.45 V/mE_0 = \sqrt{5.99} \approx 2.45\,\text{V/m}E0​=5.99​≈2.45V/m
  1. Match with options

The electric field amplitude is

2.45 V/m\boxed{2.45\,\text{V/m}}2.45V/m​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They match.

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