/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A red LED emits light at \(0.1\)... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 73

A red LED emits light at \(0.1\) watt uniformly around it. The amplitude of the electric field of the light at a distance of \(1 \mathrm{~m}\) from the diode is [2015] (A) \(2.45 \mathrm{~V} / \mathrm{m}\) (B) \(5.48 \mathrm{~V} / \mathrm{m}\) (C) \(7.75 \mathrm{~V} / \mathrm{m}\) (D) \(1.73 \mathrm{~V} / \mathrm{m}\)

Short Answer

Expert verified
The amplitude of the electric field of the light at a distance of 1 meter from the diode is (A) \(2.45 \mathrm{~V/m}\).

Step by step solution

01

Find the intensity of the light at a distance of 1m

We are given the power of the LED (\(P\)) and the distance from it (\(r\)), so we can plug in those values in the formula above to find the intensity (\(I\)). $$I = \frac{0.1\mathrm{~W}}{4\pi (1\mathrm{~m})^2}$$ Calculate the intensity: $$I = \frac{0.1}{4\pi} \approx 0.00796 \mathrm{~~W/m^2}$$
02

Relate the light intensity to the electric field amplitude

Now that we have the intensity (\(I\)) of the light at a distance of 1m, we can use the formula: $$I = \frac{1}{2}\epsilon_0 c \cdot E^2$$ Here, \(\epsilon_0\) is the vacuum permittivity whose value is approximately \(8.85 \times 10^{-12} \mathrm{~F/m}\), and \(c\) is the speed of light which is approximately \(3 \times 10^8 \mathrm{~m/s}\).
03

Solve for the amplitude of the electric field (E)

From the above formula, we can solve for the amplitude of the electric field (\(E\)) which is our desired answer. We have: $$E = \sqrt{\frac{2I}{\epsilon_0 c }}$$ Plugging in the values from the previous steps, we get: $$E = \sqrt{\frac{2(0.00796 \mathrm{~W/m^2})}{(8.85\times10^{-12}\mathrm{~F/m})(3\times10^{8}\mathrm{~m/s})}}$$ Calculate the amplitude of the electric field: $$E \approx 2.45 \mathrm{~V/m}$$ So, the answer is (A) \(2.45 \mathrm{~V/m}\).

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