A red LED emits light at 0.1 W uniformly in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The average energy density $U_{av}$ of an electromagnetic wave is given by $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$,where $E_{0}$ is the a

Vedclass · Accepted Answer

$2.45$

Vedclass · Answer

(A) The average energy density $U_{av}$ of an electromagnetic wave is given by $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$,where $E_{0}$ is the amplitude of the electric field.The intensity $I$ at a distance $r$ from a point source is $I = \frac{P}{4 \pi r^{2}}$.Also,intensity is related to energy density by $I = U_{av} 	imes c$,where $c$ is the speed of light.Equating the two expressions for intensity: $\frac{P}{4 \pi r^{2}} = \frac{1}{2} \varepsilon_{0} E_{0}^{2} c$.Rearranging for $E_{0}^{2}$: $E_{0}^{2} = \frac{2P}{4 \pi r^{2} \varepsilon_{0} c} = \frac{2P}{r^{2}} 	imes \frac{1}{4 \pi \varepsilon_{0}} 	imes \frac{1}{c}$.Given $P = 0.1 \ W$,$r = 1 \ m$,$\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \ N \ m^{2} \ C^{-2}$,and $c = 3 	imes 10^{8} \ m/s$.$E_{0}^{2} = \frac{2 	imes 0.1}{1^{2}} 	imes (9 	imes 10^{9}) 	imes \frac{1}{3 	imes 10^{8}} = 0.2 	imes 3 	imes 10 = 6$.$E_{0} = \sqrt{6} \approx 2.45 \ V \ m^{-1}$.

$A$ red $LED$ emits light at $0.1 \ W$ uniformly in all directions. The amplitude of the electric field of the light at a distance of $1 \ m$ from the diode is.... $V \ m^{-1}$

Similar Questions

An $LED$ is constructed from a $pn$ junction based on a certain semiconducting material whose energy gap is $1.9 \ eV$. The wavelength of the emitted light is:

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 \; nm$ is incident on it. The band gap in $eV$ for the semiconductor is

$A$ $p-n$ junction photodiode is fabricated from a semiconductor with a band gap of $2.5 eV$. It can detect a signal of wavelength: (Given: Planck's constant $h = 6.6 \times 10^{-34} Js$,speed of light $c = 3 \times 10^8 m/s$,elementary charge $e = 1.6 \times 10^{-19} C$)

Assertion: $A$ $p-n$ junction with reverse bias can be used as a photo-diode to measure light intensity.
Reason: In a reverse bias condition,the current is small but is more sensitive to changes in incident light intensity.

The $I-V$ characteristics for a junction diode are shown. The device is

Vedclass Test Series

Exam Paper Generator

Online Exam Module