A photodiode sensor is used to measure the output of a 300W lamp kept 10 m away. The sensor has an opening of 2 cm in diameter. How many photons enter the sensor if the wavelength of the light is 660 nm and the exposure time is 100 ms.(Assume that all the energy of the lamp is given  off as light and h=$6.6 \times 10^{-34}$ Js)

A) $3.6 \times 10^{13}$

B) $2.8 \times 10^{13}$

C) $2.5 \times 10^{13}$

D) $1.8 \times 10^{13}$


Option C


Given that, power of lamp , P=300 W

distance of lamp from sensor , d=10m

 radius of sensor opening , r=1cm, wavelength of  light emitted from  photodiode, $\lambda$=660 nm and exposure time, t= 100 ms=$100 \times 10^{-3}$ s

 Energy of photons , $E= \frac{hc}{\lambda}$

 Putting the given values , we get

 $\Rightarrow$         $E=\frac{6.6\times 10^{-34}\times 3\times10^{8}}{660 nm}=3 \times 10^{-19}$

The area of of exposure of lamp at a radius  of 10 m

 $A_{0}$ =$ 4 \pi r^{2}$= $4 \pi \times 10^{2} m^{2}$

Similarly, The area of sensor

A's=$\pi r^{2}$= $\pi  \times (10^{-2})^{2}$= $\pi \times 10^{-4} m^{2}$

Energy emitted by lamp in exposure time,

$E_{0}$ = power of lamp (p) x exposure time (t)

 $E_{0} =300 \times 100 \times 10^{-3}$ =3 J

So, the number of photons entering  to sensor is given by

 $N= \frac{E_{0}}{E}\times\frac{A'_{s}}{A_{0}}$

 =$\frac{30}{3\times 10^{-19}}\times\frac{\pi\times 10^{-4} }{4\pi\times 10^{2}}$

$\Rightarrow$       N=$2.5 \times 10^{13}$

 So, the correct option is (C)