
A sensor is exposed for time t to a lamp of power P placed at a distance l. The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is $\lambda $ is
A. $N=\dfrac{P\lambda {{d}^{2}}t}{hc{{l}^{2}}}$
B. $N=\dfrac{4P\lambda {{d}^{2}}t}{hc{{l}^{2}}}$
C. $N=\dfrac{P\lambda {{d}^{2}}t}{4hc{{l}^{2}}}$
D. $N=\dfrac{P\lambda {{d}^{2}}t}{16hc{{l}^{2}}}$
Answer
581.7k+ views
Hint:The number of photons entering the surface is proportional to the area of that surface. The basic relation between power, energy and time is a very necessary and important part of this question. Obtain the energy of light in terms of the wavelength of the light. Then we can find the number of photons entering the sensor.
Complete step by step solution:
Given that, a sensor is exposed for time t to a lamp of power P placed at distance l. We consider the energy of the lamp to be E.
Now we solve this equation step by step to reach the final expression.
We know the formula power as, $P=\dfrac{E}{t}$
The wavelength of light is $\lambda $ , then we can write above equation as,
$P=\dfrac{nhc}{\lambda t}$
$n=\dfrac{P\lambda t}{hc}\text{ }\to \text{ 1}$
where n is the total number of photons emitted by the source, h is Planck’s constant and c is the speed of light.
The sensor has an opening that is 4d in diameter. So, the area of that opening is $4\pi {{d}^{2}}$
The number of photons entering or leaving the surface is proportional to the area of that surface. So, we can write.
$\dfrac{N}{n}=\dfrac{4\pi {{d}^{2}}}{4\pi {{l}^{2}}}$
where N is the number of photons entering the sensor.
Putting the value from equation (1) in above equation, we get,
$\begin{align}
& N=\dfrac{P\lambda t}{hc}\times \dfrac{4\pi {{d}^{2}}}{4\pi {{l}^{2}}} \\
& N=\dfrac{P\lambda {{d}^{2}}t}{hc{{l}^{2}}} \\
\end{align}$
Hence, the number of photons entering the sensor is, $N=\dfrac{P\lambda {{d}^{2}}t}{hc{{l}^{2}}}$
Correct option is A.
Note: In this question, the diameter of the opening is given, so don’t forget to convert it in radius. Otherwise, the answer will be option B which is incorrect.
We can define the energy of photons in terms of frequency as, $E=nh\nu $. Again, frequency can be expressed in terms of wavelengths as, $\nu =\dfrac{c}{\lambda }$.
So, energy can be given as, $E=\dfrac{nhc}{\lambda }$
Complete step by step solution:
Given that, a sensor is exposed for time t to a lamp of power P placed at distance l. We consider the energy of the lamp to be E.
Now we solve this equation step by step to reach the final expression.
We know the formula power as, $P=\dfrac{E}{t}$
The wavelength of light is $\lambda $ , then we can write above equation as,
$P=\dfrac{nhc}{\lambda t}$
$n=\dfrac{P\lambda t}{hc}\text{ }\to \text{ 1}$
where n is the total number of photons emitted by the source, h is Planck’s constant and c is the speed of light.
The sensor has an opening that is 4d in diameter. So, the area of that opening is $4\pi {{d}^{2}}$
The number of photons entering or leaving the surface is proportional to the area of that surface. So, we can write.
$\dfrac{N}{n}=\dfrac{4\pi {{d}^{2}}}{4\pi {{l}^{2}}}$
where N is the number of photons entering the sensor.
Putting the value from equation (1) in above equation, we get,
$\begin{align}
& N=\dfrac{P\lambda t}{hc}\times \dfrac{4\pi {{d}^{2}}}{4\pi {{l}^{2}}} \\
& N=\dfrac{P\lambda {{d}^{2}}t}{hc{{l}^{2}}} \\
\end{align}$
Hence, the number of photons entering the sensor is, $N=\dfrac{P\lambda {{d}^{2}}t}{hc{{l}^{2}}}$
Correct option is A.
Note: In this question, the diameter of the opening is given, so don’t forget to convert it in radius. Otherwise, the answer will be option B which is incorrect.
We can define the energy of photons in terms of frequency as, $E=nh\nu $. Again, frequency can be expressed in terms of wavelengths as, $\nu =\dfrac{c}{\lambda }$.
So, energy can be given as, $E=\dfrac{nhc}{\lambda }$
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

