Question

SI unit of intensity of light is:
A. Newton
B. Kelvin
C. Kilogram
D. Candela

Answer 1

Hint: In this question, we will first see the definition of intensity of light. Light consists of both electric and magnetic fields. Each of these quantities has energy. So, we will find the intensity of the light using the energy concept of the wave and after this will get the SI unit of light intensity.

Complete Step-by-Step solution:
Light is an electromagnetic wave.
Let \[\vec S\] be the pointing vector then the magnitude of the pointing vector gives the power per unit area of the wave.
$S = \dfrac{1}{{{\mu _0}}}EB$
Where, S,E and B are the instantaneous values at the observation point.
We know that
E=cB
Where, c is speed of light
We can also write:
$S = \dfrac{1}{{{\mu _0}c}}{E^2}{\text{ or }}S = \dfrac{1}{{{\mu _0}c}}{B^2}$ -------(1)
The Pointing vector\[\vec S\]fluctuates very rapidly with time. The frequency is about ${10^{15}}$Hz.
Intensity of light:-
It is defined as the time average of magnitude of pointing vector ‘S’.
I = \[{S_{av}} = S = \dfrac{1}{{{\mu _0}c}}{({E^2}{\text{)}}_{av}}{\text{ = }}\dfrac{1}{{{\mu _0}c}}{E^2}_m{\left[ {{{\sin }^2}(kx - \omega t)} \right]_{av}}\]
The average value of ${\sin ^2}$ function is $\dfrac{1}{2}$ . We thus obtain:
$I = S = \dfrac{1}{{2{\mu _0}c}}{E_m}^2 = \dfrac{1}{{2{\mu _0}c}}{B_m}{E_m}$
The intensity of light may also be expressed in terms of rms magnitude of the fields. With $E = \sqrt 2 {E_{rms}}$ , we obtain
$S = \dfrac{1}{{{\mu _0}c}}{E^2}_{rms}$ .
The unit of intensity of light is Candela.
So, option D is correct.

Note- The intensity of light is also defined as the number of photons coming through a unit solid angle in unit time. The more photons emitted per unit time, the greater the intensity of the light and less is the number of photons emitted per unit time, the lesser the intensity of the light.