Question

Luminous flux and luminous intensity are related as
(A) $\phi = 4\pi I$
(B) $\phi = \dfrac{I}{{4\pi }}$
(C) $\phi = \dfrac{{4\pi }}{I}$
(D) $I = \dfrac{{4\pi }}{\phi }$

Answer 1

Hint : Luminous intensity can be defined as the luminous flux per unit solid angle. The solid angle subtended by a spherical surface is $4\pi$ . So by dividing the luminous flux by the sold angle we will get the intensity.

Complete step by step answer
Let’s consider a radiation of light from a bulb, this bulb must radiate perceivable visible light energy at a particular rate. This rate at which this visible light energy is radiated is called luminous flux. The unit of the luminous flux is given as the lumen, but can also have the same unit of power as the Watt (since they are equivalent as in they are both rate of energy change). The lumen in the base SI unit can also be written as candela-steradian.
Luminous intensity on the other hand can be defined as the luminous flux per unit solid angle. The luminous intensity is a measure of how much light is “packed” moving in a particular direction. It can be said to be the measurement of the number of rays of light within a particular solid angle (since, it is three dimensional). It has an SI unit of candela. But can be written in terms of luminous flux as lumen per steradian.
From this definition, it is clear that we can write the luminous intensity formula as
$I = \dfrac{\phi }{\theta }$ where $\phi$ is the luminous flux and $\theta$ is the solid angle. Usually since light is often diverging, the total luminous flux(light rays) is hence contained in an imaginary sphere about the source. Then, the luminous intensity would be given by
$I = \dfrac{\phi }{{4\pi }}$ (because the solid angle subtended by a square is $4\pi$ )
Rearranging, we have
$\phi = 4\pi I$
Hence, the correct option is A.

Note
Alternatively, from the unit, the lumen (unit of luminous flux) as mentioned is given as
$l = cd \cdot sr$ where $cd$ is an abbreviation of candela and $sr$ for steradian.
Hence, we can deduce, by replacing the unit with their quantity, that
$\phi = 4\pi I$.