Luminosity refers to the total amount of energy produced by different celestial bodies(stars, galaxy) per unit time and it is basically measured in joules per second or watts in SI units.

We can define luminosity as: The absolute measure of radiated electromagnetic power (light), the radiant power emitted by a light-emitting object. Values of luminosity are given in terms of the luminosity of the sun or in terms of magnitude which is called the **absolute bolometric magnitude of an object** is the measure of the total energy emission rate. A bolometer can be used to measure the radiant energy by the method of absorption and measurement of heating.

Luminosity generally depends upon two factors:

**The size of the star:**The larger a star is, the more energy it puts out and the more luminous it is.**The temperature of the star:**If two stars are of the same size but have different temperatures, then the star with a higher temperature will be more luminous than the star with lower temperature.

The brightness of a star which appears to our naked eyes mainly depends upon two factors:

- The luminosity of the star
- The distance of the star from the Earth

**The brightness of a Star**

The Apparent Brightness of a source is a consequence of geometry. As light rays emerge from a source, they spread out in an area. According to Inverse Square Law of brightness, the Apparent Brightness (B) of a source is inversely proportional to the square of its distance (d).

Mathematically,

**\(B\propto \frac{1}{d^{2}}\)**

Luminosity depends on the surface area of the star. If the radius of a star is R then,

**The surface area of the star = 4PR****2**

Two stars having the same temperature, one with radius 2R will have 4 times greater luminosity than a star with radius R.

The luminosity of a star also depends upon its temperature. When considering a star to be a completely black body, the radiation emitted per second will be according to the Stefan- Boltzmann law.

** Energy emitted per second (E) = sAT****4**

Where,

- s= Stefan’s constant with a value of
**\(5.7\times10^{-8}Wm^{-2}K^{-4}\)** - A= Surface Area of the Star
- T = absolute temperature of the star

Calculating the energy output for a star which is of the same size as the sun.

- R = 6.96x108m
- T = 6000K

So, surface area of sun = surface area of star = A = **\(4\pi r^{2}\)**

**\(= 4\times \pi \times ( 6.96\times 10^{8})\\ \\ =6.08\times 10^{18}\)**

Therefore, **\(E=5.7\times10^{-8}\times6.08\times10^{18}\times6000^{4}=4.5\times10^{26}J\)**

A star having twice the temperature of Sun would have 16 times greater energy output.

The Sun has a surface temperature of 6000K produces radiation with lmax = 420 nm. Find out the temperature of the Sirius if l max of Sirius is 72nm.

**Solution:**

**\(l_{max1}T_{1}=l_{max2}T_{2}\)**

Therefore,

**\(420\times10^{-9}\times6000=72\times10^{-9}\times T_{2}\)**

**\(T_{2}=35000K\)**