If the distance between a point source of light and a screen is doubled, then the intensity will be

Electromagnetic Waves

NEET

Two times of initial

Four times of initial

Half of initial

One fourth of initial

Solution:

For a point source of light, the intensity (I) of light at a distance (r) from the source follows the inverse square law. This means that the intensity is inversely proportional to the square of the distance from the source:

$I \propto \frac{1}{r^2}$

Let the initial distance be $r_1$ and the initial intensity be $I_1$ . So, $I_1 = \frac{k}{r_1^2}$ for some constant k.

If the distance is doubled, the new distance $r_2 = 2r_1$ .

The new intensity $I_2$ will be:

$I_2 = \frac{k}{r_2^2} = \frac{k}{(2r_1)^2} = \frac{k}{4r_1^2} = \frac{1}{4} \left(\frac{k}{r_1^2}\right) = \frac{1}{4} I_1$

Thus, the intensity will become one fourth of the initial intensity.

If the distance between a point source of light and a screen is doubled, then the intensity will be

Solution:

Related Questions