Question

Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-10 W/m2 a lit, 1000-watt light bulb mounted on it. The unpolarized light from the light bulb, upon reaching the telescope, is passed through two polarizing filters that are at an angle of 30° with respect to each other, and then the resulting light is projected onto a CCD camera. What is the intensity of the light detected by the camera?

Answer 1

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$