Question

Unpolarized light is shines upon a photocell for one hour. Then, two polarizers, whose transmission axes are offset by 60°, are placed in front of the photocell such that the light must first pass through the polarizers before reaching the cell. How much time is required for the photocell to receive the same total energy as it received in one hour’s time without the polarizers present?

Answer 1

Suppose when unpolarized light of I0 intensity shine on photocell for one hour, then energy received by photocell will be:

Intensity = Power/Area = Energy/(time*Area)

I0 = E0/(t0*A)

E0 = I0*A*t0

Now when Unpolarized light of I0 intensity passes through polarizer, then intensity of light will reduced to half (I0/2) and light will be polarized, after that when light passes through 2nd polarizer at 60 deg angle from first, then

Using Malus's law:

I2 = I1*cos2 60 deg

I2 = (I0/2)*cos2 60 deg

I2 = I0*(1/2)*(cos 60 deg)^2 = 0.125*I0

So now intensity of light received by photocell will be 0.125*I0

Now if we want photocells to receive same energy then

E0 = E2

I0*A*t0 = I2*A*t2

See that area at which light shines remains same, So

t2 = (I0/I2)*t0

t2 = (1/0.125)*t0

t0 = 1 hr, So

t2 = (1/0.125)*1 = 8 hr

So now it will take 8 hrs time for same total energy

Let me know if you've any query.