# Unpolarized light shines pon a photocell for one hour. then, two polarizers, whose transmission axies

Unpolarized light shines pon a photocell for one hour. then, two polarizers, whose transmission axies 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?

## This Post Has 3 Comments

t = 1.10 h

Explanation:

For this exercise we must calculate the intensity that reaches the cell with the polarizers, for this we use Malus's law

I = I₀ cos² θ

Let's calculate

I = I₀ cos² 60

I = I₀ 0.907

This is the intensity that reaches the cell

To find time, use an inverse proportion rule or rule of three, the intensity Io in

t = I₀ / (0.907 I₀) 1 h

t = 1.10 h

2. Expert says:

real, smaller and upside-down.   so the answer should be b.

3. Expert says:

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.