Two polarizing sheets are placed together with their transmission axes crossed so that no light is transmitted. A third sheet is inserted between them with its transmission axis at an angle of 33.0° with respect to the axis of the first sheet. Find the fraction of incident unpolarized light intensity transmitted by the three-sheet combination. (Assume each polarizing sheet is ideal.)

I3 = 0.1043 I(o)

Explanation:

When the two sheets are not allowing light, the angle between their axes is 90°

Thus when the axis of the third sheet is making angle Ө with the first the angle between the axes of

second and third will be 90° - Ө

The intensity of the unpolarized light incident on the first sheet is I(o) then half of the light is emerging from it as light polarized along its axis and hence the intensity becomes

I1= I(o)/2.

When this linearly polarized light passes through the second sheet its intensity is given by

Malus law as

I2 = [I(o)/2] cos2Ө

This light again passes through the third sheet whose axis is making an angle 90 – Ө

hence using Malus law again the intensity of the transmitted beam is given by

I3 = I1*cos² (90 - Ө) = [I(o) /2] cos²Ө*sin² Ө

Where, Ө = 33°, so

I3 = I(o)/2 * cos²33 * sin² 33

I3 = I(o)/2 * 0.7034 * 0.2966

I3 = I(o)/2 * 0.2086

I3 = 0.1043 I(o)

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s