Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to the incident polarization to get a transmission intensity that is 0.464 I0?

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 (sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees