An initially unpolarized beam of light, intensity Io , is propagating into the plane of the paper. The light travels through 4 linear polarizers. As measured counterclockwise from due East the angles of the polarizers are: 0 degrees, 45 degrees, 45 degrees, 90 degrees. If the final intensity of the light is 0.125Io what order must the polarizers be placed in (more than one right answer)?

Answer is given below.

Explanation:

1.

Solution:

Let the polarizers be P_{1} , P_{2} , P_{3} and P_{4} inclined at 0^{o} , 45^{o} ,45^{o}& 90^{o} counterclockwise from East..

(A)Consider the arrangement P_{1} P_{2}P_{3}P_{4}.

By Malu's Law,

I = lo cos

When the unpolarized light passes through P_{1} , the intensity of light transmitted is equal to half of incident light.

Thus, the intensity of light passing through P_{1} is

I_{1}=\frac{1}{2}I_{O}

 P_{1} and P_{2} are inclined at 45^{0} . The intensity of light transmitted through P_{2} is

01 = 7505 0 = 21

P_{2} and  P_{3} are inclined at 0^{0} to each other. The intensity of light transmitted through P_{3} is

 13 = -lo cos?o= =10

 P_{3} and P_{4} are inclined at 45^{0} to each other. The intensity of light transmitted through P_{4} is

 14 = -lo cos? 45 = -10=0.12510

unbalanced force

explanation: