Answer:
Answer is given below.
Step-by-step 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