Question

Hlo..... Try.. This................ ​

Answer 1

Answer:

after passing through polarized A. The intensity of light is

I1= Io/2

the intensity of light emergency from polarised be as per malus law is given by

I2=I1cos^2(45°)

(Io/2)(I/√2)^2

=Io/4

Answer 2

$\huge{\mathfrak{HEYA\:MATE\:!!}}$

Answer...

★Intensity through polariser A ,

$I_1 = \frac{I_o}{2}$

According to Law of Malus ,

★Intensity through analyser B

$I_2 = I_1 {cos}^{2} 45° \\ \\ I_2 = \frac{ I_o \: {cos}^{2}45° }{2} \\ \\ I_2 = \frac{I_o }{2 \times 2} \\ \\ I_2 = \frac{ I_o }{4}$

Option d.

$\mathscr{\large{Hope\:helps\:}}!! \\ \\ {GRACIAS!!}$

$\huge{\red{\ddot{\smile}}}$