Answer to Question #111041 in Optics for please101

Question

[img]https://upload.cc/i1/2020/04/20/y3ELI0.jpg[/img]


R=4

Answer 1

As per the given question,

We know that $I=I_o \cos^2\theta$

When any ray of light is passing through any polarizer then, it's intensity becomes, 1/2 of the initial intensity of light,

$\Rightarrow I_o/2=I_o\cos^2\theta$

$\Rightarrow \cos^2\theta=\dfrac{1}{2}$

$\Rightarrow \cos\theta=\sqrt{\dfrac{1}{2}}=\cos 45^\circ$

$\theta=45^\circ$

b) Now, $I_2=I_1\cos^2\theta$

$I_2=\dfrac{I_o}{2}\cos^2(10+8R)$

Here the value of R is not given,

c) As per the question $I_2= (R+1)\% of I_o/2$

$\Rightarrow (R+1)\dfrac{I_o}{2\times 100}=\dfrac{I_o}{2}\cos^2(10+8R)$

$\Rightarrow \dfrac{(R+1)}{100}=\cos^2(10+8R)$

value of R must be given.