Answer to Question #111041 in Optics for please101

Question #111041
[img]https://upload.cc/i1/2020/04/20/y3ELI0.jpg[/img]


R=4
1
Expert's answer
2020-04-27T10:28:18-0400

As per the given question,


We know that I=Iocos2θI=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,

Io/2=Iocos2θ\Rightarrow I_o/2=I_o\cos^2\theta

cos2θ=12\Rightarrow \cos^2\theta=\dfrac{1}{2}

cosθ=12=cos45\Rightarrow \cos\theta=\sqrt{\dfrac{1}{2}}=\cos 45^\circ

θ=45\theta=45^\circ

b) Now, I2=I1cos2θI_2=I_1\cos^2\theta

I2=Io2cos2(10+8R)I_2=\dfrac{I_o}{2}\cos^2(10+8R)

Here the value of R is not given,

c) As per the question I2=(R+1)%ofIo/2I_2= (R+1)\% of I_o/2

(R+1)Io2×100=Io2cos2(10+8R)\Rightarrow (R+1)\dfrac{I_o}{2\times 100}=\dfrac{I_o}{2}\cos^2(10+8R)

(R+1)100=cos2(10+8R)\Rightarrow \dfrac{(R+1)}{100}=\cos^2(10+8R)

value of R must be given.


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