Answer:
[tex]\frac{I_3}{I_0} = 0.12[/tex]
Explanation:
As we know that when unpolarized light passes through a polarizer then its intensity becomes half of the initial intensity
so after passing through first polarizer the intensity of light is given as
[tex]I_1 = \frac{I_0}{2}[/tex]
now By Malus law we know that when polarized light passes through a polarizer with axis inclined at some angle then intensity is given as
[tex]I_2 = I_1 cos^2\theta[/tex]
[tex]I_2 = (\frac{I_0}{2}) cos^2(17.1)[/tex]
[tex]I_2 = 0.457 I_0[/tex]
Now again it passes from another polarizer at angle 58.9 degree
[tex]I_3 = I_2cos^2\phi[/tex]
[tex]I_3 = (0.457I_0)cos^2(58.9)[/tex]
[tex]I_3 = (0.12)I_0[/tex]
so the ratio of the intensity will be
[tex]\frac{I_3}{I_0} = 0.12[/tex]
The ratio of the intensity of light emerging from the third polarizer to the intensity of light incident on the first polarizer [tex]\dfrac{I_3}{I_0} =0.12[/tex]
As we know that when unpolarized light passes through a polarizer then its intensity becomes half of the initial intensity
so after passing through the first polarizer the intensity of light is given as
[tex]I_1=\dfrac{I_0}{2}[/tex]
now By Malus law, we know that when polarized light passes through a polarizer with an axis inclined at some angle then intensity is given as
[tex]I_2=I_1cos^2\theta[/tex]
[tex]I_2=\dfrac{I_0}{2} cos^2(17.1)[/tex]
[tex]I_2=0.457I_0[/tex]
Now again it passes from another polarizer at an angle of 58.9 degree
[tex]I_3=I_2cos^2\phi[/tex]
[tex]I_3=0.457I_0cos^2(58.9)[/tex]
[tex]I_3=0.12I_0[/tex]
[tex]\dfrac{I_3}{I_0} =0.12[/tex]
Thus The ratio of the intensity of light emerging from the third polarizer to the intensity of light incident on the first polarizer [tex]\dfrac{I_3}{I_0} =0.12[/tex]
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