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Unpoiarized light falls on two polarizing sheets placed on top of the other. What be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is one third intensity of the incident beam_______ .
  • a)
    35°
  • b)
    54°
  • c)
    71°
  • d)
    10°
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Unpoiarized light falls on two polarizing sheets placed on top of the ...
Intensity of light transmitted through the first polarizer 
where I0 is the intensity of the incident unpolarized light.
Intensity of the light transmitted through the second polarizer is I2 = I0 cos2 θ
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Most Upvoted Answer
Unpoiarized light falls on two polarizing sheets placed on top of the ...
Degrees b)45 degrees c)60 degrees d)90 degrees

The intensity of the transmitted light through a polarizing sheet is given by

$I_t = I_i \cos^2 \theta$

where $I_i$ is the intensity of the incident light and $\theta$ is the angle between the polarization direction of the incident light and the characteristic direction of the polarizing sheet.

If we have two polarizing sheets placed on top of each other, the intensity of the transmitted light through both sheets is given by

$I_{t2} = I_{t1} \cos^2 \phi$

where $I_{t1}$ is the intensity of the transmitted light through the first sheet, and $\phi$ is the angle between the characteristic directions of the two sheets.

Given that the intensity of the transmitted light through both sheets is one third the intensity of the incident beam, we can write

$I_{t2} = \frac{1}{3} I_i$

$I_{t1} \cos^2 \phi = \frac{1}{3} I_i$

We want to find the angle $\phi$, so we need to first find the angle $\theta$ between the incident polarization direction and the first sheet.

Since the transmitted intensity through the first sheet is one third of the incident intensity, we have

$I_{t1} = \frac{1}{3} I_i$

$I_i \cos^2 \theta = \frac{1}{3} I_i$

$\cos^2 \theta = \frac{1}{3}$

$\cos \theta = \frac{1}{\sqrt{3}}$

$\theta = 35.3^\circ$

Now we can use the relation between the transmitted intensities through both sheets to find $\phi$.

$I_{t1} \cos^2 \phi = I_{t2} = \frac{1}{3} I_i$

$\frac{1}{3} I_i \cos^2 \phi = \frac{1}{3} I_i$

$\cos^2 \phi = 1$

$\phi = 0^\circ$ or $90^\circ$

However, we know that the two sheets are placed on top of each other, so their characteristic directions should be perpendicular. Therefore, the only valid solution is $\phi = 90^\circ$.

Answer: d) 90 degrees
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Unpoiarized light falls on two polarizing sheets placed on top of the other. What be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is one third intensity of the incident beam_______ .a)35°b)54°c)71°d)10°Correct answer is option 'A'. Can you explain this answer?
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