Test Index

VITEEE 2015 Paper

Section: Physics

Question : 36 of 120

Marks: +1, -0

Solution:

Let the intensity of unpolarised light be

I_0

, so the intensity of first polaroid is

\frac{I_0}{2}

. On rotating through 60°, the intensity of light from second polaroid

I =

\left(\frac{I_0}{2}\right)

(\cos 60)^2

\frac{I_0}{2} \frac{1}{4}

\frac{I_0}{8}

= 0.125

I_0

∴ percentage of incident light transmitted through the system = 12.5%.

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