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Unpolarised light passes through a polariser and analyser which are at an angle of 45⁰ with respect to each other. The intensity of polarised light coming from analyser is 5W/m².The intensity of unpolarised light incident on polariser is

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$5 \sqrt{3} W / m^{2}$

10W/m²

20W/m²

$5 \frac{\sqrt{3}}{4} W / m^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \frac{I_{0}}{2} \cos^{2} ϕ; \\ \Rightarrow 5 = \frac{I_{0}}{2} \times \frac{1}{2} = \frac{I_{0}}{4} \\ \Rightarrow I_{0} = 20 {Wm}^{- 2} \end{array}$

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