An unpolarized light of intensity passes through polarizer and then through a certain optically active solution and finally it goes to analyser

Question

An unpolarized light of intensity $I_{0}$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0 °$ and intensity of light emerged from analyser is $\frac{3}{8} I_{0}$ , the angle of rotation of the light by the solution with respect to analyser is _____ degrees. [2026]

Smart Achievers · Accepted Answer

(30)

$When unpolarized light of intensity I_{0} passes through a polarizer, its intensity is halved,$

$I_{1} = \frac{I_{0}}{2}$

$Angle between analyser and polarizer = 0 °$

$Final intensity = \frac{3}{8} I_{0}$

$From Malus's law,$ $\frac{3}{8} I_{0} = \frac{I_{0}}{2} \cos^{2} (ϕ)$

$\cos ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 30 °$

Solution

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Solution

Ans. (30) When unpolarized light of intensity I0 passes through a polarizer, its intensity is halved, I1=I02 Angle between analyser and polarizer=0° Final intensity=38I0 From Malus's law, 38I0=I02cos2(ϕ) cosϕ=32⇒ϕ=30°

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