Wave Optics | Physics JEE previous year questions with Solutions

(D) Let $I_{0}$ be intensity of unpolarised light incident on first polaroid.

$I_{1} =$ Intensity of light transmitted from $1^{s t}$ polaroid $= \frac{I_{0}}{2}$

$θ$ be the angle between $1^{s t}$ and $2^{n d}$ polaroid

$ϕ$ be the angle between $2^{n d}$ and $3^{r d}$ polaroid

$θ + ϕ = 90^{o}$ (as $1^{s t}$ and $3^{r d}$ polaroid are crossed)

$ϕ = 90 ° - θ$

$I_{2} =$ Intensity from second polaroid

$I_{2} = I_{1} \cos^{2} θ = \frac{I_{0}}{2} \cos^{2} θ$

$I_{3} =$ Intensity from $3^{r d}$ polaroid

$I_{3} = I_{2} \cos^{2} ϕ = I_{1} \cos^{2} θ \cos^{2} ϕ$

$I_{3} = \frac{I_{0}}{2} \cos^{2} θ \cos^{2} ϕ$

$ϕ = 90 - θ$

$I_{3} = \frac{I_{0}}{2} {[\frac{2 \sin θ \cos θ}{2}]}^{2}$

$I_{3} = \frac{I_{0}}{8} \sin^{2} 2 θ$

$I_{3}$ will be maximum when sin $2 θ = 1$

$2 θ = 90$

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