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Wave Optics Part 2

Section: Physics

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Question : 66 of 77

Marks: +1, -0

When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

[27-Jan-2024 Shift 2]

$60^{\circ}$
$30^{\circ}$
$90^{\circ}$
$45^{\circ}$

Solution: 👈: Video Solution

Let

I_0

be intensity of unpolarised light incident on first polaroid.

I_1=

Intensity of light transmitted from

1^{\;\text{st}\;}

polaroid

=\;\frac{I_0}{2}

\theta

be the angle between

1^{\;\text{st}\;}

and

2^{\;\text{nd}\;}

polaroid

\phi

be the angle between

2^{\;\text{nd}\;}

and

3^{\;\text{rd}\;}

polaroid

\theta+\phi=90^{\circ}

(as

1^{\;\text{st}\;}

and

3^{\;\text{rd}\;}

polaroid are crossed)

\phi=90^{\circ}-\theta

I_2=

Intensity from

2^{\;\text{nd}\;}

polaroid

I_2= I_1 \cos^2 \theta =\;\frac{I_0}{2} \cos^2 \theta

I_3=

Intensity from

3^{\;\text{rd}\;}

polaroid

I_3= I_2 \cos^2 \phi

I_3= I_1 \cos^2 \theta \cos^2 \phi

I_3=\;\frac{I_0}{2} \cos^2 \theta \cos^2 \phi

\phi=90-\theta

\;I_3=\;\frac{I_0}{2} \cos^2 \theta \sin^2 \theta

\;I_3=\;\frac{I_0}{2} \left[\;\frac{2 \sin\;\theta \cos\theta}{2}\right]^2

\;I_3=\;\frac{I_0}{8} \sin^2 2\theta

I_3

will be maximum when

\sin\;2\theta=1

\;2\theta=90^{\circ}

\;\theta=45^{\circ}

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