A beam of unpolarised light with intensity I_0 falls on a system of three polaroids P_1 , P_2 , and P_3 arranged in series. The transmission axes of P_1 and P_3 are perpendicular to each other. If the transmission axis

Question

Quantrex Academy · Accepted Answer

Correct answer: B. (I_0)/(32)

Intensity of light after passing through

P_1

is

I_1 = \dfrac{I_0}{2}

Angle between transmission axes of

P_1

and

P_2

is

\theta_1 = 15^{\circ}

Intensity of light after passing through

P_2

is

I_2 = I_1 \cos^2(15^{\circ}) = \dfrac{I_0}{2} \cos^2(15^{\circ})

Angle between transmission axes of

P_1

and

P_3

is

90^{\circ}

Angle between transmission axes of

P_2

and

P_3

is

\theta_2 = 90^{\circ} - 15^{\circ} = 75^{\circ}

Intensity of light after passing through

P_3

is

I_3 = I_2 \cos^2(75^{\circ}) = \dfrac{I_0}{2} \cos^2(15^{\circ}) \cos^2(75^{\circ})

Since

\cos(75^{\circ}) = \sin(15^{\circ})

, we get:

I_3 = \dfrac{I_0}{2} (\sin(15^{\circ}) \cos(15^{\circ}))^2

I_3 = \dfrac{I_0}{2} \left( \dfrac{\sin(30^{\circ})}{2} \right)^2

I_3 = \dfrac{I_0}{2} \left( \dfrac{1/2}{2} \right)^2 = \dfrac{I_0}{2} \left( \dfrac{1}{4} \right)^2 = \dfrac{I_0}{32}

Answer:

\dfrac{I_0}{32}

A beam of unpolarised light with intensity I_0 falls on a system of...

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