Question

What will be the fraction in intensity with which it reduced wen analyzer is rotated through 60 degree?​

Answer 1

If light of intensity I₀ is pa-ssing through the polαrizer, then it becФmes I₀/2.

Now, the light of intensity I₀/2 will pαss from analyzer then it's intensity will become I_a.

From Malus law,

Intensity of light in analyzer can be given as:

I_a = I₀ cos²θ

where, θ = angle between trαnsmission ax-es of polαrizer and analyzer

here, θ = 60°

Substituing the values,

I_a = I₀ cos² 60°

I_a = I₀ (1/2)²

I_a =  I₀ / 4

I_a = 0.25 I₀     (←Answer)

Answer 2

$\gray{ \huge \: answer}$

If light of intensity I₀ is pa-ssing through the polαrizer, then it becФmes I₀/2.

Now, the light of intensity I₀/2 will pαss from analyzer then it's intensity will become I_a.

From Malus law,

Intensity of light in analyzer can be given as:

I_a = I₀ cos²θ

where, θ = angle between trαnsmission ax-es of polαrizer and analyzer

here, θ = 60°

Substituing the values,

I_a = I₀ cos² 60°

I_a = I₀ (1/2)²

I_a =  I₀ / 4

I_a = 0.25 I₀