Question

A beam of unpolarised light of intensity i0i0 is passed through a polaroid a and then through another polaroid b which is oriented so that its principal plane makes an angle of 6060 relative to that of
a. The intensity of the emergent light is:

Answer 1

Explanation:

Iα A

2

where AA is the amplitude and II is the intensity

For unpolarised light, it can be resolved into two components of equal magnitude in any pair of perpendicular directions. So the amplitude that is along the plane of polarisation of the polaroid will always be \dfrac{A}{\sqrt{2}}

2

A

So, the intensity which will pass through it will always be \dfrac{I_{0}}{2}

2

I

0

Now when this polarised light falls on polaroid B whose plane of polarisation makes an angle \thetaθ with the plane of polarisation of incident light, the component of amplitude along the plane of polarisation of the polaroid is A cos \thetaAcosθ.

Therefore, intensity of light that passes through is I_{i} cos^{2} \thetaI

i

cos

2

θ where I_{i}I

i

is the intensity of incident light.

Here, I_{i}=\dfrac{I_{0}}{2}I

i

=

2

I

0

and \theta = 45^{\circ}θ=45

∘

So intensity of light emerging through B is

\dfrac{I_{0}}{2}\times \dfrac{1}{2} = \dfrac {I_{0}}{4}

2

I

0

×

2

1

=

4

I

0