Wave Optics

JEE Main level, single correct, medium

169 of 191

Question 169 of 191

‘nn’ polarizing sheets are arranged such that each makes an angle 45∘45^{\circ} with the proceeding sheet. An unpolarized light of intensity II is incident into this arrangement. The output intensity is found to be I64\dfrac{I}{64} . The value of nn will be:

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Answer: (B) 66

When an unpolarized light passes through a polariser, the intensity of the emergent light from the polariser depends on the square of the cosine of the angle between the plane of the polariser and the direction of the incident light.
The formula to calculate the intensity (I1)\left({I}_{1}\right) of the emergent light from a polariser which makes an angle θ\theta with the direction of the incident light is given by
I1=Icos⁡2θ..................(1){I}_{1}=I{\cos }^{2}\theta ..................(1)
where, II is the intensity of the incident light.
Substitute 45∘45^{\circ} for θ\theta into equation (1) to obtain the intensity of the first emergent light.
I1=Icos⁡245∘=I2\begin{matrix}{I}_{1} & = & I{\cos }^{2}45^{\circ} \\ & = & \dfrac{I}{2}\end{matrix}
In a similar manner, the intensity (I2)\left({I}_{2}\right) of the second emergent light can be written as
I2=I1cos⁡245∘=I12=I4=I22\begin{matrix}{I}_{2} & = & {I}_{1}{\cos }^{2}45^{\circ} \\ & = & \dfrac{{I}_{1}}{2} \\ & = & \dfrac{I}{4} \\ & = & \dfrac{I}{{2}^{2}}\end{matrix}
Thus, for nn number of such polarisers, the net intensity (In)\left({I}_{n}\right) of the emergent light can be written as
In=I2n......................(1){I}_{n}=\dfrac{I}{{2}^{n}}......................(1)
Comparing equation (1) with the given expression for the intensity, it can be concluded that
n=6n=6

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