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  • A beam of natural light falls on a system of 5 polaroids, which are arranged in succession such that the pass axis of each polaroid is turned through <img alt="\mathrm{60^{\circ}}" src="https://ent

A beam of natural light falls on a system of 5 polaroids, which are arranged in succession such that the pass axis of each polaroid is turned through \mathrm{60^{\circ}} with respect to the preceding one. The fraction of the incident light intensity that passes through the system is:

Option: 1

1 / 64


Option: 2

1 / 32


Option: 3

1 / 256


Option: 4

1 / 512


Answers (1)

Let \mathrm{\mathrm{I}_0} be the intensity of incident light. Then the intensity of light from the \mathrm{1^{\text {st }}} polaroid is \mathrm{\mathrm{I}_1=\frac{\mathrm{I}_0}{2}}

Intensity of light from the \mathrm{2^{\text {nd }}} polaroid is \mathrm{\mathrm{I}_2=\mathrm{I}_1 \cos ^2 60^{\circ}=\frac{\mathrm{I}_0}{2}\left(\frac{1}{2}\right)^2=\frac{\mathrm{I}_0}{8}}

Intensity of light from the \mathrm{3^{\text {rd }}} polaroid is \mathrm{\mathrm{I}_3=\mathrm{I}_2 \cos ^2 60^{\circ}=\frac{\mathrm{I}_0}{8}\left(\frac{1}{2}\right)^2=\frac{\mathrm{I}_0}{32}}

Intensity of light from the \mathrm{4^{\text {th }}} polaroid is\mathrm{ \mathrm{I}_4=\mathrm{I}_3 \cos ^2 60^{\circ}=\frac{\mathrm{I}_0}{32}\left(\frac{1}{2}\right)^2=\frac{\mathrm{I}_0}{128}}

Intensity of light from \mathrm{5^{\text {th }}} polaroid is \mathrm{\mathrm{I}_5=\mathrm{I}_4 \cos ^2 60^{\circ}=\frac{\mathrm{I}_0}{128}\left(\frac{1}{2}\right)^2=\frac{\mathrm{I}_0}{512}}

Therefore, the fraction of the incident light that passes through the system is \mathrm{\frac{\mathrm{I}_5}{\mathrm{I}_0}=\frac{1}{512}}

Posted by

Ramraj Saini

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