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  • (a) Explain, using a suitable diagram, how unpolarized light gets linearly polarized by scattering. (b) Describe briefly the variation of the intensity of transmitted light when a polaroid sheet kept between two crossed polaroids is rotated.

(a) Explain, using a suitable diagram, how unpolarized light gets linearly polarized by scattering.

(b) Describe briefly the variation of the intensity of transmitted light when a polaroid sheet kept between two crossed polaroids is rotated. Draw the graph depicting the variation of intensity with the angle of rotation. How many maxima and minima would be observed when \theta varies from 0 to \pi?

Answers (1)

(a) When the light encounters the molecules of the atmosphere, the electrons in molecules acquire components of motion under the influence of the electric field.

Charges accelerated parallel to the double arrows do not radiate energy towards the observes since their acceleration has no transverse component.

The Radiation scattered by the molecules is thus polarized light.

           

(b) Let us, suppose I_{0} be the intensity of polarised light after passing through polarized P_{1}/

Therfore intensity of polarized light after passing through P_{2}.

    I=I_{0}\cos ^{2}\theta

Since, polarised P_{1} and P_{2} are crossed, hence, the angle between their pass axis will be (90-\theta )

    I=I_{0}\cos ^{2}\theta \cdot \cos ^{2}(90-\theta )

    I=I_{0}\cos ^{2}\theta \cdot \sin ^{2}\theta                        \left [ \because \sin 2\theta =2\sin \theta \cdot \cos \theta \right ]

    I=\frac{I_{0}}{4}\sin ^{2}2\theta

(i) When \theta =0

                                                                \left [ Such\; that\; \sin 0^{\circ}=0 \right ]    I=\frac{I_{0}}{4}\sin ^{2}2\cdot O=0

ii) When \theta =\frac{\pi }{4}

        I=\frac{I_{0}}{4}\sin ^{2}2\frac{\pi }{4}

I=\frac{I_{0}}{4}

(iii) When \theta =\frac{\pi }{2}

        I=\frac{I_{0}}{4}\sin ^{2}2\frac{\pi }{2}

        I=0

(iv) When \theta =\frac{3\pi}{4}

        I=\frac{I_{0}}{4}\sin ^{2}\times \frac{3\pi }{2}

        I=\frac{I_{0}}{4}

    and when 

(v)     \theta =\pi

then, I=\frac{I_{0}}{4}\sin 2\pi

        I=0

The graph depicting the variation of intensity with the angle of rotation:-

There are two maxima and two minima, when \theta varies from 0 to \pi.

If the light is unpolarised then maximum intensity is \frac{I_{0}}{8}

And

If the light is polarised then maximum intensity is \frac{I_{0}}{4}

Posted by

Safeer PP

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