Question: Q. 5. (i) How does one demonstrate, using a suitable diagram, that unpolarized light when passed through a polaroid gets polarised?

(ii) A beam of unpolarized light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarised, when $\mu=\tan i_{\mathrm{B}}$, where $\mu$ is the refractive index of glass with respect to air and $i_{\mathrm{B}}$ is the Brewster’s angle. U [Delhi I, II, III 2014]

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Solution:

Ans.(i)

Unpolarised light

Polaroid

Plane polarised light The components of electric vector associated with light wave, along the direction of aligned molecules of a polaroid, get absorbed. As a result after passing through it, the components perpendicular to the direction of aligned molecules will be obtained in the form of plane polarised light. (ii) Try yourself similar to Q. 2 SATQ-II

[CBSE Marking Scheme 2014]

[II Q. 4. (i) How does an unpolarised light incident on a polaroid get polarised?

Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection from a transparent medium.

(ii) Two polaroids ’ $A$ ’ and ’ $B$ ’ are kept in crossed position. How should a third polaroid ’ $C$ ’ be placed between them so that the intensity of polarised light transmitted by polaroid $B$ reduces to $1 / 8^{\text {th }}$ of the intensity of unpolarised light incident on $A$ ?

[OD I, II, III 2012]

Ans. (i) When an unpolarised light falls on a polaroid, it lets only those of its electric vectors that are oscillating along a direction perpendicular to its aligned molecules to pass through it. The incident light thus gets linearly polarised.

Alternatively,

1

1

Whenever an unpolarised light is incident on a transparent surface, the reflected light gets completely polarised when the reflected and refracted light are perpendicular to each other. $\mathbf{1}$

(ii) Let $\theta$ be the angle between the pass-axis of $A$ and $C$.

Intensity of light passing through $A=I_{0} / 2$

Intensity of light passing through $B=\left(I_{0} / 2\right) \cos ^{2} \theta$ Intensity of light passing through $C$

The third polaroid is placed at $\theta=45^{\circ}$.

$1 / 2$

[CBSE Marking Scheme, 2012]

OSWAAL LEARNING TOOIS

UNIT - VII



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