Question:

The graph showing the dependence of intensity of transmitted light on the angle between polariser and analyser, is

Updated On: Aug 1, 2022
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Solution and Explanation

According to law of Malus, when a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light (I) transmitted from the analyser varies directly as the square of the cosine of the angle $(\theta)$ between planes of transmission of analyser and polariser. $ I =cos^2 \, \theta$ and $ I = I_0 cos^2 \, \theta ....(i)$
where $I_0$ = intensity of the light from polariser. From E (i), we note that if the transmission axes of polariser and analyser are parallel (i.e., \theta = 0$^{\circ}$ or 180"), then $I = I_0$. It means that intensity of transmitted light is maximum. When the transmission axes of polariser and analyser are perpendicular (ie, $\theta$ = 90$^{\circ}$), then I = $I_0 cos^2 \, 90^\circ = 0$ It means the intensity of transmitted light is minimum. On plotting a graph between I and $\theta$ as given by relation (i), we get the curve as shown in figure.
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Concepts Used:

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When two or more waves meet each other then interference happens . Interference is a phenomenon in which 2 or more waves superpose to form a resultant wave of greater, lower or the same amplitude.

There are two types of wave interference:

The principle of superposition of waves refers that when two or more waves of the same type are incident on the same point, the resultant amplitude at that point is equal to the vector sum of the amplitudes of the individual waves. If the crest of a wave meets the crest of another wave of the same frequency at the same point,  sum of individual amplitudes is called as constructive interference.The destructive interference occurs when the maxima of the two waves are at 180 degrees out of phase and a positive displacement of one wave is cancelled exactly by a negative displacement of the other wave.