Two linear polarisers are placed coaxially. T | vedclass

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(A) Since the beam incident on the first polariser is unpolarised,its intensity is reduced to half upon passing through the first polariser. Malus' la

Vedclass · Accepted Answer

$I_{1}=10.0 \,W/m^{2}$ and $I_{2}=7.5 \,W/m^{2}$

Vedclass · Answer

(A) Since the beam incident on the first polariser is unpolarised,its intensity is reduced to half upon passing through the first polariser. Malus' law is not applicable here.Therefore,the intensity $I_{1}$ after the first polariser is:$I_{1} = \frac{I_{0}}{2} = \frac{20}{2} = 10 \,W/m^{2}$As the light emerging from the first polariser is linearly polarised,Malus' law is applicable for the second polariser.The intensity $I_{2}$ obtained after the second polariser is given by:$I_{2} = I_{1} \cdot \cos^{2} 	heta$where $	heta$ is the angle between the transmission axes of the first and second polarisers.Given the angles are $30^{\circ}$ and $60^{\circ}$ from the vertical,the angle between them is $	heta = 60^{\circ} - 30^{\circ} = 30^{\circ}$.Thus,$I_{2} = 10 	imes \cos^{2} 30^{\circ}$$I_{2} = 10 	imes \left(\frac{\sqrt{3}}{2}ight)^{2} = 10 	imes \frac{3}{4} = 7.5 \,W/m^{2}$

$A$. Angle of reflection	$(i)$ $2 \sin^{-1} \left(\sqrt{\frac{2}{3}}\right)$
$B$. Angle of refraction	$(ii)$ $\sin^{-1} \left(\sqrt{\frac{2}{3}}\right) - \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$C$. Angle between incident and completely polarised light	$(iii)$ $\sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$D$. Angle of deviation of the incident ray	$(iv)$ $\cos^{-1} \left(\frac{1}{\sqrt{3}}\right)$

Similar Questions

$A$ beam of unpolarised light of intensity $I_0$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$. The intensity of the emergent light is

Explain Malus's law and state its mathematical expression.

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