Intensity of polarized lightexperiment

(i) An unpolarised light of intensity (I0) passes through two successive polaroids (P1 and P2), and corresponding intensities of light coming out from them (I1 and I2) clearly distinguish the difference between (I1) and (I2).

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3. An unpolarised light passes through two successive polaroids (P1 and P2); the polaroid P1 makes an angle θ with the axis of the polaroid P2. Find out the intensity final outcoming light. And if θ is varied from 0 to 27, plot the intensity variation graph.

Intensity of polarized lightin physics

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After observing the results, he further put forth a concept that natural light consisted of the s- and p-polarisation and that they were perpendicular to each other. Today, this law is used to define the intrinsic connection between optics and electromagnetism as well as demonstrate the transverse nature of electromagnetic waves.

Intensity of polarized lightgraph

2. An unpolarised light with intensity (I) is passing through a polariser. What happens to the intensity of incident light?

5. An Unpolarised The light of intensity (I0) passes through three successive polaroids P1, P2 and P3. The polarisation axes of P1 make an angle of 60° with P2, and P2 makes an angle of 30° with P3. Then, what are the corresponding intensity outcomes from P1, P2 and P3 (I1, I2 and I3)?

Malus law is crucial if we want to learn or understand the polarisation properties of light. The law helps us to study the light intensity relation of the polariser-analyser. Malus law is named after Étienne-Louis Malus, who, in the year 1808, discovered that natural incident light could be polarised when it was reflected by a glass surface. He used calcite crystal for his experiment.

Intensity of polarized lightformula

Lightwaves can be polarised. The phenomenon of polarisation takes place only in the transverse nature of waves. So, sound waves cannot be polarised.

The orientation of electric field vectors \(\begin{array}{l}\left( \overrightarrow{E} \right)\ \text{will be in a direction on a plane}\end{array} \) perpendicular to the direction of propagation of light. 2. An unpolarised light with intensity (I) is passing through a polariser. What happens to the intensity of incident light? Solution: If an unpolarised light of intensity (I) passes through a polariser, outcoming light intensity becomes half of its initial value (I/2). 3. An unpolarised light passes through two successive polaroids (P1 and P2); the polaroid P1 makes an angle θ with the axis of the polaroid P2. Find out the intensity final outcoming light. And if θ is varied from 0 to 27, plot the intensity variation graph. Solution: An unpolarised light passes through two successive polaroids. We know that when an unpolarised light passes through a polaroid, its intensity becomes half of its initial intensity. If it passes again through any other polaroids, then its intensity given by Malus law is, \(\begin{array}{l}I=\frac{I}{2}{{\cos }^{2}}\theta\end{array} \) Here, θ is the angle between the axis of the polaroid’s intensity variation with respect to 0 to 2π. It is nothing (cos2 θ) curve. 4. How does an unpolarised light of intensity (I0) get plane-polarised when passing through a polaroid? (i) An unpolarised light of intensity (I0) passes through two successive polaroids (P1 and P2), and corresponding intensities of light coming out from them (I1 and I2) clearly distinguish the difference between (I1) and (I2). (ii) What is the necessary condition to get maximum intensity after passing through two successive polaroids? Solution: \(\begin{array}{l}\text{As we know from Malus law, an unpolarised light passes through a polaroid, electric field}\ \left( \overrightarrow{E} \right)\end{array} \) vector of unpolarised light gets polarised along a direction in a plane perpendicular to the direction of propagation of light. The intensity of the plane polarised light becomes half its incident light intensity (I/2). Then, this plane polarised light passes through another polaroid called the analyser, the outcoming light, intensity as a function of cosine square angle between the plane of polariser and analyser.

We know that when an unpolarised light passes through a polaroid, its intensity becomes half of its initial intensity. If it passes again through any other polaroids, then its intensity given by Malus law is,

The angle (θ) between the plane of transmission of the polariser and analyser must be zero (or) the polariser and analyser must be parallel to each other.

Here, θ is the angle between the axis of the polaroid’s intensity variation with respect to 0 to 2π. It is nothing (cos2 θ) curve.

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Intensity of polarized lightcalculator

Since the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave, the ratio of transmitted to incident amplitude is cos ϕ, so the ratio transmitted to incident intensity is cos2ϕ

Point 1 – When unpolarised light is incident on an ideal polariser, the intensity of the transmitted light is exactly half that of the incident unpolarised light, no matter how the polarising axis is oriented.

When (I0) intensity of unpolarised light passes through a polaroid (P1), its intensity becomes (I1), and unpolarised light has an electric field vector in all possible directions; when they pass through a polaroid, they get polarised along a direction from Malus law \(\begin{array}{l}I={{I}_{0}}{{\cos }^{2}}\theta\end{array} \)   \(\begin{array}{l}{{I}_{1}}=\frac{{{I}_{0}}}{2}\end{array} \) Then, the same unpolarised light of intensity (I0) after passing through a polaroid, its intensity becomes (I1), and if this (I1) passes through the second polaroid (Analyser), then its intensity becomes (I2). \(\begin{array}{l}{{I}_{2}}={{I}_{1}}{{\cos }^{2}}\theta\end{array} \) It is a function of the square of the cosine angle (θ) between the arms of polarisation of the polaroid and analyser. 5. An Unpolarised The light of intensity (I0) passes through three successive polaroids P1, P2 and P3. The polarisation axes of P1 make an angle of 60° with P2, and P2 makes an angle of 30° with P3. Then, what are the corresponding intensity outcomes from P1, P2 and P3 (I1, I2 and I3)?  Solution: \(\begin{array}{l}{{I}_{1}}=\frac{{{I}_{0}}}{2}\ (\text{corresponding to}\ P_1)\end{array} \) \(\begin{array}{l}{{I}_{2}}={{I}_{1}}{{\cos }^{2}}\theta\ (\text{corresponding to}\ P_2)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{2}\left( {{\cos }^{2}}60{}^\circ \right)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{2}{{\left( \frac{1}{2} \right)}^{2}}=\frac{{{I}_{0}}}{8}\end{array} \) \(\begin{array}{l}{{I}_{3}}={{I}_{2}}{{\cos }^{2}}\theta\ (\text{corresponding to}\ P_3)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{8}{{\cos }^{2}}\left( 30{}^\circ \right)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{8}{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{3{{I}_{0}}}{32}\end{array} \) Frequently Asked Questions on Malus LawQ1 What is the use of Malus law? The law helps us understand the polarising properties of light. Q2 State Malus law. The Malus law states that the intensity of the plane-polarised light that passes through the analyser is directly proportional to the square of the cosine of the angle between the plane of the polariser and the transmission axis of the analyser. Q3 Which wave can be polarised? Lightwaves can be polarised. The phenomenon of polarisation takes place only in the transverse nature of waves. So, sound waves cannot be polarised. Q4 Who is the Malus law named after? The Malus law is named after Etienne Malus, who published this law in 1809.

We know the average value of \(\begin{array}{l}{{\cos }^{2}}\theta \simeq \frac{1}{2}\end{array} \) \(\begin{array}{l}<{{\cos }^{2}}\theta >=\frac{1}{2}\end{array} \) When (I0) intensity of unpolarised light passes through a polaroid (P1), its intensity becomes (I1), and unpolarised light has an electric field vector in all possible directions; when they pass through a polaroid, they get polarised along a direction from Malus law \(\begin{array}{l}I={{I}_{0}}{{\cos }^{2}}\theta\end{array} \)   \(\begin{array}{l}{{I}_{1}}=\frac{{{I}_{0}}}{2}\end{array} \) Then, the same unpolarised light of intensity (I0) after passing through a polaroid, its intensity becomes (I1), and if this (I1) passes through the second polaroid (Analyser), then its intensity becomes (I2). \(\begin{array}{l}{{I}_{2}}={{I}_{1}}{{\cos }^{2}}\theta\end{array} \) It is a function of the square of the cosine angle (θ) between the arms of polarisation of the polaroid and analyser. 5. An Unpolarised The light of intensity (I0) passes through three successive polaroids P1, P2 and P3. The polarisation axes of P1 make an angle of 60° with P2, and P2 makes an angle of 30° with P3. Then, what are the corresponding intensity outcomes from P1, P2 and P3 (I1, I2 and I3)?  Solution: \(\begin{array}{l}{{I}_{1}}=\frac{{{I}_{0}}}{2}\ (\text{corresponding to}\ P_1)\end{array} \) \(\begin{array}{l}{{I}_{2}}={{I}_{1}}{{\cos }^{2}}\theta\ (\text{corresponding to}\ P_2)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{2}\left( {{\cos }^{2}}60{}^\circ \right)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{2}{{\left( \frac{1}{2} \right)}^{2}}=\frac{{{I}_{0}}}{8}\end{array} \) \(\begin{array}{l}{{I}_{3}}={{I}_{2}}{{\cos }^{2}}\theta\ (\text{corresponding to}\ P_3)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{8}{{\cos }^{2}}\left( 30{}^\circ \right)\end{array} \) \(\begin{array}{l}=\frac{{{I}_{0}}}{8}{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{3{{I}_{0}}}{32}\end{array} \) Frequently Asked Questions on Malus LawQ1 What is the use of Malus law? The law helps us understand the polarising properties of light. Q2 State Malus law. The Malus law states that the intensity of the plane-polarised light that passes through the analyser is directly proportional to the square of the cosine of the angle between the plane of the polariser and the transmission axis of the analyser. Q3 Which wave can be polarised? Lightwaves can be polarised. The phenomenon of polarisation takes place only in the transverse nature of waves. So, sound waves cannot be polarised. Q4 Who is the Malus law named after? The Malus law is named after Etienne Malus, who published this law in 1809.

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What happens when the linearly polarised light emerging from a polariser passes through a second polariser (analyser)? In general, the polarising axis of the second polariser (analyser) makes an angle (d) with the polarising axis of the first polariser.

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Intensity of polarized lightpdf

Then, the same unpolarised light of intensity (I0) after passing through a polaroid, its intensity becomes (I1), and if this (I1) passes through the second polaroid (Analyser), then its intensity becomes (I2).

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The orientation of electric field vectors \(\begin{array}{l}\left( \overrightarrow{E} \right)\ \text{will be in all possible directions}\end{array} \)

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To determine the direction of polarisation, we need one polariser, which is known as analyser oriented, making an angle (p) with the polariser.

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If an unpolarised light of intensity (I) passes through a polariser, outcoming light intensity becomes half of its initial value (I/2).

The Malus law states that the intensity of the plane-polarised light that passes through the analyser is directly proportional to the square of the cosine of the angle between the plane of the polariser and the transmission axis of the analyser.