
An unpolarized light of intensity $32W{m^{ - 2}}$ passes through three polarizes is crossed with that of the first. The intensity of final emerging light is $3W{m^{ - 2}}$ . The intensity of light transmitted by first polarizer will be
(A)$32W{m^{ - 2}}$
(B) $16W{m^{ - 2}}$
(C) $8W{m^{ - 2}}$
(D) $4W{m^{ - 2}}$
Answer
495.3k+ views
Hint: malus law is, when a plane-polarized light incident i.e. falls on the analyzer, then the intensity I of the transmitted light is found to be directly proportional to the square of the cosine of the angle between the axis of transmission of polarizer and analyzer.
Formula used:
$I = {I_0}{\cos ^2}\theta $
Where,
$I$=intensity of transmitted light
${I_0}$= intensity of incident light
$\theta $=angle between transmission axis of polarizer and analyzer.
Complete step by step solution:
Malus law states that the intensity of a beam of plane-polarized light after passing through a rotatable polarizer varies as the square of the cosine of the angle.
Mathematically given as below.
$I = {I_0}{\cos ^2}\theta $ (1)
Now, let us first write the information given in the question.
${I_0}$ =$32W{m^{ - 2}}$, intensity of final emerging light =$3W{m^{ - 2}}$.
We have to find the intensity of light transmitted by the first polarizer given below using Malus law from equation (1).
$I = {I_0}{\cos ^2}\theta $ (2)
Average value of cos2θ is ½.
Hence, equation (2) becomes,
$I = {I_0} \times \dfrac{1}{2}$
Now let us substitute the value of ${I_0}$ in the above equation.
$I = 32W{m^{ - 2}} \times \dfrac{1}{2}$
$ = 16W{m^{ - 2}}$
Therefore, the intensity of light transmitted by first polarizer is 16Wm-2.
Hence, option (B) is correct option.
Additional information:
When ‘
Θ=00 or 1800,
In such case, $I = {I_0}{\cos ^2}\theta $
⟹${I_0}{\cos ^2}{0^0}$
⇒${I_0}$
I.e. when the transmission axes of analyzer and polarizer are parallel to each other, then intensity of light transmitted from the analyzer is maximum.
When ‘
Θ=900,
In such case, $I = {I_0}{\cos ^2}\theta $
⟹${I_0}{\cos ^2}{90^0}$
⇒$0$
I.e. when the transmission axes of analyzer and polarizer are perpendicular to each other, then intensity of light transmitted from the analyzer is minimum.
Note:
Unpolarized light wave vibrates in more than one plane whereas polarized light wave vibrates in a single plane. The process of converting unpolarized light into polarized light is called polarization of light.
A polarizer is an optical device that converts unpolarized light into polarized light in some form.
The intensity of light is the power radiated per unit area where the area is measured plane perpendicular to the direction of propagation of energy. Its SI unit is \[W{m^{ - 2}}\].
Formula used:
$I = {I_0}{\cos ^2}\theta $
Where,
$I$=intensity of transmitted light
${I_0}$= intensity of incident light
$\theta $=angle between transmission axis of polarizer and analyzer.
Complete step by step solution:
Malus law states that the intensity of a beam of plane-polarized light after passing through a rotatable polarizer varies as the square of the cosine of the angle.
Mathematically given as below.
$I = {I_0}{\cos ^2}\theta $ (1)
Now, let us first write the information given in the question.
${I_0}$ =$32W{m^{ - 2}}$, intensity of final emerging light =$3W{m^{ - 2}}$.
We have to find the intensity of light transmitted by the first polarizer given below using Malus law from equation (1).
$I = {I_0}{\cos ^2}\theta $ (2)
Average value of cos2θ is ½.
Hence, equation (2) becomes,
$I = {I_0} \times \dfrac{1}{2}$
Now let us substitute the value of ${I_0}$ in the above equation.
$I = 32W{m^{ - 2}} \times \dfrac{1}{2}$
$ = 16W{m^{ - 2}}$
Therefore, the intensity of light transmitted by first polarizer is 16Wm-2.
Hence, option (B) is correct option.
Additional information:
When ‘
Θ=00 or 1800,
In such case, $I = {I_0}{\cos ^2}\theta $
⟹${I_0}{\cos ^2}{0^0}$
⇒${I_0}$
I.e. when the transmission axes of analyzer and polarizer are parallel to each other, then intensity of light transmitted from the analyzer is maximum.
When ‘
Θ=900,
In such case, $I = {I_0}{\cos ^2}\theta $
⟹${I_0}{\cos ^2}{90^0}$
⇒$0$
I.e. when the transmission axes of analyzer and polarizer are perpendicular to each other, then intensity of light transmitted from the analyzer is minimum.
Note:
Unpolarized light wave vibrates in more than one plane whereas polarized light wave vibrates in a single plane. The process of converting unpolarized light into polarized light is called polarization of light.
A polarizer is an optical device that converts unpolarized light into polarized light in some form.
The intensity of light is the power radiated per unit area where the area is measured plane perpendicular to the direction of propagation of energy. Its SI unit is \[W{m^{ - 2}}\].
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Phenol on treatment with conc HNO3 gives A Picric acid class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which sugar is called invert sugar Why is it called class 12 chemistry CBSE

a Give chemical tests to distinguish between 1 Propanal class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE
