SATHEE: Optics Question 186

Optics Question 186

Question: A beam of plane polarized light falls normally on a polarizer of cross sectional area $3 \times 10^{- 4} m^{2}$ . Flux of energy of incident ray in 10?3 W. The polarizer rotates with an angular frequency of 31.4 rad/sec. The energy of light passing through the polarizer per revolution will be

Options:

A) 10?4 Joule

B) 10?3 Joule

C) 10?2 Joule

D) 10?1 Joule

Show Answer

Answer:

Correct Answer: A

Solution:

Using Matus law, $I = I_{0} \cos^{2} θ$

As here polariser is rotating i.e. all the values of q are possible.

$I_{a v} = \frac{1}{2 π} \int_{0}^{2 π} I d θ = \frac{1}{2 π} \int_{0}^{2 π} I_{0} \cos^{2} θ d θ$

On integration we get $I_{a v} = \frac{I_{0}}{2}$

where $I_{0} = \frac{Energy}{Area \times Time} = \frac{p}{A} = \frac{10^{- 3}}{3 \times 10^{- 4}} = \frac{10}{3} \frac{W a t t}{m^{2}}$

$I_{a v} = \frac{1}{2} \times \frac{10}{3} = \frac{5}{3} W a t t$

and Time period $T = \frac{2 π}{ω} = \frac{2 \times 3.14}{31.4} = \frac{1}{5} s e c$

Energy of light passing through the polariser per revolution $= I_{a v} \times Area \times T$

$= \frac{5}{3} \times 3 \times 10^{- 4} \times \frac{1}{5}$

$= 10^{- 4} J .$

Optics Question 186

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Optics Question 186

Question: A beam of plane polarized light falls normally on a polarizer of cross sectional area 3×10−4m2 . Flux of energy of incident ray in 10?3 W. The polarizer rotates with an angular frequency of 31.4 rad/sec. The energy of light passing through the polarizer per revolution will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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