A beam of plane polarized light of large cross-sectional area and uniform intensity of <math><mn>3.3</mn><msup><mrow><mrow><mi mathvariant="normal">W</mi><mi mathvariant="normal">m</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> falls normally on a polarizer (cross sectional area <math><mn>3</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup><msup><mrow><mrow><mtext></mtext><mi mathvariant="normal">m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> ) which rotates about its axis with an angular speed of <math><mn>31.4</mn><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">d</mi><mo>/</mo><mi mathvariant="normal">s</mi></math> . The energy of light passing through the polariser per revolution, is close to:

E= (I) (t) (A)cos2? θ⇒ (3.3)2π31.43×10-4×12⇒0.99×10-4

A beam of plane polarized light of large cross-sectional area and uniform intensity of&nbsp;<math><mn>3.3</mn><msup><mrow><mrow><mi mathvariant="normal">W</mi><mi mathvariant="normal">m</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math>&nbsp;falls normally on a polarizer (cross sectional area&nbsp;<math><mn>3</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup><msup><mrow><mrow><mtext></mtext><mi mathvariant="normal">m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>&nbsp;) which rotates about its axis with an angular speed of&nbsp;<math><mn>31.4</mn><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">d</mi><mo>/</mo><mi mathvariant="normal">s</mi></math>&nbsp;. The energy of light passing through the polariser per revolution, is close to:

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Physics Class 12th Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3 {W m}^{- 2}$ falls normally on a polarizer (cross sectional area $3 \times 10^{- 4} m^{2}$ ) which rotates about its axis with an angular speed of $31.4 r a d / s$ . The energy of light passing through the polariser per revolution, is close to:

Option 1 -

$1.0 * 10^{- 5} J$

Option 2 -

$1.5 * 10^{- 4} J$

Option 3 -

^{$5.0 * 10^{- 4} J$}

Option 4 -

$1.0 * 10^{- 4} J$

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$E = (I) (t) (A) ?{c o s}^{2} ? θ?$
$\Rightarrow (3.3) [\frac{2 π}{31.4}] [3 \times 10^{- 4}] \times \frac{1}{2}$

$\Rightarrow 0.99 \times 10^{- 4}$

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