Question

 

 

 

Rank the light intensity, from largest to smallest, at the point P in the figures.

 

Answer 1

Concepts and reason

The concept required to solve the given problem is intensity of light at a given distance.

Initially, write the expression for intensity of light source at a given distance. Then, use this expression for to find the intensity of light at given distance. Finally, take intensity of light at given point as multiple of number of sources to rank light intensity from greatest to smallest.

Fundamentals

The intensity I of a light source at a distance r from the source is given by following expression.

$I = \frac{P}{{4\pi {r^2}}}$

Here, P is the power.

The intensity at a given distance is proportional to number of light sources. Thus, the intensity at given point if there are n number of sources is,

$I = n\left( {\frac{P}{{4\pi {r^2}}}} \right)$

The intensity of light I due to n number of sources at a distance r from the sources is given by following expression.

$I = n\left( {\frac{P}{{4\pi {r^2}}}} \right)$

Substitute 1 for n, and 1.0 m for r in the above equation to solve for intensity at point P in figure A.

$\begin{array}{c}\\{I_A} = \left( 1 \right)\left( {\frac{P}{{4\pi {{\left( {1.0\,{\rm{m}}} \right)}^2}}}} \right)\\\\ = \left( 1 \right)\frac{P}{{4\pi }}\\\end{array}$

Substitute 2 for n, and 0.5 m for r in the above equation to solve for intensity at point P in figure B.

$\begin{array}{c}\\{I_B} = \left( 2 \right)\left( {\frac{P}{{4\pi {{\left( {0.5\,{\rm{m}}} \right)}^2}}}} \right)\\\\ = \left( 8 \right)\frac{P}{{4\pi }}\\\end{array}$

Substitute 4 for n, and 2.0 m for r in the above equation to solve for intensity at point P in figure C.

$\begin{array}{c}\\{I_C} = \left( 4 \right)\left( {\frac{P}{{4\pi {{\left( {2.0\,{\rm{m}}} \right)}^2}}}} \right)\\\\ = \left( 1 \right)\frac{P}{{4\pi }}\\\end{array}$

Substitute 3 for n, and 1.0 m for r in the above equation to solve for intensity at point P in figure D.

$\begin{array}{c}\\{I_D} = \left( 3 \right)\left( {\frac{P}{{4\pi {{\left( {1.0\,{\rm{m}}} \right)}^2}}}} \right)\\\\ = \left( 3 \right)\frac{P}{{4\pi }}\\\end{array}$

Substitute 2 for n, and 1.5 m for r in the above equation to solve for intensity at point P in figure D.

$\begin{array}{c}\\{I_E} = \left( 2 \right)\left( {\frac{P}{{4\pi {{\left( {1.5\,{\rm{m}}} \right)}^2}}}} \right)\\\\ = \left( {0.88} \right)\frac{P}{{4\pi }}\\\end{array}$

The rank of intensity from largest to smallest is ${I_B} > {I_D} > {I_A} = {I_C} > {I_E}.$

Ans:

The rank of intensities from largest to smallest is ${I_B} > {I_D} > {I_A} = {I_C} > {I_E}.$