Referring to the electric room heater considered in the first example in this section, calculate the intensity of IR radiation in \({\bf{W}}/{{\bf{m}}^{\bf{2}}}\) projected by the concave mirror on a person \({\bf{3}}.{\bf{00}}{\rm{ }}{\bf{m}}\) away. Assume that the heating element radiates \({\bf{1500}}{\rm{ }}{\bf{W}}\) and has an area of \({\bf{100}}{\rm{ }}{\bf{c}}{{\bf{m}}^{\bf{2}}}\), and that half of the radiated power is reflected and focused by the mirror.

The radiation intensity is defined as the power per unit solid angle or the power incident on that portion of a sphere's surface that subtends an angle of one radian in both the horizontal and vertical planes at the sphere's center.

• Image distance of the IR radiation\({d_i} = 3.0\;m\).
• Radiating power of the heating element\(P = 1500\;\;W\).
• Area of the heating element\(A = 100\;\;c{m^2}\).
• Half of the power is reflected and focused by the mirror.

The formula for the radius of curvature of a thin mirror is,

\(f = \dfrac{R}{2} \ldots \ldots (1)\)

Here,

\(R\)is the radius of curvature.

The formula for the focal length of a thin mirror is,

\(\dfrac{1}{f} = \dfrac{1}{{{d_0}}} + \dfrac{1}{{{d_i}}}\)

Here,

-\(f\)is the focal length.

-\({d_0}\)is the object distance.

- \({d_i}\) is the image distance.

The formula for magnification is,

\(\dfrac{1}{{{d_0}}} = \dfrac{1}{f} - \dfrac{1}{{{d_i}}} \ldots (2)\)

The formula for magnification is,

\(\begin{aligned}= \dfrac{{{A^\prime }}}{A}\\ = - \dfrac{{{d_i}}}{{{d_m}}}\end{aligned}\)

Here;

-\(m\)is the magnification.

-\({A^\prime }\)is the area of the image.

-\(A\)is the area of the object.

Rearranging the formula,

\({A^\prime } = |m|A \ldots (3)\)

The formula for intensity of the\(\mathbb{R}\)radiation is,

\(I = \dfrac{P}{{{A^\prime }}}\)

Here,

-\(P\)is the power of the mirror.

- \(I\) is the intensity of the mirror.

Substitute\(50\;{\rm{cm}}\)for\(R\)in equation\((1)\)to calculate the focal length.

\(\begin{aligned}f = \dfrac{{50}}{2}\\ = 25\;cm\\ = 0.25\;m\end{aligned}\)

Substitute\(0.25\;m\) for \(f\)and\(3\;m\)for\({d_i}\)in equation\((2)\).

\(\begin{aligned}\dfrac{1}{{{d_0}}} = \dfrac{1}{{0.25}} - \dfrac{1}{3}\\ = \dfrac{1}{{3.667}}\\ = 0.27m\end{aligned}\)

The magnification will be,

\(\begin{aligned}m = - \dfrac{3}{{0.27}}\\ = - 11\end{aligned}\)

Substitute\( - 11\)for\(m\)and\(100\;c{m^2}\)for\(A\)in equation\((3)\).

\(\begin{aligned}{A^\prime } = 11 \times 100\;c{m^2}\\ = 11 \times 100 \times {10^{ - 4}}\;{m^2}\\ = 0.11\;{m^2}\end{aligned}\)

As half of the power is reflected, effective power will be,

\(\begin{aligned}P = \dfrac{{1500}}{2}\\ = 750\;W\end{aligned}\)

So, the intensity of the mirror is given by,

\(\begin{aligned}\dfrac{{750}}{{0.11}}\\ = 6818.18\;W/{m^2}\\ = 6.82\;kW/{m^2}\end{aligned}\)

Conclusion:Therefore, the intensity of radiation is \(6.82\;kW/{m^2}\).