Chapter 22, Problem 14PEB

In the evening, a stick measuring 0.75 m casts a 2.3 m shadow. At noon the same stick casts a 0.30 m shadow. What is the insolation in each case? (Assume the maximum insolation on a clear day is 1.0 $\times$ 10³ W/m².)

To determine

The insolation of stick measuring $0.75\text{ m}$ at noon and in the evening.

Answer to Problem 14PEB

Solution:

$9.25\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ and $3.11\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$

Explanation of Solution

Given data:

In the evening, a stick measuring $0.75\text{m}$ casts a $2.3\text{ m}$ shadow.

At noon, a stick measuring $0.75\text{m}$ casts a $0.30\text{ m}$ shadow.

Maximum insolation on a day is $1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}$.

Formula used:

Write the formula for insolation.

$I=I_{\text{max}}\left(\frac{L_{\text{stick}}}{\sqrt{L_{\text{stick}}^2+L_{\text{shadow}}^2}}\right)$

Here, $I_{\text{max}}$ is the maximum insolation, $L_{\text{stick}}$ is the length of the stick and $L_{\text{shadow}}$ is the length of the shadow.

Explanation:

Recall the formula for insolation.

$I=I_{\text{max}}\left(\frac{L_{\text{stick}}}{\sqrt{L_{\text{stick}}^2+L_{\text{shadow}}^2}}\right)$…… (1)

At noon:

Substitute $1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for $I_{\text{max}}$, $0.75\text{m}$ for $L_{\text{stick}}$ and $0.30\text{ m}$ for $L_{\text{shadow}}$ in equation (1).

$\begin{array}{c}I=1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{\sqrt{{\left(0.75\text{m}\right)}^2+{\left(0.30\text{ m}\right)}^2}}\right)\\ =1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{\sqrt{0.5625\text{}{\text{ m}}^2+0.09{\text{ m}}^2}}\right)\\ =1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{0.81\text{ m}}\right)\\ =9.25\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

At evening:

Substitute $1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for $I_{\text{max}}$, $0.75\text{m}$ for $L_{\text{stick}}$ and $2.3\text{ m}$ for $L_{\text{shadow}}$ in equation (1).

$\begin{array}{c}I=1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{\sqrt{{\left(0.75\text{m}\right)}^2+{\left(2.3\text{ m}\right)}^2}}\right)\\ =1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{\sqrt{0.5625\text{}{\text{ m}}^2+5.29{\text{ m}}^2}}\right)\\ =1.0\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\left(\frac{0.75\text{m}}{2.41\text{ m}}\right)\\ =3.11\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Conclusion:

The insolation of stick measuring $0.75\text{ m}$ at noon is $9.25\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ and in the evening is $3.11\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$.