Q55PP - University Physics with Modern Physics

Question

Doubling the frequency of a wave in the range of $\mathbf{24}\mathit{H}\mathit{z}$ to represents what change in the maximum allowed electromagnetic-wave intensity? (a) A factor of 2; (b) A factor of $\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}$; (c) A factor of 1/2; (d) A factor of 1/4.

Answer:

Step-by-Step Solution

Answer

The intensity will decrease by four times (d) A factor of $\frac{\mathbf{1}}{\mathbf{4}}$.

Step 1: Formula for maximum electric field

${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{350}}{\mathbf{f}}$

Step 2: Calculate the change in intensity

The maximum electric field is given as: ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{350}}{\mathbf{f}}$ 

When an electromagnetic wave strikes a surface, it exerts radiation pressure, and the maximum intensity of the electric field is given as:

${\mathit{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{{\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}^{\mathbf{2}}$ 

Substitute ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$,

${\mathit{I}}_{\mathbf{E}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{\mathbf{\left(}\frac{\mathbf{350}}{\mathbf{f}}\mathbf{\right)}}^{\mathbf{2}}$ 

From the above equation, the maximum intensity is inversely proportional to the square of the frequency.

So, if the frequency is doubled, the intensity will decrease 4 times.

Hence, the correct answer is (a) A factor of $\frac{\mathbf{1}}{\mathbf{4}}$.