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22P

Expert-verifiedFound in: Page 1002

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A black, totally absorbing piece of cardboard of area ${\mathbf{\text{A = 2.0}}}{\mathbf{}}{{\mathbf{\text{cm}}}}^{{\mathbf{\text{2}}}}$ **** intercepts light with an intensity of **** ${\mathbf{110}}{\mathbf{}}{{\mathbf{\text{W/m}}}}^{{\mathbf{\text{2}}}}$ from a camera strobe light. What radiation pressure is produced on the cardboard by the light?**

The radiation pressure is, ${P}_{r}=3.3\times {10}^{-8}Pa$_{.}

Area of cardboard, $\mathrm{A}=2{\text{cm}}^{\text{2}}$.

Intensity, I = 10 W/m^{2}.

**The radiation pressure formula is for the total absorption of radiation. First, find the force on the surface due to radiation, and substituting this value and the area of the surface, calculate the radiation pressure of the light bulb. When electromagnetic radiation of intensity $\mathrm{I}$**** is incident on a perfectly reflective surface, the radiation pressure $\mathrm{P}$**** exerted on the surface is given by, ****, where $\mathrm{c}$**** is the speed of light.**

If the radiation is totally absorbed by the surface, the force is,

$F=\frac{IA}{c}$

Here I is the intensity of radiation, and $A$ is the area of the surface perpendicular to the path of the radiation.

The intensity $I$ is related to the power of the bulb P as,

$I=\frac{P}{A}$

Where, A is the area of the cardboard.

The radiation pressure P_{r} is the force per unit area, and it can be calculated as,

${p}_{r}=\frac{IA}{AC}\phantom{\rule{0ex}{0ex}}{p}_{r}=\frac{I}{C}$

Substitute the values in the expression, and we get,

${P}_{r}=\frac{10w}{3\times {10}^{8}m/s}\phantom{\rule{0ex}{0ex}}=3.3\times {10}^{-8}Pa$

Therefore, the radiation pressure ${\mathrm{p}}_{\mathrm{r}}$ is $3.3\times {10}^{-8}\text{Pa}$.

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