91影视

The photon energy can be expressed as:-

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{位}}$

Here c is the speed of light h is the Plank鈥檚 constant and 位 is the wavelength.

Substituting the values and solving the above equation for E, we get,

$\begin{array}{rcl}\mathit{E}& \mathbf{=}& \left(6.63脳{10}^{-34}J路s\right)\left(\frac{3脳{10}^{8}m/s}{550脳{10}^{-9}m}\right)\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{61}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{J}\end{array}$

And then to get the number of photons per second, we decide the power P by the energy per photon E.

$\begin{array}{rcl}\frac{no.ofphotons}{time}& =& \frac{P}{E}\\ n& =& \frac{2.5J/s}{3.6脳{10}^{-19}J}\\ n& =& 6.9脳{10}^{18}photons/second\end{array}$

To find that we can start with Newton鈥檚 relation:-

$F=\frac{dp}{dt}$鈥︹赌︹赌︹赌︹赌︹赌�..(1)

For a given momentum p and using the expression for the momentum of a photon for a given energy E.

$P=\frac{E}{C}$

This can be used to rewrite (1)

$$\begin{gathered} F=\frac{dp}{dt} \\ F=\frac{d}{dt}\left(\frac{E}{c}\right) \\ F=\frac{1}{c}\frac{dE}{dt} \\ F=\frac{1}{c}P \end{gathered}$$

That though is the force for a surface that鈥檚 perfectly absorbing the force on a perfect mirror would be double that, since the mirror is effectively stopping, then re-emitting them back the way they came:-

$F=\frac{2P}{c}$

substitute 2.5 J/s for P and 3x10⁸m/s for the speed of light in the above equation and solve for F

$\begin{array}{rcl}F& =& \frac{2脳\left(2.5J/s\right)}{3脳{10}^{8}m/s}\\ & =& 1.7脳{10}^{-8}N\end{array}$

Hence, the total force exerted on the perfect mirror by the flashlight would be

$F=1.7脳{10}^{-8}N$