The momentum of light, as it is for particles, is exactly reversed when a photon is reflected straight back from a mirror, assuming negligible recoil of the mirror. the change in momentum is twice the photon’s incident momentum, as it is for the particles. suppose that a beam of light has an intensity 1.0kw/m2 and falls on a −2.0−m2 area of a mirror and reflects from it.

(a) calculate the energy reflected in 1.00 s.

(b) what is the momentum imparted to the mirror?

(c) use newton’s second law to find the force on the mirror.

(d) does the assumption of no-recoil for the mirror seem reasonable?

a)   E = 2.00 10³ J , b)   I = 6.66 10⁻⁶ N s , c)   F = 1.66 10⁻⁶ N

Explanation:

a) The intensity is defined as the power per unit area

          I = P / A

          P = I A

Power is energy for time

         P = E / t

We replace

        E / t = I A

        E = I A t

        E = 1.0 10³ 2.0 1.00

        E = 2.00 10³ J

b) The moment is

       p = U / c

In the case of a reflection the speed is reversed, so the moment

      Δp = 2 U / c

       I = Δp

       I = 2 U / c

       I = 2.00 10³/3 10⁸

       I = 6.66 10⁻⁶ N s

c) The defined impulse is

        I = F t

       F = I / t

For a time of 1 s

       F = 6.66 10⁻⁶ / 1

       F = 1.66 10⁻⁶ N

d) Suppose n small mass mirror m = 10 10⁻³ kg, we write Newton's second law

        F = ma

        a = F / m

        a = 1.66 10⁻⁶ / 10 10⁻³

         a = 1.66 10⁻⁴ m / s

We see that the acceleration is very small and attended to increase the mass of the mirror will be less and less, so the assumption of no twisting of the mirror is very reasonable

no sign of dia of fuel cell