Question

8. A parallel beam of light of intensity I is
incident normally on a plane surface which
absorbs 50% of the incident light. The
reflected light falls on B which is a perfect
reflector, the light reflected by B is again
partly reflected and partly absorbed and this
process continues. For all absorption by A,
absorption coefficient is 0.5. The pressure
experienced by A due to light is​

Answer 1

Answer:

$Radiation~Pressure={3I\over c}$

Explanation:

Radiant pressure formula goes like-

(i) When light is completely absorbed,

$P_{rad.}=\frac{I}{c} \times a$

(ii) When light is completely reflected,

$P_{rad.}=\frac{2I}{c} \times r$

 where, $I$ = Intensity of incident light

             $c$ = speed of light

             $a$ = absorption coefficient (0.5)

             $r$ = reflection coefficient (1 - a = 0.5)

Now, we only have to find radiation pressure on surface A.

$a=0.5;~r=0.5$

• At first time,

              $I=I$

              For the light which is absorbed,

              $P_{rad.}=\frac{I}{2c}$

              For the light which is reflected,

              $P_{rad.}=\frac{2I}{2c}=\frac{I}{c}$

• At second time,

              $I = I/2$ because intensity halved after first reflection with $a$ = 0.5

              For the light which is absorbed,

              $P_{rad.}=\frac{I}{4c}$

              For the light which is reflected,

              $P_{rad.}=\frac{2I}{4c}=\frac{I}{2c}$

And hence so on...

Now addition of all the radiation pressure by light absorbed-

$P_{rad.}=\frac{I}{2c}+\frac{I}{4c}+\frac{I}{8c}+...\\\\P_{rad.}=\frac{I}{2c}(1+{1\over 2} + {1\over 4}+...)\\\\P_{rad.}={I\over 2c}[{1\over(1-{1\over 2})}]\\\\P_{rad.}={I\over c}$

Now addition of all the radiation pressure by light reflected-

$P_{rad.}=\frac{I}{c}+\frac{I}{2c}+\frac{I}{4c}+...\\\\P_{rad.}=\frac{I}{c}(1+{1\over 2} + {1\over 4}+...)\\\\P_{rad.}={I\over c}[{1\over(1-{1\over 2})}]\\\\P_{rad.}={2I\over c}$

Hence,

Total radiation pressure,

$P_{rad.}={I\over c}+{2I\over c}\\\\P_{rad.}={3I\over c}$