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a point source of light is placed at the center of curvature of a hemispherical surface. radius r&inner surface is totally reflecting.find force on the hemisphere due 2 lightfalling on it if source emits power w MY APPROACH ENERGY EMITED by source in 4pie solid angle per sec = W for hemisphere solid angle=2pie so, energy falling/sec on hemisphere =(w/4pie)*(2pie)=w/2 momentum=e/c=w/2c as the surface is reflecting so total momentumchange=2*(w/2c)=w/c HOWEVER THE ANSER IS GIVEN W/2C IN H.C.VERMA2 PG NO.360 QN NO.6 TELL HOW I THINK H.C.VERMA HAS WRONGLY WRITEN THE AREA AS 4(pie)R(square) it should be 2 instead of 4


 

a point source of light is placed at the center of curvature of a hemispherical surface. radius r&inner surface is totally reflecting.find force on the hemisphere due 2 lightfalling on it if source emits power w
 MY APPROACH
ENERGY EMITED by source in 4pie solid angle per sec = W
for hemisphere solid angle=2pie
so, energy falling/sec on hemisphere =(w/4pie)*(2pie)=w/2
momentum=e/c=w/2c
as the surface is reflecting so total momentumchange=2*(w/2c)=w/c
HOWEVER THE ANSER IS GIVEN W/2C IN H.C.VERMA2 PG NO.360 QN NO.6
TELL HOW
I THINK H.C.VERMA HAS WRONGLY WRITEN THE AREA AS 4(pie)R(square) it should be 2 instead of 4
 
 


Grade:upto college level

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