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In the previous question instead of spherical wall there is a vertical wall at a perpendicular distance d from the point & where the light is incident. |
Answer» SOLUTION : `TAN theta=x/d` `RARR x=d tan theta` `(DX)/(dt)=d sec^(2) theta (d theta)/(dt)` `=2 omega d sec^(@) theta""( :' (d theta)/(dt) =2 omega)` OR Considering an instantaneus cricle of radius `d sec theta`.  `v_(t)= 2 omega d sec theta` (`2 omega s cos theta` is a component of V.) `v cos theta=2 omega d sec theta` `rArr v= (2 omega sec theta)/(cos theta)=2 omega d sec^(2) theta` |
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