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Find the velocity of image w.r.t ground shown in figure. |
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Answer» Solution :For the velocity component along the principal axis `(V_(II))_(II)=+v^2/u^2 (V_(OL))_s`. Applying lens equation `1/v-1/u=1/f IMPLIES 1/v-1/(-20)=1/(+30) implies v=-60cm` `m=+ v/u=(-60)/(-20)=3, (V_(IM))_(II)=+((-60)/(-20))^2 (5)=+45m//s` `vecV_(RG)=vecV_(IL)+vecV_(LG)=+45+0=45m//s` Hence, IMAGE aproaches the lens with a velocity `45m//s` |
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