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In Figure ., L is a converging lens of focal length 10cm and M Iis a concave mirror of radius of curvature 20cm. A point object O is placed in frontof the lens at a distance of 15cm. Ab and CD are optical axes of the lens andmirror, respectively. Find the distance of the final image formed by this system from the optical center of the lens. The distance between CD and AB is 1cm. |
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Answer» `(1)/(v_(1))-(1)/(u_(1))=(1)/(f) ` `u_(1)=-15` and `f_(1)=10` Solving, we get `v_(1)=30cm` `I_(1)` acts as source for the mirror `:. u_(2)=-(45-v_(1))=-15cm` `I_(2)` is the image formed by the mirror `:.(1)/(v_(1))=(1)/(f_(m))-(1)/(u_(2))=-(1)/(10)-(1)/(15)` `:. v_(2)=-30cm` The height of `I_(2)` above principal axis fo lens ` =(v_(2))/(u_(2))xx1+1=3cm` `I_(2)` acts as a COURCE for the lens `:. u_(3)=-(45-v_(2))=-15cm` HENCE, the lens forms ANIMAGE `I_(3)` at a distance `v_(3)=30cm` to the left of the lens and at a distance The height of `I_(2)` above principal axis of lens `=(v_(2))/(u_(2))xx1+1=3cm` `:. ` Required distance `=sqrt(30^(2)+6^(2))=6sqrt(26)CM` 
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