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In the figure shown L is a converging lens of foacl length 10 cm and M is a concave mirror of radius of curvature 20 cm. A point object O is placed in front of the lens at a distance 15 cm.AB and CD are optical axes of the lens and mirror repectively. Find the distance of the final image formed by this system from the optical centre of the lens. The distance between CD & AB is 1 cm |
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Answer» `(1)/(v_(1))-(1)/(u_(1))=(1)/(f) u_(1)=-15 f_(1)=10` SOLVING we get `v_(1)=30cm` `I_(1)` acts as source for MIRROR `therefore u_(2)=-(45-v_(1))=-15 cm` `I_(2)` is the iamge formed by the mirror `therefore (1)/(v_(2))=(1)/(f_(m))-(1)/(u_(2))=-(1)/(10)+(1)/(1\5) thereforev_(2)=-30 cm`  The height of `I_(2)` above pricnipal axis of lins is `=(v_(2))/(u_(2))xx1+1=3cm` `I_(2)` acts a source for lens`therefore u_(3)=-(45-v_(2))=-15 cm` Hence the lens forms an image `I_(3)` at a distance `v_(3)=30 cm` to the left of lens and at a distance The height of `I_(2)` above principal axis of lens is `|(v_(3))/(u_(3))|xx 3 cm =6 cm` `therefore` required distance `=sqrt(30^(2)+6^(2))=6sqrt(26) cm` |
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