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Two converging lenses have focal length ...


Two converging lenses have focal length `f_1` and `f_2` `(f_1gtgtf_2` ). The optical axis of the two lenses coincide. This lens system is used to from an image of real object. It is observed that final magnification of the image does not depend on the distance x. Whole arrangement is shown in figure. Final magnification is :

A

`-(f_(1))/(f_(2))`

B

`-(f_(2))/(f_(1))`

C

`-(f_(1))/(f_(1)+f_(2))`

D

`-(f_(2))/(f_(1)-f_(2))`

Text Solution

Verified by Experts


Image-1
`(1)/(v_(1))-(1)/(-x)=(1)/(f_(1))`
`v_(1)=(xf_(1))/(x-f_(1))`
`m_(1)=(v_(1))/(u_(1))=(v_(1))/(x)=-((f_(1))/(x-f_(1)))`

Image -2
`u_(2)=-(d-v_(1))`
`(1)/(v_(2))-(1)/(-(d-v_(1)))=(1)/(f_(2))`
`v_(2)=((d-v_(1))f_(2))/(d-v_(1)-f_(2))`
`m_(2)=(v_(2))/(-(d-v_(1)))=-((f_(2))/(d-v_(1)-f_(2)))` ltbr. `m_(1)m_(2)=((f_(1))/(x-f_(1)))((f_(2))/(d-(xf_(1))/((x-f_(1)))-f_(2)))=(f_(1)f_(2))/(x(d-f_(1)-f_(2))-df_(1)+f_(1)f_(2))`
Since m is independent of x
`implies(d-f_(1)-f_(2))=0impliesd=f_(1)+f_(2)`
`impliesm=-(f_(2))/(f_(1))`
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