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a) Determine the 'eefective focal length' of the combination of the two lenses in Exercise 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ? b) A object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens is 40 cm. |
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Answer» Solution :b) Here, `h_(1)=1.5cm, u_(1)=40cm,m=?` `h_(2)=?` for the 1st lens, `(1)/(upsilon_(1))-(1)/(u_(1))=(1)/(f_(1))` `(1)/(upsilon_(1))=(1)/(P_(1))+(1)/(u_(1))=(1)/(30)-(1)/(40)=(1)/(120)` `upsilon_(1)=120cm` Magnitude of magnification produced by FIRST lens, `m_(1)=(upsilon_(1))/(u_(1))=(120)/(40)=3` The image formed by 1st lens as VIRTUAL object for the 2nd lens. `u_(2)=120-8=112cm, f_(2)=-20cm` `upsilon_(2)=?` As `(1)/(upsilon_(2))-(1)/(u_(2))=(1)/(f_(2)) rArr (1)/(v_(2))=(1)/(f_(2))+(1)/(u_(2))` `(1)/(v_(2))=(-1)/(20)+(1)/(112)` `v_(2)=(-112xx20)/(92)cm` Magnitude of Magnification produced by second lens `m_(2)=(v_(2))/(u)=(112xx20)/(92xx112)=(20)/(92)` Net magnification produced by the combination `m=m_(1)xxm_(2)` `=3xx(20)/(92)=(60)/(92)=0.652` `therefore"size of image "h_(2)="mh"_(1)` `=0.652xx1.5` `=0.98cm` |
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