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A thin plano-convex lens of focal length f is split into two halves,one of the halves is shifted along the optical axis. The separation between object and image plane is 1.8m. The magnification of theimage formed by one of the half lenses is 2. Find the focal length of the lens and separation between the helves. Draw the ray diagram for image formation. |
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Answer» SOLUTION :Let MAGNIFICATION caused by the first lens be 2 and distance `OL_(1)=x.`Distance v of image from first lens`L_(1)` is given by `m=(v)/(u)=2 rArr v=2u=2x.` Clearly, `u+v=1.8rArr x+2x=1.8m ` or `3x = 1.8m rArr x=(1.8)/(3)=0.6m` By sign convention, `u=-x=-0.6m, upsilon=2x=1.2m` Lens formula `(1)/(F)=(1)/(v)-(1)/(u)` GIVES `(1)/(f)=(1)/(1.2)+(1)/(0.6)=(1+2)/(1.2)` `:.` Focal length `f=(1.2)/(3) =0.4m` For real image, lens formula takes the form `(1)/(f)=(1)/(v)+(1)/(u)` Clearly, u and v are interchangeable. THEREFORE, for lens `L_(2)` `u^(')=v=1.2m` and `v^(')=0.6m` `OL_(1)=L_(2)I_(2)=x` If d is the separation between the lenses, then `x+d+x=1.8m` `:. d=1.8-2x=1.8-2xx0.6=0.6m` Method-2 Since the magnification for `L_(1)` is 2 `rArr (v)/(u)=-2rArr ((D+d)/(2))/(-(D-d)/(2))=-2` `(D+d)/(D-d)=2rArr D=1.8m,d=0.6m.` `f=(D^(2)-d^(2))/(4D)=((1.8+0.6)(1.8-0.6))/(4xx1.8)=0.4m` 
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