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If separation between an object and a screen is D, then prove that D must be greater or equal to four times of focal length of convex lens. Also prove that there are two positions of lens for which real image of object can be projected on screen. If d is separation between these two positions of lens then what will be the focal length of lens in terms of D and d. |
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Answer» Solution :Let D be the separation between object and screen and x be the distance of lens from object. Then we can use lens formula as follows : `(1)/(f)=(1)/(v)-(1)/(u)` `RARR""(1)/(f)=((1)/(+D-x))-(1)/(-x)` `rArr""(1)/(f)=(1)/((D-x))+(1)/(x)=(x+D-x)/((D-x)x)` `rArr""(1)/(f)=(D)/(Dx-x^(2))` `rArr""Dx-x^(2)=fD` `rArr""x^(2)-Dx+fD=0"....(1)"` For real solution of above quadratic equation its discriminant must be greater or equal to zero. `rArr""D^(3)-4fD ge 0` `rArr""D(D-4f)ge0` `rArr""Dge4f` Hence we have proved that D must be greater or equal to four times of focal length of convex lens. Now from equation (i) we can see that it is a quadratic equation. Hence there are two ROOTS and hence there are two positions of lens for which we can project real image of object on screen. Two solutions of the quadratic equation can be written as follows : `x=(Dpm SQRT(D^(2)-4fD))/(2)` Let `x_(1) and x_(2)` be the zero roots then we can write as follows : `x_(1)=(D+sqrt(D^(2)-4fD))/(2)` `x_(2)=(D-sqrt(D^(2)-4fD))/(2)` Distance between these two position of lens is given to us as d. `x_(1)-d_(2)=d` `rArr""sqrt(D^(2)-4fD)=d` `rArr""D^(2)-4fD=d^(2)` `rArr""f=(D^(2)-d^(2))/(4D)` |
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