a convex lens is placed betwn an object and d screen wich r a fix d distnce apart for one position of d elns the magnification of d image is m1 wich is obtaind on d screen , when d lens is moved by a idstnce d the magnification of d image is m2 obtained on d same screen . d focal lenght of d lens is ?
(m2>m1)
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2 Answers
62see i will give a hint..
let x be the total separation between the object and the screen
let the positions of the lens be "a" from the object..
the object is first then lens then the screen..
so 1/v-1/u=1/f
now v=x-a
u=-a
substitute to get a quadratic in a
so we know that the difference in the roots of the quadratic is "d"
we also know V/U bcos we know the magnifications..
so u know enuf.. i hope it helps..
62now v=x-a
u=-a
we get a quadratic in a
a2-ax+xf=0
a=(x±D)/2 where D2=x2-4xf
magnification is m=v/u = 1-x/a
now m2>m1 => a2>a1
so m2-m1=(1-x/a2)-(1-x/a1) (find this )
d=difference of roots = D
how it is easy to solve..
just 1 step remains..
