A source of light is located at a disatnce `l = 90 cm` from a screen. A thin converging lens provides the sharp image of the source when placed between the source of light and the screen at two positions. Determine the focal length o fthe lens if
(a) the distance between the two positions of the lens is `Delta l = 30 cm`,
(b) the transverse diamensions of the image at one position o fthe lens are `eta = 4.0` greater than those at the pther position.

Explanation

The distance between the object and the image is `l`. Let `x =` distance between the object and the lens. Then , since the image is real, we have in our convention, `u =-x, v =l - x`
so `(1)/(x) + (1)/(l - x) = (1)/(f)`
or `x(l - x) = if` or `x^(2) - xl + lf = 0`
Solving we get the roots
`x = (1)/(2) [1+= sqrt(l^(2) - 4lf)]`
(We must have `l gt 4f` fro real roots)
(a) If the distance between the two positions of the lens is `Deltal`, then clearly `Delta l = x_(2) - x_(1) =` difference between roots `= sqrt(l^(2) - 4lf)`
so `f = (l^(2) - Deltal^(2))/(4l) = 20 cm`.
(b) The two roots are conjugate in the sence that if one gives the object distance the other gives the corresponding image distance (in both cases ). Thus the magnifications are
`-(l + sqrt(l^(2) - 4lf))/(l - sqrt(l^(2) - 4lf))` (enlarged) and `-(l - sqrt(l^(2) - 4lf))/(l + sqrt(l^(2) - 4lf))` (diminished).
The ratio of these magnification being `eta` we have
`-(l - sqrt(l^(2) - 4lf))/(l - sqrt(l^(2) - 4lf)) sqrt(eta)` or `-(l + sqrt(l^(2) - 4lf))/(l ) = (sqrt(eta) -1)/(sqrt(eta)+1)`
or `1 - (4f)/(l) = ((sqrt(eta) -1)/(sqrt(eta)+1))^(2) =1-4 (sqrt(eta))/((1 + sqrt(eta))^(2))`
Hence `f = l (sqrt(eta))/((1+ sqrt(eta))^(2)) = 20 cm`.