Determine the focal length of a concave mirror if:
(a) with the distance between an object and its image being equal to `l = 15 cm`, the transverse magnification `beta =02.0`,
(b) in a certain position of the object the transverse magnification is `beta_(1) =-0.50` and in another position displaced with respect to the former by a distance `l = 5.0 cm` the transverse magnification `beta_(2) = - 0.25`.

Explanation

From the mirror formula,
`(1)/(s') + (1)/(s) = (1)/(f)` we get `f = (s's)/(s' + s)` (1)
In accordance with the problem `s - s' = l(s')/(s) = beta`,
From these two relations, we get : `s = (l)/(1 - beta), s' =-(l beta)/(1 + beta)`
Substituting it in the Eqn. (1),
`f = (beta((l)/(1 - beta))^(2))/(l((1 - beta)/(1 - beta))) = (l beta)/((1 - beta^(2))) = -10cm`
(b) Again we have, `(1)/(s') + (1)/(s) = (1)/(F)` or,`(s)/(s') + 1= (s)/(f)`
or, `(1)/(beta_(1)) = (s)/(f)-1 = (s - f)/(f)`
or, `beta_(1)= (f)/(s - f)` (2)
Now,it is clear from the above equation, that fro smaller `beta, s` must be large, so the object is displaced away from the mirror in second position.
i.e `beta_(2) = (f)/(s + l - f)` (3)
Eliminating `s` from the Eqn. (2) and (3), we get
`f = (l beta_(1) beta_(2))/((beta_(2) - beta_(1))) =-2.5 cm`.