A Keplerian telescope with magnification `T = 15` was submerged into water which filled up the inside of the telescope. To make the system work as a telescope again within the former dimensions, the objective was replaced. What has the magnification of the telescope become equal to ? the refractive index of the glass of which the ocular is made is equal to `n = 1.50`.

Explanation

When a glass lens is immersed in water its focal length increase approximately four times. We check this as follows as :
`(1)/(f'_(a)) = (n - 1) ((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(f_(w)) = ((1)/(n_(0)) - 1) ((1)/(R_(1)) - (1)/(R_(2))) = ((1)/(n_(0)) - 1)/(n - 1). (1)/(f_(a)) = (n - n_(0))/(n_(0)(n - 1)) (1)/(f_(a))`
Now back to the problem. originally in air
`T = (f_(0))/(f_(e)) = 15` so `l = f_(0) + f_(e) = f_(e) (T + 1)`
In water, `f'_(e) = (n_(0)(n - 1))/(n- n_(0))f_(e)`
and the focal length of the replaced objective is given by the condition
`f'_(0) + f'_(e) = l = (T + 1) f_(e)`
or `f'_(0) = (T + 1) f_(e) - f'_(e)`
Hence `T' = (f'_(0))/(f'_(0)) = (T + 1) (n - n_(0))/(n_(0)(n - 1)) - 1`
Subsituation gives `(n = 1.5, n_(0) = 1.33), T' = 3.09`