A point object is placed 60 cm from the pole of a concave mirror of focal length 10 cm on the principal axis.
Find:
a. the position of image.
b. If the object is shifted 1 mm towards the mirror along principal axis, find the shift in image. Explain the result.

Explanation

(a) `u = -60cm`
`f = -10cm`
`v = (fu)/(u-f) = (-10(-60))/(-60-(-10)) = -12cm`
(b) `(1)/(v) + (1)/(u) = (1)/(f)`
differentiating, we get `dv = -(v^(2))/(u^(2)) du = -((-12)/(-60))^(2) [1mm] = -(1)/(25)mm`
[`:' du = 1mm` sign of `du` is `+` becuase it is shifted in `+ve` direction defined by sign convention.]
(A) `-ve` sign of `dv` indicates that the image will shift towards negative direction.
(B) The sign of `v` is negative. Which implies the image is formed on negative side of poie `(A)` and `(B)` together imply that the image will shift away from pole.
Note that diffentials `dv` and `du` denote small changes only.
(h) Newton's formula `XY = f^(2)`
`X` and `Y` are the distances (along the principle axis) of the object the image respectively form the principle focus. This formula can be used when the distances are mentioned or asked form the focus.