(a)The focal length of a diverging lens is
(1/f) = (1/u) + (1/v) --------(1)
Here,f = -18cm and u = 36cm.
Substituting the above values in equation (1),we get
(1/v) = (1/f) - (1/u) = -(1/18) - (1/36) = -(3/36)
or v = - (36/3) = - 12cm
The image is virtual and erect.
The magnification is
m = (v/u) = (-12/36) = (1/3) = 0.33
(b)When u = 18cm,then
(1/v) = (1/f) - (1/u) = - (1/18) - (1/18)
or (1/v) = - (2/18)
or v = - 9cm
or v = 9cm
The image is virtual and erect.
The magnification is m = (v/u) = (9/18) = (1/2) = 0.5
(c)When u = 9.0cm,then
(1/v) = - (1/18) - (1/9) = - (3/18)
or v = 6cm
The image is virtual and erect.
The magnification is
m = (v/u) = (6/9) = 0.67.