Question

Use the thin lens equation and the equation for magnification, to show that for a single diverging lens that the image of a real object is always virtual, upright, and reduced. (Hint: Solve the lens equation for di and substitute into the equation for magnification. Discuss the value and sign of magnification for negative values of f.

Answer 1

f = focal length

do = object distance

di = image distance

Using thin lens equation

1/f = 1/di + 1/do

1/di = 1/f - 1/do

di = do f(do - f)                                                Eq-1

Using the magnification equation

m = - di/do

m = - (do f(do - f))/do

m = - f(do - f)

focal length for diverging lens is always negative , hence

do - f > 0

magnification is always positive. hence the image is always upright.

m < 1     Since value at numerator is smaller than denominator.

So the image is reduced.

From Eq-1

di < 0

hence the image is virtual