Question

A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 20.4 cm to the left of the diverging lens. How far is the object from the converging lens?

d_o1 =

Answer 1

Given that final image produced by two lens combination is 20.4 cm to the left of the diverging lens, So that means final image is virtual and

image distance from diverging lens will be negative, v = -20.4 cm

f = focal length of divergins lens = -28.0 cm

Now Using lens equation for diverging lens:

1/f = 1/u + 1/v

u = object distance from diverging lens = ?

u = v*f/(v - f)

u = (-20.4)*(-28.0)/((-20.4) - (-28.0)) = 75.16 cm

Since u is positive, So object for diverging lens is on the left side of diverging lens,

u = 75.16 cm left of diverging lens

Now remember that object of 2nd lens was image produced by 1st lens

v0 = image of 1st converging lens = d = u

v0 = 56.0 - 75.16 = -19.16 cm

So image produced by 1st converging lens was on the left of converging lens at a distance of 19.16 cm

f0 = focal length of converging lens = 24.0 cm

So now using lens equation for converging lens

1/f0 = 1/u0 + 1/v0

u0 = v0*f0/(v0 - f0)

u0 = (-19.16)*(24.0)/((-19.16) - 24.0)

u0 = object distance from converging lens = 10.65 cm

let me know if you've any query.