The path of the light rays is in the figure below. The convention signs for distances object-lens, image-lens and height of the object and image are also depicted in the attached figure.

The equation of the lens is

`1/x_1 +1/x_2 =1/f` , with `f >0` for convergent lenses.

From similar...

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The path of the light rays is in the figure below. The convention signs for distances object-lens, image-lens and height of the object and image are also depicted in the attached figure.

The equation of the lens is

`1/x_1 +1/x_2 =1/f` , with `f >0` for convergent lenses.

From similar triangles `ABO` and `A'B'O` one can write for the magnification:

`|M| =|-y_2/y_1| = x_2/x_1`

Thus `2x_1 =x_2`

and from the original condition given in the problem text that

`x_1+x_2 =45`

one obtains

`x_2/2 +x_2 =45`

`3/2*x_2 =45`

`x_2 =(2*45)/3 =90/3 =30 cm`

`x_1 =45-30 =15 cm`

**Answer: the object distance is `x_1 =15 cm` **