An object is 30 cm to the left of a concave lens of focal length −15 cm. Find image distance and describe the image.

Solution & explanation

1. Use real-is-positive lens distances: u = +30 cm for the real object and f = −15 cm for the diverging lens.

2. Start with the thin-lens equation.

   $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$

3. Subtract 1/u from both sides.

   $$\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$$

4. Substitute signed values.

   $$\frac{1}{v}=\frac{1}{-15}-\frac{1}{30}$$

5. Use denominator 30.

   $$\frac{1}{v}=-\frac{2}{30}-\frac{1}{30}=-\frac{3}{30}=-\frac{1}{10}$$

6. Take reciprocal.

   $$v=-10$$

7. Calculate magnification to identify orientation and size.

   $$m=\frac{10}{30}=\frac{1}{3}$$

8. State the result with units: v = −10 cm; virtual, upright and one-third size, on the object side.

v = −10 cm; virtual, upright, diminished