An object 30 cm tall is 10.5 cm from a concave mirror of focal length 16 cm. Find (a) image position and (b) image height. Use real-is-positive mirror signs.

Solution & explanation

1. Given: h₀ = 30 cm; u = +10.5 cm; f = +16 cm.

2. Rearrange the mirror equation.

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

3. Substitute the distances in the same unit.

   $$\frac{1}{v}=-1\times \frac{1}{10.5}+\frac{1}{16}$$

4. Use a common denominator 336.

   $$\frac{1}{v}=-1\times \frac{32}{336}+\frac{21}{336}=-\frac{11}{336}$$

5. Take the reciprocal. Negative means virtual behind the mirror.

   $$v=-\frac{336}{11}$$

6. Calculate magnification.

   $$m=\frac{-v}{u}$$

7. Substitute the unrounded image distance.

   $$m=\frac{\frac{336}{11}}{\frac{21}{2}}=\frac{32}{11}$$

8. Multiply by object height. Positive height means upright.

   $$h_i=\frac{960}{11}=\frac{960}{11}$$

9. State the result with units: v ≈ −30.55 cm; hᵢ ≈ +87.27 cm.

v ≈ −30.55 cm; hᵢ ≈ 87.27 cm, virtual and upright