Question

An object placed in front of a convex mirror of radius 40cm produces an erect image which is one-sixth the size of the object. How far is the object from the mirror?

Answer 1

Answer on Question 68639, Physics, Optics

Question:

An object placed in front of a convex mirror of radius $40~\mathrm{cm}$ produces an erect image which is one-sixth the size of the object. How far is the object from the mirror?

Solution:

We can find the position of the object from the mirror equation:

$$\frac{1}{d_o} + \frac{1}{d_i} = -\frac{1}{f},$$

here, $d_o$ is the distance from the object to the mirror, $d_i$ is the distance from the image to the mirror and $f$ is the focal length (since we have the convex mirror, the focal length will be with sign minus).

By the definition, the focal length of the curved mirror is half a radius of curvature:

$$f = \frac{R}{2}.$$

Substituting $f$ into the mirror equation we get:

$$\frac{1}{d_o} + \frac{1}{d_i} = -\frac{2}{R}.$$

From the initial condition of the question we know that the size of the image is one-sixth the size of the object:

$$h_i = \frac{1}{6} h_o.$$

Also, we know that:

$$\frac{h_i}{h_o} = \frac{-d_i}{d_o} = \frac{1}{6}.$$

From this equation we can express $d_o$ in terms of $d_i$:

$$d_o = -6 d_i.$$

Let's first substitute $d_{o}$ into the mirror equation and find the distance from the image to the mirror:

$$- \frac {1}{6 d _ {i}} + \frac {1}{d _ {i}} = - \frac {2}{R},$$ $$\frac {5}{6 d _ {i}} = - \frac {2}{R},$$ $$d _ {i} = - \frac {5 R}{12} = - \frac {5 \cdot 40 \text{ cm}}{12} = - 16.6 \text{ cm}.$$

The sign minus indicates that the image is located behind the mirror.

Finally, using the same mirror equation we can find the distance from the object to the mirror:

$$\frac {1}{d _ {o}} = - \frac {2}{R} - \frac {1}{d _ {i}},$$ $$d _ {o} = \frac {1}{- \frac {2}{40 \text{ cm}} - \frac {1}{- 16.6 \text{ cm}}} = 97.6 \text{ cm}.$$

Answer:

$$d _ {o} = 97.6 \text{ cm}.$$

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