Reflection Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

medium

A concave mirror has a focal length of

15 \text{ cm}

. An object is placed

30 \text{ cm}

from the mirror. Use the mirror equation

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

to find the image distance. Is the image real or virtual?

1
$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ . $\frac{1}{15} = \frac{1}{30} + \frac{1}{d_i}$ .
2
$\frac{1}{d_i} = \frac{1}{15} - \frac{1}{30} = \frac{2 - 1}{30} = \frac{1}{30}$ . So $d_i = 30 \text{ cm}$ .
3
Since $d_i$ is positive, the image is real (formed in front of the mirror). The object is at $2f$ , so the image is also at $2f$ — same size, inverted.

Answer

d_i = 30 \text{ cm (real, inverted, same size)}

When an object is placed at twice the focal length (

2f

) of a concave mirror, the image forms at

2f

on the same side — real, inverted, and the same size. This is the special case used in certain optical instruments.

About Reflection

The change in direction of a wave at a boundary so that it returns into the original medium.

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