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The confusion between \( d_i = 15 \, \text{cm} \) and \( d_i = 0.066 \) likely arises from a misunderstanding of the reciprocal format used in the mirror equation. Let’s break it down:

The mirror equation is:

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

where:
- \( f \) is the focal length of the mirror,
- \( d_o \) is the object distance, and
- \( d_i \) is the image distance.

In this case:
- The focal length \( f = 10 \, \text{cm} \),
- The object distance \( d_o = 30 \, \text{cm} \).

The calculation shown in the image applies the mirror equation correctly. By substituting and rearranging terms, the equation becomes:

\[
\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}
\]

To find \( d_i \), take the reciprocal of \(\frac{1}{15}\):

\[
d_i = 15 \, \text{cm}
\]

The answer is not \( 0.066 \), which you might have gotten if you misunderstood \(\frac{1}{15}\) as a decimal. However, taking the reciprocal of \(\frac{1}{15}\) results in \( d_i = 15 \, \text{cm}\), not 0.066.

Would you like further details on this explanation or any other aspect?

Here are some questions for further exploration:

1. Why is the mirror equation structured as reciprocals?
2. How would the image distance change if the object distance were less than the focal length?
3. What would be the effect of using a convex mirror instead of a concave mirror?
4. How does the nature of the image (real or virtual) relate to the sign of the image distance \( d_i \)?
5. How would the equation change if the focal length were negative?

**Tip:** When using reciprocal equations like the mirror formula, always take the final reciprocal if you want the actual value rather than the inverse.

Explore the solution to finding the image distance for an object placed 30 cm in front of a concave mirror with a focal length of 10 cm. Learn why the correct answer is 15 cm instead of 0.066, with a breakdown of the mirror equation and reciprocal calculations. Suitable for high school students in Grades 10-12.

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