An object is placed on the principal axis of a concave mirror at a distance of $1.5f$ ($f$ is the focal length

Question

An object is placed on the principal axis of a concave mirror at a distance of $1.5f$ ($f$ is the focal length). The image will be at:

Accepted Answer

1. **Identify Formula:** Use the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
2. **Sign Convention:** According to the Cartesian sign convention for a concave mirror:
   - Object distance ($u$) is taken as negative: $u = -1.5f$.
   - Focal length ($f$) is taken as negative for a concave mirror: Focal length = $-f$.
3. **Calculation:** Substitute the values into the formula:
   $$ \frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f} $$
   $$ \frac{1}{v} = \frac{1}{1.5f} - \frac{1}{f} $$
   $$ \frac{1}{v} = \frac{10}{15f} - \frac{1}{f} = \frac{2}{3f} - \frac{3}{3f} $$
   $$ \frac{1}{v} = \frac{2 - 3}{3f} = \frac{-1}{3f} $$
   $$ v = -3f $$
4. **Conclusion:** The image is formed at a distance of $3f$ from the mirror. The negative sign indicates that the image is formed on the same side as the object (real image).

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