Question

An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the position and nature of the image.

• A. \(20\) cm in front, real and inverted
• B. \(30\) cm in front, real and inverted
• C. \(30\) cm behind, virtual and erect
• D. \(15\) cm behind, virtual and erect

Hint:

Always apply the Cartesian sign convention in mirror problems: distances measured in front of the mirror are negative, and distances behind the mirror are positive.

Answer 1

The Correct Option is B

Concept: The mirror formula is \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] where \(f\) = focal length, \(u\) = object distance, \(v\) = image distance. For a concave mirror, distances measured in front of the mirror are negative according to the Cartesian sign convention.

Step 1: Substitute the given values. \[ u=-15\text{ cm}, \quad f=-10\text{ cm} \] Using the mirror formula: \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] \[ \frac{1}{-10}=\frac{1}{v}+\frac{1}{-15} \]

Step 2: Solve for \(v\). \[ \frac{1}{v}=\frac{1}{-10}+\frac{1}{15} \] \[ \frac{1}{v}=-\frac{1}{30} \] \[ v=-30 \text{ cm} \]

Step 3: Interpret the result. The negative sign indicates that the image forms in front of the mirror. Hence, the image is real and inverted.