Question

An object is placed 10 cm in front of a concave mirror of focal length 15 cm. The image formed is:

• A. Virtual and enlarged
• B. Real and enlarged
• C. Virtual and diminished
• D. Real and diminished

Hint:

For concave mirrors:
- Object beyond center → real, inverted, diminished
- Object at center → real, inverted, same size
- Object between center and focus → real, inverted, enlarged
- Object at focus → image at infinity
- Object between focus and pole → virtual, erect, enlarged

Answer 1

The Correct Option is A

We are given:
- Object distance \( u = -10 \, \text{cm} \) (object is in front of mirror, so it’s negative)
- Focal length of concave mirror \( f = -15 \, \text{cm} \) (concave mirrors have negative focal length)
We use the mirror equation:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substitute the known values:
\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{-15} + \frac{1}{10} \] Take LCM of 30:
\[ \frac{1}{v} = \frac{-2 + 3}{30} = \frac{1}{30} \Rightarrow v = 30 \, \text{cm} \] Interpretation: The image distance \( v = +30 \, \text{cm} \), which means the image is on the same side as the object (behind the mirror), indicating a virtual image.
To understand the size, calculate magnification:
\[ m = -\frac{v}{u} = -\frac{30}{-10} = +3 \] So, the image is enlarged (magnification \(>1 \)) and upright (positive sign).
Conclusion: A concave mirror forms a virtual and enlarged image when the object is placed between the pole and the focus (which is the case here since object is at 10 cm and focus is at 15 cm).