Question

An object is placed 15 cm from a concave mirror of focal length 10 cm, then the nature of the image formed will be

• A. magnified and inverted
• B. magnified and erect
• C. small in size and inverted
• D. small in size and erect

Hint:

Use the mirror formula with the object distance negative, then find the magnification and read its sign and size.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a concave mirror the focal length and object distance are both negative in the sign convention. The mirror formula \(\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\) gives the image position, and \(m=-\dfrac{v}{u}\) gives its nature.

Step 2: Calculation
\(u=-15\) cm and \(f=-10\) cm.
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30}\), so \(v=-30\) cm.

Step 3: Magnification
\(m=-\dfrac{v}{u}=-\dfrac{-30}{-15}=-2\).
The negative sign means the image is inverted and the magnitude 2 means it is magnified.

Step 4: Conclusion
The object lies between F and C, so the image is real, inverted and magnified, formed beyond C.

Final Answer:
The image is magnified and inverted. \[ \boxed{\text{(A) }\text{magnified and inverted}} \]