Question:
An object is placed at 9 cm in front of a concave mirror of radius of curvature 12 cm. The following statement is true:
An object is placed at 9 cm in front of a concave mirror of radius of curvature 12 cm. The following statement is true:
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Object between F and C ⇒ magnified real image.
Updated On: May 1, 2026
- Image 36 cm behind
- Image 36 cm in front
- Virtual, erect
- Real, erect
- Real, inverted, magnified
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The Correct Option is
Solution and Explanation
Concept: Mirror formula and image formation
For a spherical mirror: \[ f = \frac{R}{2} \]
Step 1: Find focal length
Given: \[ R = 12 \text{ cm} \Rightarrow f = \frac{12}{2} = 6 \text{ cm} \]
Step 2: Apply mirror formula
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Given object distance: \[ u = -9 \text{ cm (sign convention)} \] \[ \frac{1}{v} = \frac{1}{6} - \frac{1}{9} \]
Step 3: Simplify
\[ \frac{1}{v} = \frac{3 - 2}{18} = \frac{1}{18} \] \[ v = 18 \text{ cm} \]
Step 4: Determine nature of image
- $v$ is positive ⇒ image is real - Real images formed by mirrors are inverted - Image distance (18 cm) > object distance (9 cm) ⇒ magnified Final Answer: \[ \boxed{\text{Real, inverted, magnified}} \]
For a spherical mirror: \[ f = \frac{R}{2} \]
Step 1: Find focal length
Given: \[ R = 12 \text{ cm} \Rightarrow f = \frac{12}{2} = 6 \text{ cm} \]
Step 2: Apply mirror formula
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Given object distance: \[ u = -9 \text{ cm (sign convention)} \] \[ \frac{1}{v} = \frac{1}{6} - \frac{1}{9} \]
Step 3: Simplify
\[ \frac{1}{v} = \frac{3 - 2}{18} = \frac{1}{18} \] \[ v = 18 \text{ cm} \]
Step 4: Determine nature of image
- $v$ is positive ⇒ image is real - Real images formed by mirrors are inverted - Image distance (18 cm) > object distance (9 cm) ⇒ magnified Final Answer: \[ \boxed{\text{Real, inverted, magnified}} \]
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