Question

An object is placed at a distance of 10 cm from a convex mirror of radius of curvature 30 cm. Where is the position of the image ?

• A. 12 cm
• B. -12 cm
• C. 6 cm
• D. -6 cm

Hint:

1. Sign convention for convex mirror: \(u\) is negative, \(R\) and \(f\) are positive. 2. Given: \(u = -10\) cm, Radius of curvature \(R = +30\) cm. 3. Calculate focal length: \(f = R/2 = +30/2 = +15\) cm. 4. Use mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). \(\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}\) \(\frac{1}{v} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10}\) 5. Solve for \(v\): \(\frac{1}{v} = \frac{2+3}{30} = \frac{5}{30} = \frac{1}{6}\). So, \(v = +6\) cm. The image is 6 cm behind the mirror (virtual and erect).

Answer 1

The Correct Option is C

Concept: This problem uses the mirror formula and sign conventions for spherical mirrors. Mirror Formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) Where \(f\) is focal length, \(v\) is image distance, and \(u\) is object distance. For a spherical mirror, focal length \(f = \frac{R}{2}\), where \(R\) is the radius of curvature. Sign Conventions (for mirrors, object to the left):
Object distance (\(u\)): Negative.
Focal length (\(f\)) for a convex mirror: Positive.
Radius of curvature (\(R\)) for a convex mirror: Positive.
Image distance (\(v\)) for a convex mirror: Always positive (image is virtual and behind the mirror). Step 1: Identify given values and apply sign conventions
Object distance, \(u = -10 \text{ cm}\) (object is placed in front of the mirror).
Radius of curvature, \(R = +30 \text{ cm}\) (for a convex mirror, R is positive). Step 2: Calculate the focal length (\(f\)) \[ f = \frac{R}{2} = \frac{+30 \text{ cm}}{2} = +15 \text{ cm} \] Step 3: Apply the mirror formula to find image distance (\(v\)) \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Rearrange to solve for \(\frac{1}{v}\): \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Substitute the values of \(f\) and \(u\): \[ \frac{1}{v} = \frac{1}{15} - \frac{1}{-10} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{10} \] Step 4: Solve for \(v\) To add the fractions, find a common denominator for 15 and 10, which is 30. \[ \frac{1}{15} = \frac{2}{30} \] \[ \frac{1}{10} = \frac{3}{30} \] So, \[ \frac{1}{v} = \frac{2}{30} + \frac{3}{30} = \frac{2+3}{30} = \frac{5}{30} \] \[ \frac{1}{v} = \frac{1}{6} \] Therefore, \(v = 6 \text{ cm}\). The positive sign for \(v\) indicates that the image is formed behind the mirror, which is characteristic of a virtual image formed by a convex mirror. The position of the image is 6 cm behind the mirror. This matches option (3).