Question

If the distance between an object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be:

• A. \( \frac{10}{3} \) cm
• B. \( -12 \) cm
• C. \( -10 \) cm
• D. \( 15 \) cm

Hint:

For a concave mirror, a magnified virtual image is always formed when the object is placed between the focal point and the mirror.

Answer 1

The Correct Option is C

Step 1: {Understanding the given data}
The magnification formula for a mirror is: \[ m = -\frac{v}{u} \] Since the image is virtual and magnified two times: \[ m = 2 \] which gives: \[ 2 = -\frac{v}{u} \] Step 2: {Using the given object-image distance}
The total distance between the object and image is: \[ u + v = 15 \] Substituting \( v = -2u \), \[ u + (-2u) = 15 \] \[ - u = 15 \Rightarrow u = 5 { cm} \] \[ v = -2(5) = -10 { cm} \] Step 3: {Using the mirror formula}
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ = \frac{1}{-10} + \frac{1}{5} = -\frac{1}{10} \] \[ f = -10 { cm} \] Thus, the correct answer is \( -10 \) cm.

Answer 2

Step 1: Understanding the Problem
We are given that a **virtual image** is formed by a  curved mirror and the image is 2 times magnified. That means: \[ M = \frac{v}{u} = -2 \] Since the image is virtual and magnified, the mirror must be a **concave mirror**, and virtual images formed by concave mirrors occur when the object is placed between the **pole** and **focus**, resulting in: - \( v > 0 \) (virtual image, on the same side as object) - \( u < 0 \) (real object) - \( M = -2 \)

Step 2: Using the Magnification Formula
\[ M = \frac{v}{u} = -2 \Rightarrow v = -2u \]

Step 3: Total Distance Between Object and Image
Given: Distance between object and image = 15 cm
Since \( u \) is negative and \( v \) is positive: \[ |v - u| = 15 \Rightarrow |-2u - u| = 15 \Rightarrow |-3u| = 15 \Rightarrow u = -5 \, \text{cm} \] Now substitute \( u = -5 \) into \( v = -2u \Rightarrow v = 10 \, \text{cm} \)

Step 4: Using the Mirror Formula
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-5} = \frac{1}{10} - \frac{2}{10} = -\frac{1}{10} \Rightarrow f = -10 \, \text{cm} \]

Step 5: Final Answer
\[ \boxed{f = -10 \, \text{cm}} \] So the correct option is: Option 3: −10 cm