Question

Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ________ cm.

• A. \( -\dfrac{15}{2} \)
• B. \( -10 \)
• C. \( -20 \)
• D. \( -15 \)

Hint:

For real images formed by mirrors, magnification is always negative.

Answer 1

The Correct Option is D

To find the focal length of the mirror, we need to understand and apply the mirror formula and magnification concept.

Step-by-Step Solution

1. Given that the image formed is real, and it is magnified three times (i.e., with a magnification, \( m = -3 \)). For mirror images, magnification \( m \) is given by: \(m = \dfrac{-v}{u}\) Here, \( v \) is the image distance and \( u \) is the object distance.
2. The distance between the object and the image is given as 40 cm: \(|v - u| = 40 \, \text{cm}\)
3. Since the magnification \( m = -3 \), we have: \(-3 = \dfrac{-v}{u} \Rightarrow v = 3u\)
4. Substitute \( v = 3u \) into the distance formula: \(|3u - u| = 40 \Rightarrow 2u = 40 \Rightarrow u = 20 \, \text{cm}\)
5. Now, substitute \( u = 20 \) cm in the expression \( v = 3u \): \(v = 3 \times 20 = 60 \, \text{cm}\)
6. The mirror formula is: \(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\) Substitute the values of \( v \) and \( u \) into the mirror formula: \(\dfrac{1}{f} = \dfrac{1}{60} + \dfrac{1}{(-20)}\)

\(\Rightarrow \dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20}\)

\(\Rightarrow \dfrac{1}{f} = \dfrac{1 - 3}{60}\)

\(\Rightarrow \dfrac{1}{f} = \dfrac{-2}{60}\)

\(\Rightarrow f = -30/2 = -15 \, \text{cm}\)

Thus, the focal length of the mirror is \(-15\) cm, which corresponds to the correct option. Real images in concave mirrors have negative focal lengths, confirming that we used the right mirror type.