Question

A concave mirror forms a real image of an object kept at a distance of 9 cm from it. If the object is taken away further from the mirror by 6 cm, the image size is reduced to \((1/4)^{th}\) of its previous size. The focal length of the mirror is:

• A. -5 cm
• B. -7 cm
• C. 15 cm
• D. 18 cm

Hint:

Use \(m = f/(f-u)\) for both object positions (9 cm and 15 cm) and set the ratio of sizes to 1/4.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A concave mirror makes a real image of an object at 9 cm. The object is then moved to 15 cm, and the image size becomes one fourth. We must find the focal length.

Step 2: Key Formula or Approach:
The magnification is \(m = \dfrac{f}{f-u}\), with distances measured from the mirror and the object distance \(u\) negative. For a concave mirror \(f\) is negative.

Step 3: Write Both Magnifications:
First position: \(u_1 = -9\). \[ m_1 = \frac{f}{f+9} \] Second position: \(u_2 = -15\). \[ m_2 = \frac{f}{f+15} \] The image size is one fourth, so \(m_2 = \tfrac{1}{4}m_1\).

Step 4: Solve for f:
Both images are real, so both magnifications are negative and their ratio is positive. \[ \frac{f+9}{f+15} = \frac{1}{4} \] \[ 4f + 36 = f + 15 \quad\Rightarrow\quad 3f = -21 \quad\Rightarrow\quad f = -7\ \text{cm} \]

Step 5: Check:
With \(f = -7\): \(m_1 = -7/2 = -3.5\) and \(m_2 = -7/8 = -0.875\). The ratio is \(0.875/3.5 = 1/4\). The object at 9 cm is beyond the 7 cm focal length, so the image is real.

Step 6: Checking Each Option:
-5 cm gives \(m_1 = -5/4\) and \(m_2 = -5/10\), so \(m_2/m_1 = 0.4\), not one fourth. Positive focal lengths (15 cm, 18 cm) mean a convex mirror, which cannot form a real image. So option 2 is correct.

Final Answer:
The focal length of the mirror is -7 cm, so option 2 is correct. \[ \boxed{f = -7\ \text{cm}} \]