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: Use the Mirror Equation:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ and $m = -v/u$. Let the focal length have size $F$, so $f = -F$.

Step 2: Find the Magnification Sizes:
For $u = -9$, the image distance works out so that $|m_1| = \dfrac{F}{9 - F}$. Similarly for $u = -15$: $|m_2| = \dfrac{F}{15 - F}$.

Step 3: Apply the Size Condition:
\[ \frac{|m_2|}{|m_1|} = \frac{9 - F}{15 - F} = \frac{1}{4} \] \[ 36 - 4F = 15 - F \Rightarrow 3F = 21 \Rightarrow F = 7 \]

Step 4: Sign of f:
A concave mirror has negative focal length in the standard convention, so $f = -7$ cm.

Step 5: Test with Numbers:
Object at 9 cm: $v = 31.5$ cm, magnification size 3.5. Object at 15 cm: $v = 13.125$ cm, magnification size 0.875. The ratio is 1/4, so the answer holds.

Final Answer:
\[\boxed{f = -7\ \text{cm (option 2)}}\]