Question

An object placed in front of a concave mirror at a distance of $x\text{ cm}$ from the pole gives a 3 times magnified real image. If it is moved to a distance of $(x + 5)\text{ cm}$, the magnification of the image becomes 2. The focal length of the mirror is:

• A. $15\text{ cm}$
• B. $20\text{ cm}$
• C. $25\text{ cm}$
• D. $30\text{ cm}$

Hint:

For real images formed by a concave mirror:
$F = \frac{m_1 m_2 (u_2 - u_1)}{m_1 - m_2}$.
Here, $u_2 - u_1 = (x+5) - x = 5\text{ cm}$, $m_1 = 3$, and $m_2 = 2$.
Thus, $F = \frac{3 \times 2 \times 5}{3 - 2} = 30\text{ cm}$.
This simple formula avoids setting up systems of linear equations.

Answer 1

The Correct Option is D

Step 1: Build a direct shortcut formula linking f, u and m.
From the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ and $m=\frac{v}{u}$, replacing $v = mu$ gives \[ \frac{1}{f} = \frac{1}{u}\left(1+\frac{1}{m}\right) \quad\Rightarrow\quad f = \frac{mu}{m+1} \]
Step 2: Apply it to the first position. \[ f = \frac{3x}{3+1} = \frac{3x}{4} \]
Step 3: Apply it to the second position and equate. \[ f = \frac{2(x+5)}{2+1} = \frac{2(x+5)}{3} \] Since both expressions equal the same $f$: \[ \frac{3x}{4} = \frac{2(x+5)}{3} \quad\Rightarrow\quad 9x = 8x+40 \quad\Rightarrow\quad x = 40 \]
Step 4: Find the focal length. \[ f = \frac{3(40)}{4} \] \[ \boxed{f = 30\text{ cm}} \]