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:
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:
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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.
$F = \frac{m_1 m_2 (u_2 - u_1)}{m_1 - m_2}$.
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.
Updated On: Jul 22, 2026
- $15\text{ cm}$
- $20\text{ cm}$
- $25\text{ cm}$
- $30\text{ cm}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given two different object distances and their corresponding real-image magnifications for a concave mirror.
We need to calculate the focal length of the mirror.
Step 2: Key Formula and Approach:
The formula relating magnification $m$, focal length $f$, and object distance $u$ is:
\[ m = \frac{f}{f - u} \] Since a concave mirror forms real, inverted images, the magnification $m$ must be negative.
We will write equations for both positions using proper sign conventions and solve for the focal length.
Step 3: Detailed Explanation:
• Sign Conventions:
Let the focal length of the concave mirror be $f = -F$ (where $F \gt 0$).
The object distance is measured along the negative axis.
• Case 1: Object at $u_1 = -x$, real image with magnification $m_1 = -3$:
\[ m_1 = \frac{f}{f - u_1} \] \[ -3 = \frac{-F}{-F - (-x)} = \frac{-F}{-F + x} \] \[ 3 = \frac{F}{x - F} \] \[ 3x - 3F = F \implies 3x = 4F \implies x = \frac{4}{3}F \quad \text{--- (Equation 1)} \]
• Case 2: Object at $u_2 = -(x+5)$, real image with magnification $m_2 = -2$:
\[ -2 = \frac{-F}{-F - (-(x+5))} = \frac{-F}{-F + x + 5} \] \[ 2 = \frac{F}{x + 5 - F} \] \[ 2x + 10 - 2F = F \implies 2x + 10 = 3F \quad \text{--- (Equation 2)} \]
• Substitute Equation 1 into Equation 2:
\[ 2\left(\frac{4}{3}F\right) + 10 = 3F \] \[ \frac{8}{3}F + 10 = 3F \] Subtract $\frac{8}{3}F$ from both sides:
\[ 10 = 3F - \frac{8}{3}F = \frac{1}{3}F \] \[ F = 30\text{ cm} \] Thus, the focal length is $f = -30\text{ cm}$.
Step 4: Final Answer:
The focal length of the mirror is $30\text{ cm}$, which corresponds to Option (D).
We are given two different object distances and their corresponding real-image magnifications for a concave mirror.
We need to calculate the focal length of the mirror.
Step 2: Key Formula and Approach:
The formula relating magnification $m$, focal length $f$, and object distance $u$ is:
\[ m = \frac{f}{f - u} \] Since a concave mirror forms real, inverted images, the magnification $m$ must be negative.
We will write equations for both positions using proper sign conventions and solve for the focal length.
Step 3: Detailed Explanation:
• Sign Conventions:
Let the focal length of the concave mirror be $f = -F$ (where $F \gt 0$).
The object distance is measured along the negative axis.
• Case 1: Object at $u_1 = -x$, real image with magnification $m_1 = -3$:
\[ m_1 = \frac{f}{f - u_1} \] \[ -3 = \frac{-F}{-F - (-x)} = \frac{-F}{-F + x} \] \[ 3 = \frac{F}{x - F} \] \[ 3x - 3F = F \implies 3x = 4F \implies x = \frac{4}{3}F \quad \text{--- (Equation 1)} \]
• Case 2: Object at $u_2 = -(x+5)$, real image with magnification $m_2 = -2$:
\[ -2 = \frac{-F}{-F - (-(x+5))} = \frac{-F}{-F + x + 5} \] \[ 2 = \frac{F}{x + 5 - F} \] \[ 2x + 10 - 2F = F \implies 2x + 10 = 3F \quad \text{--- (Equation 2)} \]
• Substitute Equation 1 into Equation 2:
\[ 2\left(\frac{4}{3}F\right) + 10 = 3F \] \[ \frac{8}{3}F + 10 = 3F \] Subtract $\frac{8}{3}F$ from both sides:
\[ 10 = 3F - \frac{8}{3}F = \frac{1}{3}F \] \[ F = 30\text{ cm} \] Thus, the focal length is $f = -30\text{ cm}$.
Step 4: Final Answer:
The focal length of the mirror is $30\text{ cm}$, which corresponds to Option (D).
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