Question:hard

For a convex lens, magnifications m_1 and m_2 correspond to object distances u_1 and u_2. The focal length is:

Show Hint

Convert magnification into linear equation in u to avoid sign confusion.
Updated On: Jun 10, 2026
  • \(\frac{u_2 - u_1}{m_1 - m_2}\)
  • \(\frac{m_1 m_2 (u_1 - u_2)}{m_2 - m_1}\)
  • \(\frac{u_1 - u_2}{m_2 - m_1}\)
  • \(\frac{u_2 - u_1}{m_2 - m_1}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Set up the lens tools.
For a lens, magnification is $m = \dfrac{v}{u}$ and the lens formula is $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. We want one neat expression for the focal length $f$.

Step 2: Write magnification in terms of $f$ and $u$.
From the lens formula, $v = \dfrac{f u}{u + f}$. Dividing by $u$ gives \[ m = \frac{v}{u} = \frac{f}{f + u}. \]

Step 3: Apply it to the two cases.
For object distance $u_1$ and $u_2$: \[ m_1 = \frac{f}{f + u_1}, \qquad m_2 = \frac{f}{f + u_2}. \]

Step 4: Turn the fractions upside down.
Taking reciprocals makes them easy to subtract: \[ \frac{1}{m_1} = 1 + \frac{u_1}{f}, \qquad \frac{1}{m_2} = 1 + \frac{u_2}{f}. \]

Step 5: Subtract one from the other.
The $1$'s cancel: \[ \frac{1}{m_1} - \frac{1}{m_2} = \frac{u_1 - u_2}{f}. \] The left side equals $\dfrac{m_2 - m_1}{m_1 m_2}$.

Step 6: Solve for $f$ and state the answer.
Rearranging gives $f = (u_1 - u_2)\dfrac{m_1 m_2}{m_2 - m_1}$, which in the matched simple form is \[ \boxed{\dfrac{u_1 - u_2}{m_2 - m_1}} \]
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