Question

A convex lens forms a real image of an object with magnification \(m_1\). The lens is moved towards the object to obtain another real image of magnification \(m_2\). The image distance is increased by \(x\). The focal length of the lens is

• A. \( \left(\frac{m_2}{m_1}\right)x \)
• B. \( \left(\frac{m_1}{m_2}\right)x \)
• C. \( \frac{x}{m_2 - m_1} \)
• D. \( x(m_2 - m_1) \)

Hint:

For lens shift problems, combine \(m = v/u\) with lens formula and relate change in image distance directly.

Answer 1

The Correct Option is C

Step 1: Magnification relation.
For a lens, magnification is:
\[ m = \frac{v}{u} \]
Thus, \(v = mu\).

Step 2: Lens formula.
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substitute \(v = mu\):
\[ \frac{1}{f} = \frac{1}{mu} + \frac{1}{u} \]
\[ \frac{1}{f} = \frac{1 + m}{mu} \]

Step 3: Write for two positions.
For first position:
\[ \frac{1}{f} = \frac{1 + m_1}{m_1 u_1} \]
For second position:
\[ \frac{1}{f} = \frac{1 + m_2}{m_2 u_2} \]

Step 4: Image distance difference.
Given image distance increases by \(x\):
\[ v_2 - v_1 = x \]
\[ m_2 u_2 - m_1 u_1 = x \]

Step 5: Using lens relation consistency.
Since both expressions equal \(\frac{1}{f}\), simplifying gives relation between \(u_1\) and \(u_2\), leading to:
\[ x = f(m_2 - m_1) \]

Step 6: Solve for focal length.
\[ f = \frac{x}{m_2 - m_1} \]

Step 7: Final conclusion.
\[ \boxed{\frac{x}{m_2 - m_1}} \] Hence, correct answer is option (C).