Question:
A convex lens forms a real image \(4\,\text{cm}\) long on a screen. When the lens is shifted to a new position without disturbing the object, again a real image is formed on the screen which is \(16\,\text{cm}\) tall. The length of the object must be
A convex lens forms a real image \(4\,\text{cm}\) long on a screen. When the lens is shifted to a new position without disturbing the object, again a real image is formed on the screen which is \(16\,\text{cm}\) tall. The length of the object must be
Show Hint
In the lens displacement method,
\[
m_1=\frac{v_1}{u_1},
\qquad
m_2=\frac{v_2}{u_2},
\]
and the two magnifications satisfy
\[
m_1m_2=1.
\]
Hence,
\[
\left(\frac{h_{i1}}{h_o}\right)
\left(\frac{h_{i2}}{h_o}\right)=1.
\]
This provides a quick way to determine the object height.
Updated On: Jul 9, 2026
- \(4\,\text{cm}\)
- \(8\,\text{cm}\)
- \(12\,\text{cm}\)
- \(20\,\text{cm}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
In the displacement method of a convex lens, for the two lens positions producing sharp images on the same screen, the magnifications are reciprocals of each other.
Thus,
\[
m_1m_2=1.
\]
Also,
\[
m=\frac{\text{Image height}}{\text{Object height}}.
\]
Step 1: Let the object height be \(h\). For the first position, \[ m_1=\frac{4}{h}. \] For the second position, \[ m_2=\frac{16}{h}. \]
Step 2: Use the reciprocal magnification property. \[ m_1m_2=1. \] Therefore, \[ \frac{4}{h}\cdot\frac{16}{h}=1. \] \[ \frac{64}{h^2}=1. \] \[ h^2=64. \] \[ h=8\,\text{cm}. \]
Step 3: Write the final answer. \[ \boxed{h=8\,\text{cm}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Let the object height be \(h\). For the first position, \[ m_1=\frac{4}{h}. \] For the second position, \[ m_2=\frac{16}{h}. \]
Step 2: Use the reciprocal magnification property. \[ m_1m_2=1. \] Therefore, \[ \frac{4}{h}\cdot\frac{16}{h}=1. \] \[ \frac{64}{h^2}=1. \] \[ h^2=64. \] \[ h=8\,\text{cm}. \]
Step 3: Write the final answer. \[ \boxed{h=8\,\text{cm}} \] \[ \boxed{\text{Answer = (B)}} \]
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