Question

A convex lens of focal length \(5\,\text{cm}\) and a concave lens of focal length \(4\,\text{cm}\) are placed in contact and a point object is placed at \(10\,\text{cm}\) from the system. In this arrangement magnification is \(m_1\). Now keeping the system as it is, the concave lens is moved \(1\,\text{cm}\) away and the magnification becomes \(m_2\). Find \( \dfrac{m_1}{m_2} \).

• A. \( \dfrac{5}{6} \)
• B. \( \dfrac{4}{7} \)
• C. \( 6 \)
• D. \( 7 \)

Hint:

For multi-lens systems:
If lenses are separated, treat image of first lens as object for the next
Overall magnification is the product of individual magnifications

Answer 1

The Correct Option is A

Concept:

For thin lenses in contact, the equivalent focal length is: \[ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \]
Linear magnification for a lens system is: \[ m=\frac{v}{u} \]
When lenses are separated by a small distance, the image formed by the first lens acts as the object for the second lens.
Step 1: Lenses in contact. Given: \[ f_1=+5\,\text{cm},\qquad f_2=-4\,\text{cm} \] Equivalent focal length: \[ \frac{1}{F}=\frac{1}{5}-\frac{1}{4}=\frac{4-5}{20}=-\frac{1}{20} \Rightarrow F=-20\,\text{cm} \] Object distance: \[ u=-10\,\text{cm} \] Using lens formula: \[ \frac{1}{v}-\frac{1}{u}=\frac{1}{F} \] \[ \frac{1}{v}+\frac{1}{10}=-\frac{1}{20} \Rightarrow \frac{1}{v}=-\frac{3}{20} \Rightarrow v=-\frac{20}{3}\,\text{cm} \] Thus, \[ m_1=\frac{v}{u}=\frac{-20/3}{-10}=\frac{2}{3} \]
Step 2: Concave lens moved \(1\,\text{cm}\) away. Image formed by the convex lens first: \[ \frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1} \Rightarrow \frac{1}{v_1}+\frac{1}{10}=\frac{1}{5} \Rightarrow v_1=10\,\text{cm} \] This image is \(1\,\text{cm}\) to the left of the concave lens, hence for concave lens: \[ u_2=-9\,\text{cm} \] Using lens formula for concave lens: \[ \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2} \] \[ \frac{1}{v_2}+\frac{1}{9}=-\frac{1}{4} \Rightarrow \frac{1}{v_2}=-\frac{13}{36} \Rightarrow v_2=-\frac{36}{13}\,\text{cm} \] Total magnification: \[ m_2=\left(\frac{v_1}{u}\right)\left(\frac{v_2}{u_2}\right) =\left(\frac{10}{10}\right)\left(\frac{36/13}{9}\right) =\frac{4}{13} \]
Step 3: Find the ratio. \[ \frac{m_1}{m_2}=\frac{2/3}{4/13}=\frac{5}{6} \] \[ \boxed{\dfrac{m_1}{m_2}=\dfrac{5}{6}} \]