Question

An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is _____

• A. virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB
• B. real, formed at 7.5 cm right of lens Q, with a size same as that of AB
• C. formed at infinity
• D. real, formed at 7 cm right of lens Q, with a size smaller than that of AB

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We analyze image formation by lens \(P\) first. The image formed by lens \(P\) acts as the object for lens \(Q\). The final position and magnification are obtained by applying lens and magnification formulas sequentially.

Step 2: Key Formula or Approach:
Lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Magnification: \[ m = \frac{v}{u} \] Total magnification: \[ M = m_1 \times m_2 \] 
Step 3: Detailed Explanation:
For Lens \(P\):
\[ u_1 = -15 \text{ cm}, \quad f_1 = +10 \text{ cm} \] \[ \frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \Rightarrow \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \] \[ v_1 = +30 \text{ cm} \] Image \(I_1\) is formed 30 cm to the right of lens \(P\).
\[ m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2 \] 
For Lens \(Q\):
Lens \(Q\) is 15 cm to the right of lens \(P\).
Object distance for \(Q\): \[ u_2 = 30 - 15 = +15 \text{ cm} \] (This is a virtual object for lens \(Q\)).
\[ f_2 = +15 \text{ cm} \] \[ \frac{1}{v_2} - \frac{1}{15} = \frac{1}{15} \Rightarrow \frac{1}{v_2} = \frac{2}{15} \] \[ v_2 = +7.5 \text{ cm} \] Final image is real and formed 7.5 cm to the right of lens \(Q\).
\[ m_2 = \frac{v_2}{u_2} = \frac{7.5}{15} = 0.5 \] 
Total Magnification:
\[ M = m_1 \times m_2 = (-2)(0.5) = -1 \] \[ |M| = 1 \] So, the image size is equal to the object size.

Step 4: Final Answer:
The final image is real, formed 7.5 cm to the right of lens \(Q\), and is of the same size as the object.