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 the image formation by lens P first. The image formed by lens P then serves as the object for lens Q. The final position and magnification are determined by the lens formula and magnification formula applied sequentially.
: 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 2: Detailed Explanation:
For Lens P:
\( u_1 = -15 \text{ cm}, f_1 = +10 \text{ cm} \).
\[ \frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} \implies v_1 = +30 \text{ cm} \).
The image \( I_1 \) is formed 30 cm to the right of lens P.
Magnification \( m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2 \).
For Lens Q:
Lens Q is placed 15 cm to the right of lens P.
The image \( I_1 \) (which is 30 cm right of P) acts as the object for Q.
Its position relative to Q is \( u_2 = 30 - 15 = 15 \text{ cm} \) to the right of Q.
Since the rays were converging toward this point, it is a virtual object for Q: \( u_2 = +15 \text{ cm} \).
Given \( f_2 = +15 \text{ cm} \):
\[ \frac{1}{v_2} - \frac{1}{+15} = \frac{1}{15} \implies \frac{1}{v_2} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \implies v_2 = +7.5 \text{ cm} \).
The final image is real (since \( v_2>0 \)) and 7.5 cm to the right of Q.
Magnification \( m_2 = \frac{v_2}{u_2} = \frac{7.5}{15} = +0.5 \).
Total Magnification \( M = m_1 \times m_2 = (-2) \times (0.5) = -1 \).
Since \( |M| = 1 \), the size of the final image is the same as that of object AB.
Step 3: Final Answer:
The final image is real, formed at 7.5 cm right of lens Q, with a size same as that of AB.