Question

A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?

Hint:

In a system of two lenses, the image formed by the first lens acts as the object for the second lens, and the distances must be carefully calculated using the lens formula for both lenses.

Answer 1

Step 1: Understanding the setup.
Let the diverging lens be \( L_1 \) and the converging lens be \( L_2 \). The focal lengths of \( L_1 \) and \( L_2 \) are \( f_1 = -20 \, \text{cm} \) (diverging lens) and \( f_2 = +30 \, \text{cm} \) (converging lens). The distance between the two lenses is \( d = 15 \, \text{cm} \). The object should be placed in such a way that the rays coming out of \( L_1 \) (diverging lens) and converging at \( L_2 \) form an image at infinity.
Step 2: Using the lens formula.
The lens formula for a single lens is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where: - \( f \) is the focal length of the lens, - \( v \) is the image distance, - \( u \) is the object distance. For the diverging lens \( L_1 \), the image formed will be virtual, and the rays will diverge, but the image formed by \( L_1 \) will act as a virtual object for \( L_2 \). Let the object distance for \( L_1 \) be \( u_1 \), and the image distance for \( L_1 \) be \( v_1 \). Then, the object distance for \( L_2 \) is: \[ u_2 = d - v_1 \] Now, applying the lens formula for \( L_1 \) and \( L_2 \), we can calculate \( u_1 \) to find the position where the object should be placed.
Step 3: Conclusion.
After solving the lens equations, the object should be placed at \( u_1 \approx 60 \, \text{cm} \) from the diverging lens \( L_1 \) so that the image is formed at infinity.