Question

The lens combination as shown consists of two thin lenses \(L_1\) and \(L_2\) of focal lengths \(+10\ \text{cm}\) and \(-10\ \text{cm}\), respectively. The object is placed \(30\ \text{cm}\) to the left of \(L_1\), and the distance between the two lenses is \(3\ \text{cm}\). The position of the image formed is:

• A. 60 cm to the right of the concave lens
• B. 20 cm to the left of the concave lens
• C. 60 cm to the left of the concave lens
• D. 30 cm to the right of the concave lens

Hint:

In multi-lens systems, always solve one lens at a time. The image formed by the first lens becomes the object for the second lens.

Answer 1

The Correct Option is C

Concept:

• Use the lens formula successively for both lenses.

• The image formed by the first lens acts as the object for the second lens.

• Sign convention must be applied carefully.

Step 1: Find image formed by the convex lens \(L_1\)
Given, \[ f_1=+10\ \text{cm} \] \[ u_1=-30\ \text{cm} \] Using lens formula, \[ \frac{1}{f} = \frac{1}{v} -\frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v_1} +\frac{1}{30} \] \[ \frac{1}{v_1} = \frac{1}{10} -\frac{1}{30} = \frac{1}{15} \] \[ v_1=15\ \text{cm} \] Thus the first image is formed \(15\ \text{cm}\) to the right of \(L_1\).

Step 2: Locate this image with respect to \(L_2\)
Distance between lenses \[ =3\ \text{cm} \] Therefore image formed by \(L_1\) lies \[ 15-3=12\ \text{cm} \] to the right of \(L_2\). Hence for \(L_2\), \[ u_2=+12\ \text{cm} \] (virtual object).

Step 3: Apply lens formula for the concave lens
\[ f_2=-10\ \text{cm} \] Using \[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \] \[ -\frac{1}{10} = \frac{1}{v_2} - \frac{1}{12} \] \[ \frac{1}{v_2} = -\frac{1}{10} +\frac{1}{12} \] \[ = -\frac{1}{60} \] Therefore, \[ v_2=-60\ \text{cm} \]

Step 4: Interpret the sign
Negative sign indicates that the image lies to the left of the concave lens. Hence, \[ \boxed{\text{Image position }=60\ \text{cm to the left of the concave lens}} \] Therefore, \[ \boxed{\text{Option (C)}} \]