Question

Find the distance of the final image from the second lens. Given \(f_1 = 10\,\text{cm}\), \(f_2 = 15\,\text{cm}\). The object is placed \(15\,\text{cm}\) to the left of the first lens and the separation between the lenses is \(15\,\text{cm}\). 

• A. \(15\,\text{cm}\)
• B. \(\infty\)
• C. \(7.5\,\text{cm}\)
• D. \(30\,\text{cm}\)

Hint:

In multiple lens problems, always solve sequentially: 1. Find image from the first lens. 2. Treat that image as the object for the second lens. 3. Apply the lens formula again.

Answer 1

The Correct Option is C

To find the distance of the final image from the second lens, we will use the lens formula for each lens individually and take into account the separation between the lenses.

The lens formula 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.

 

Let's solve this step-by-step for both lenses:

Step 1: Find the image distance for the first lens.

The object is placed 15 cm to the left of the first lens, so \(u_1 = -15 \, \text{cm}\) and \(f_1 = 10 \, \text{cm}\).

Using the lens formula:

\(\frac{1}{10} = \frac{1}{v_1} - \frac{1}{-15}\)

Simplifying,

\(\frac{1}{v_1} = \frac{1}{10} - \frac{1}{15}\)

\(\frac{1}{v_1} = \frac{3 - 2}{30}\)

\(\frac{1}{v_1} = \frac{1}{30}\)

Thus, \(v_1 = 30 \, \text{cm}\).

Step 2: Consider the position of the image from the second lens.

The image formed by the first lens acts as the object for the second lens. The distance of this image from the second lens is:

\(s = v_1 - \text{distance between the lenses} = 30 \, \text{cm} - 15 \, \text{cm} = 15 \, \text{cm}\).

Step 3: Find the final image distance from the second lens.

For the second lens, the object distance \(u_2 = -15 \, \text{cm}\) (since the object is on the left, we consider it as negative) and \(f_2 = 15 \, \text{cm}\).

Using the lens formula again:

\(\frac{1}{15} = \frac{1}{v_2} - \frac{1}{-15}\)

Simplifying,

\(\frac{1}{v_2} = \frac{1}{15} + \frac{1}{15}\)

\(\frac{1}{v_2} = \frac{2}{15}\)

\(v_2 = \frac{15}{2} = 7.5 \, \text{cm}\)

Therefore, the distance of the final image from the second lens is \(7.5 \, \text{cm}\).