1. First refraction in L1 (lens 1):

When considering the first lens (L₁), the object is at infinity, so the image I₁ formed by this lens is at its focal point. Using the lensmaker's equation:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Given that the object is at infinity, $$d_o = \infty$$, and the focal length of the first lens is $$f = 20 \mathrm{~cm}$$. Plugging in these values, we get:

$$\frac{1}{20} = \frac{1}{\infty} + \frac{1}{d_i}$$

As $$\frac{1}{\infty}$$ is essentially 0, we have:

$$\frac{1}{20} = \frac{1}{d_i}$$

This implies that $$d_i = 20 \mathrm{~cm}$$, meaning that the image I₁ is formed 20 cm from the first lens L₁.

2. Second refraction in L2 (lens 2):

Now, the image I₁ formed by L₁ becomes the object for L₂. The distance between the lenses is 60 cm, so the object distance for L₂ (u) is -40 cm, because the object is to the left of the lens (u is negative). The focal length of L₂ (f) is 20 cm. We can use the lensmaker's equation to find the image distance (v) for L₂:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Plugging in the values:

$$\frac{1}{v} - \frac{1}{(-40)} = \frac{1}{20}$$

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40}$$

$$\frac{1}{v} = \frac{1}{40}$$

Therefore, $$v = 40 \mathrm{~cm}$$

Since the image distance (v) is positive, the image I₂ is formed on the right side of L₂ at a distance of 40 cm. To find the distance of the final image from L₁, we add the distance between the lenses (60 cm) and the image distance (v) from L₂.

Final image distance from L₁ = 60 cm + 40 cm = 100 cm

Thus, the correct answer is 100 cm.