Given data

• Two convex lenses, each of focal length: $f_1=f_2=20\text{ cm}$
• Separation between lenses: $d=60\text{ cm}$
• Object is at infinity.

Image formed by the first lens

For an object at infinity, a convex lens forms its image at its focal plane.

So, the first lens forms an image at: $v_1=f_1=20\text{ cm}$ to the right of the first lens.

This image acts as object for the second lens

The second lens is 60 cm to the right of the first lens. Since the first image is 20 cm to the right of the first lens, its distance from the second lens is: $60-20=40\text{ cm}$

Thus, for the second lens, the object is 40 cm to its left. So: $u_2=-40\text{ cm}$

Apply lens formula to the second lens

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

For the second lens: $\frac{1}{20}=\frac{1}{v_2}-\frac{1}{(-40)}$ $\frac{1}{20}=\frac{1}{v_2}+\frac{1}{40}$ $\frac{1}{v_2}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$ $v_2=40\text{ cm}$

So the final image is 40 cm to the right of the second lens.

Distance of final image from the first lens

The second lens is 60 cm from the first lens, and the final image is 40 cm beyond the second lens: $60+40=100\text{ cm}$

Final answer The image of the distant object is formed at: $\boxed{100\text{ cm}}$ from the first lens.