A diverging lens having a focal length of 20 ext cm and a...

Considering the case where light passes through the diverging lens first, then the converging lens.

For the final image to be at infinity, the image formed by the diverging lens must lie at the first principal focus of the converging lens.

The focal length of the converging lens is $f_2 = +30 \text{ cm}$.

The focus of the converging lens is $30 \text{ cm}$ in front of it. Since the diverging lens is placed $15 \text{ cm}$ in front of the converging lens, the image formed by the diverging lens must be at $30 - 15 = 15 \text{ cm}$ in front of the diverging lens.

For the diverging lens, $v_1 = -15 \text{ cm}$ and $f_1 = -20 \text{ cm}$.

Using the lens formula:

$\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

$\dfrac{1}{-15} - \dfrac{1}{u_1} = \dfrac{1}{-20}$

$\dfrac{1}{u_1} = \dfrac{1}{20} - \dfrac{1}{15} = \dfrac{3 - 4}{60} = -\dfrac{1}{60}$

$u_1 = -60 \text{ cm}$

The object must be placed $60 \text{ cm}$ from the diverging lens.

Considering the case where light passes through the converging lens first, then the diverging lens.

For the final image to be at infinity, the image formed by the converging lens must act as a virtual object at the focus of the diverging lens.

The focal length of the diverging lens is $f_2 = -20 \text{ cm}$.

The focus is $20 \text{ cm}$ behind the diverging lens. Since the diverging lens is $15 \text{ cm}$ behind the converging lens, the image formed by the converging lens must be at $15 + 20 = 35 \text{ cm}$ behind the converging lens.

For the converging lens, $v_1 = +35 \text{ cm}$ and $f_1 = +30 \text{ cm}$.

Using the lens formula:

$\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

$\dfrac{1}{35} - \dfrac{1}{u_1} = \dfrac{1}{30}$

$\dfrac{1}{u_1} = \dfrac{1}{35} - \dfrac{1}{30} = \dfrac{6 - 7}{210} = -\dfrac{1}{210}$

$u_1 = -210 \text{ cm}$

The object must be placed $210 \text{ cm}$ from the converging lens.

Answer: $60 ext{ cm}$ from the diverging lens; $210 ext{ cm}$ from the converging lens