A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is _____.

For two thin lenses kept in contact, the equivalent focal length $$F$$ is given by the formula
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$

The convex lens has $$f_1 = +30\ \text{cm}$$ and the concave lens has $$f_2 = -20\ \text{cm}$$ (negative because it is concave).

Substituting,
$$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = \frac{2}{60} - \frac{3}{60} = -\frac{1}{60}$$
Hence,
$$F = -60\ \text{cm}$$

The combination therefore behaves like a single concave lens of focal length $$60\ \text{cm}$$.

Using the lens formula for a thin lens,
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad -(1)$$
where
• $$u$$ = object distance (negative when the object is on the left of the lens)
• $$v$$ = image distance (positive to the right, negative to the left)
• $$F$$ = focal length of the equivalent lens

The object is placed $$20\ \text{cm}$$ to the left, so $$u = -20\ \text{cm}$$. The focal length is $$F = -60\ \text{cm}$$. Substituting in $$(1)$$:

$$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60}$$
$$\frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$$

Move the second term to the right side:
$$\frac{1}{v} = -\frac{1}{60} - \frac{1}{20}$$

Convert to a common denominator of $$60$$:
$$\frac{1}{v} = -\frac{1}{60} - \frac{3}{60} = -\frac{4}{60} = -\frac{1}{15}$$

Therefore,
$$v = -15\ \text{cm}$$

The negative sign shows the image forms on the same side as the object (to the left of the lens system). The distance from the lens is the magnitude:

Image distance = $$15\ \text{cm}$$.

Hence, the correct option is Option D (15 cm).