A point object is positioned at a distance of 15 cm from a convex...

For the convex lens, the object distance is $u = -15 \text{ cm}$ and the image distance is $v = +30 \text{ cm}$.

Using the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_1}$:

$\dfrac{1}{30} - \dfrac{1}{-15} = \dfrac{1}{f_1}$

$\dfrac{1}{f_1} = \dfrac{1}{30} + \dfrac{2}{30} = \dfrac{3}{30} = \dfrac{1}{10}$

$f_1 = 10 \text{ cm}$

When the concave lens is placed in contact, the image shifts further away by $30 \text{ cm}$. The new image distance is $v' = 30 + 30 = +60 \text{ cm}$.

Using the lens formula for the combination:

$\dfrac{1}{v'} - \dfrac{1}{u} = \dfrac{1}{F}$

$\dfrac{1}{60} - \dfrac{1}{-15} = \dfrac{1}{F}$

$\dfrac{1}{F} = \dfrac{1}{60} + \dfrac{4}{60} = \dfrac{5}{60} = \dfrac{1}{12}$

$F = 12 \text{ cm}$

The equivalent focal length of two thin lenses in contact is given by:

$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

$\dfrac{1}{12} = \dfrac{1}{10} + \dfrac{1}{f_2}$

$\dfrac{1}{f_2} = \dfrac{1}{12} - \dfrac{1}{10} = \dfrac{5 - 6}{60} = -\dfrac{1}{60}$

$f_2 = -60 \text{ cm}$

The focal length of the convex lens is $10 \text{ cm}$ and the focal length of the concave lens is $60 \text{ cm}$.