The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is _______ cm.

For a thin lens we use the Cartesian sign convention: distances measured to the left of the lens are negative, those to the right are positive.
  • Object positions: $$u_1 = -8\text{ cm}$$ and $$u_2 = -24\text{ cm}$$ (both on the left of the lens).
  • Corresponding image distances are $$v_1$$ and $$v_2$$ (to be found).
  • Lens formula: $$\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$.

The images are stated to be of equal size, so their magnifications have equal magnitudes.
  Magnification $$m = \dfrac{v}{u}$$, hence
$$\left|\dfrac{v_1}{u_1}\right| = \left|\dfrac{v_2}{u_2}\right|$$.

Because one image is erect (virtual) and the other inverted (real), the two magnifications have opposite signs.
Therefore
$$\dfrac{v_1}{u_1} = -\,\dfrac{v_2}{u_2}$$.

Substituting $$u_1 = -8$$ and $$u_2 = -24$$ gives
$$\dfrac{v_1}{-8} = -\,\dfrac{v_2}{-24} \quad\Longrightarrow\quad v_2 = -3\,v_1$$ $$-(1)$$.

Apply the lens formula to each object position.

For $$u_1 = -8\text{ cm}$$:
$$\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{v_1} - \frac{1}{-8} = \frac{1}{v_1} + \frac{1}{8}$$ $$-(2)$$.

For $$u_2 = -24\text{ cm}$$:
$$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{v_2} - \frac{1}{-24} = \frac{1}{v_2} + \frac{1}{24}$$ $$-(3)$$.

Equate the right-hand sides of $$(2)$$ and $$(3)$$ and substitute $$v_2 = -3v_1$$ from $$(1)$$:

$$\frac{1}{v_1} + \frac{1}{8} = \frac{1}{-3v_1} + \frac{1}{24}$$
Re-arrange:
$$\frac{1}{v_1} + \frac{1}{3v_1} = \frac{1}{24} - \frac{1}{8}$$
$$\frac{4}{3v_1} = -\frac{1}{12}$$
$$v_1 = -16\text{ cm}$$.

Insert $$v_1 = -16\text{ cm}$$ in $$(2)$$ to obtain the focal length:

$$\frac{1}{f} = \frac{1}{-16} + \frac{1}{8} = -\frac{1}{16} + \frac{2}{16} = \frac{1}{16}$$
$$\therefore \; f = 16\text{ cm}$$.

The positive sign of $$f$$ shows the lens is converging (convex).
Hence the focal length of the lens is 16 cm.