The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is ______

A convex lens forms images of the same size when the object is placed at 20 cm or at 10 cm from the lens. We need to find the focal length $$f$$.

The thin lens equation is $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$, where $$u$$ is the object distance (negative for a real object) and $$v$$ is the image distance. The linear magnification is $$m = \frac{v}{u}$$.

Case 1: Object at $$u_1 = -20\,\text{cm}$$. Using the lens equation: $$\frac{1}{v_1} = \frac{1}{f} + \frac{1}{u_1} = \frac{1}{f} - \frac{1}{20}$$. This gives $$v_1 = \frac{20f}{20 - f}$$. The magnification is $$m_1 = \frac{v_1}{u_1} = \frac{20f/(20 - f)}{-20} = \frac{-f}{20 - f}$$. Since we expect $$f < 20$$, this object is beyond the focal point, and $$v_1$$ is positive (real image). The magnification $$m_1$$ is negative (inverted image), and $$|m_1| = \frac{f}{20 - f}$$.

Case 2: Object at $$u_2 = -10\,\text{cm}$$. Using the lens equation: $$\frac{1}{v_2} = \frac{1}{f} + \frac{1}{u_2} = \frac{1}{f} - \frac{1}{10}$$. This gives $$v_2 = \frac{10f}{10 - f}$$. The magnification is $$m_2 = \frac{v_2}{u_2} = \frac{10f/(10 - f)}{-10} = \frac{-f}{10 - f}$$. If $$f > 10$$, then $$10 - f < 0$$, so $$v_2 < 0$$ (virtual image) and $$m_2 > 0$$ (erect image). The magnitude is $$|m_2| = \frac{f}{f - 10}$$.

For the images to be of the same size: $$|m_1| = |m_2|$$. Substituting: $$\frac{f}{20 - f} = \frac{f}{f - 10}$$. Since $$f \neq 0$$, we can cancel $$f$$ from both sides to get $$\frac{1}{20 - f} = \frac{1}{f - 10}$$.

Cross-multiplying: $$f - 10 = 20 - f$$, which gives $$2f = 30$$, so $$f = 15\,\text{cm}$$.

Verification: For $$u_1 = -20\,\text{cm}$$, $$|m_1| = \frac{15}{20 - 15} = \frac{15}{5} = 3$$. For $$u_2 = -10\,\text{cm}$$, $$|m_2| = \frac{15}{15 - 10} = \frac{15}{5} = 3$$. Both magnifications are equal, confirming equal image sizes.

Therefore, the focal length of the convex lens is $$15$$ cm.