Solution for (b)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
For a combination of thin lenses placed in contact, the equivalent focal length $F$ is given by the formula:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$$Given data (using sign convention):
- $f_1 = +10\text{ cm}$ (Convex lens)
- $f_2 = -15\text{ cm}$ (Concave lens)
- $F = +12\text{ cm}$ (Combination)
- $f_3 = ?$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Substituting the values into the combination formula:
$$\frac{1}{12} = \frac{1}{10} + \frac{1}{-15} + \frac{1}{f_3}$$Calculating the sum of the first two terms:
$$\frac{1}{12} = \left( \frac{3 - 2}{30} \right) + \frac{1}{f_3}$$ $$\frac{1}{12} = \frac{1}{30} + \frac{1}{f_3}$$Isolating $\frac{1}{f_3}$:
$$\frac{1}{f_3} = \frac{1}{12} - \frac{1}{30}$$ $$\frac{1}{f_3} = \frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20}$$Step 3 (Final Concluded Answer): [0.5 Mark]
The focal length of the third lens is $f_3 = +20\text{ cm}$. Since the focal length is positive, the third lens is a convex lens.
Effect of thickness: Yes, the answer would change if the lenses were thick. The formula for the combination of thin lenses assumes the distance between them is negligible; for thick lenses, the separation distance and the principal planes must be considered, making the effective focal length different.
💡 EL10 Concept Insight: Think of lenses like filters for light; adding them together is like combining their powers. Because the third lens has a positive focal length, it acts as a "converging" lens that helps bring the light rays together to reach the desired total focus of $12\text{ cm}$.