Solution for Part (a)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The focal length of a lens is given by the Lens Maker's Formula:
$$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$For the plano-convex lens ($L_1$): $R_1 = R$, $R_2 = \infty$.
For the biconcave lens ($L_2$): $R_1 = -R$, $R_2 = R$.
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Calculating focal length for $L_1$ ($f_1$):
$$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu_1 - 1}{R}$$Calculating focal length for $L_2$ ($f_2$):
$$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = (\mu_2 - 1) \left( -\frac{2}{R} \right) = -\frac{2(\mu_2 - 1)}{R}$$For lenses in contact, the equivalent focal length $F$ is:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$ $$\frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{2(\mu_2 - 1)}{R}$$Step 3 (Final Concluded Answer): [0.5 Mark]
$$\frac{1}{F} = \frac{\mu_1 - 2\mu_2 + 1}{R} \implies F = \frac{R}{\mu_1 - 2\mu_2 + 1}$$💡 EL10 Concept Insight: Think of the lenses as "light benders." The first lens bends light inward to converge it, while the second lens (being concave) acts like a spreader; by adding their individual "bending powers" (reciprocal of focal length), we find the net effect on the light beam.
Solution for Part (b)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Given: $u_1 = -2f$, $f_1 = f$, distance between lens and mirror $d = 5f$.
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
For the convex lens:
$$\frac{1}{v_1} - \frac{1}{-2f} = \frac{1}{f} \implies \frac{1}{v_1} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f} \implies v_1 = 2f$$The image is formed $2f$ to the right of the lens. Since the mirror