(a) (i) Compound Microscope: Image Formation and Magnification
Ray Diagram Requirements: The diagram must show the objective lens forming a real, inverted, and magnified image ($A'B'$) of the object ($AB$). This image acts as an object for the eyepiece, which forms the final virtual, inverted, and highly magnified image at infinity.
Step 1 (Governing Formula): [1 Mark]
The total magnification $M$ of a compound microscope is the product of the linear magnification of the objective ($m_o$) and the angular magnification of the eyepiece ($m_e$):
$$M = m_o \times m_e$$
For final image at infinity (normal adjustment), $m_o = \frac{v_o}{u_o}$ and $m_e = \frac{D}{f_e}$.
Step 2 (Derivation): [1 Mark]
Since the image $A'B'$ is formed very close to the focal point of the eyepiece ($f_e$), $v_o \approx L$ (tube length) and $u_o \approx f_o$.
Thus, $m_o = \frac{L}{f_o}$.
Substituting into the total magnification formula:
$$M = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right)$$
Where $D = 25\text{ cm}$ (least distance of distinct vision).
Step 3 (Numerical Calculation): [1 Mark]
Given: $u_o = -1.5\text{ cm}$, $f_o = 1.25\text{ cm}$, $f_e = 5\text{ cm}$.
Using lens formula for objective: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$
$$\frac{1}{v_o} = \frac{1}{1.25} + \frac{1}{-1.5} = \frac{1}{1.25} - \frac{1}{1.5} = \frac{6-5}{7.5} = \frac{1}{7.5}$$
$$v_o = 7.5\text{ cm}$$
For final image at infinity, the intermediate image must be at the focus of the eyepiece, so $u_e = f_e = 5\text{ cm}$.
Total tube length $L = v_o + u_e = 7.5 + 5 = 12.5\text{ cm}$.
💡 EL10 Concept Insight: Think of the objective lens as a "magnifying glass" that creates a zoomed-in picture, and the eyepiece as a second "magnifying glass" that zooms in even further on that picture so your eye can see the tiny details clearly.
(b) (i) Refraction using Huygens' Principle
Step 1 (Governing Principle): [1.5 Marks]
According to Huygens' principle, every point on a wavefront acts as a source of secondary wavelets. When a plane wavefront $AB$ hits the interface $XY$, the secondary wavelets from $A$ travel in glass with speed $v_2$ while wavelets from $B$ travel in air with speed $v_1$.
Step 2 (Mathematical Verification): [1.5 Marks]
In time $t$, distance $BC = v_1 t$ and $AE = v_2 t$.
In $\triangle ABC$, $\sin i = \frac{BC}{AC} = \frac{v