OPTION 1: Compound Microscope
Step 1: Why two lenses. A single convex lens (simple microscope) cannot give a very high magnification because a short focal length is hard to grind. A compound microscope solves this by magnifying twice: first with a short-focus \(objective\) and again with the \(eyepiece\). Both are converging lenses fitted in a tube of length \(L\).
Step 2: Path of light. The specimen \(AB\) is fixed a little beyond the objective focus \(F_o\). The objective throws a real, inverted, enlarged image \(A'B'\) inside the tube. This image is deliberately made to lie just within the eyepiece focus, so that the eyepiece treats \(A'B'\) as its object and creates a large virtual image \(A''B''\) seen comfortably by the eye at the near point \(D = 25\) cm.
Step 3: Magnification as a product. Since the object is magnified in two stages, the overall magnifying power is the product of the two stage magnifications:
\[M = m_o \, m_e\]
Step 4: First stage (objective). Using \(m_o = v_o/u_o\) and taking the object almost at \(F_o\) (so \(u_o \approx f_o\)) with the image near the eyepiece end (so \(v_o \approx L\)):
\[m_o \approx \frac{L}{f_o}\]
Step 5: Second stage (eyepiece). The eyepiece is a simple magnifier. Its angular magnification for the image at the near point is
\[m_e = 1 + \frac{D}{f_e}\]
Step 6: Result. Combining the two stages,
\[M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\]
and for a relaxed eye (image at infinity) \(M = \dfrac{L\,D}{f_o f_e}\). Keeping both focal lengths small makes \(M\) very large.
\[\boxed{M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)}\]
OPTION 2: Binding Energy of Nucleus
Step 1: Idea of binding. Protons repel each other electrically, yet nuclei hold together. The strong nuclear force binds the nucleons, and this binding shows up as a loss of mass. The energy tied up in holding the nucleus together is the binding energy.
Step 2: Mass defect and Einstein relation. The nucleons weigh more when free than when bound. The difference is
\[\Delta m = Z m_p + (A-Z)m_n - M_{nucleus}\]
and by \(E = mc^2\) the binding energy is \(B.E. = \Delta m\, c^2\). Per nucleon it is \(B.E./A\), a direct measure of stability.
Step 3: Shape of the \(B.E./A\) vs \(A\) graph. Starting from light nuclei the curve climbs quickly, flattens into a plateau around \(8.5\) to \(8.8\) MeV between mass numbers \(30\) and \(120\) (peak at iron, \(A \approx 56\)), and then droops gently to roughly \(7.6\) MeV for the heaviest nuclei. The top of the hump marks the most stable nuclei.
Step 4: Reading fission off the curve. Heavy nuclei sit on the right, below the peak. Splitting one into two mid-mass fragments moves the products up the curve to a higher \(B.E./A\). Since the fragments are more tightly bound, the surplus energy (about \(200\) MeV) is liberated. This is nuclear fission.
Step 5: Reading fusion off the curve. Light nuclei sit on the far left, far below the peak. Joining them into a heavier nucleus climbs the steep left side of the curve to a much higher \(B.E./A\), so a large amount of energy is released. This is nuclear fusion, the source of solar energy.
Both fission and fusion release energy precisely because each carries the system toward the peak of the binding-energy curve, where nucleons are bound most tightly.
\[\boxed{\text{Energy released} = (\text{final } B.E.) - (\text{initial } B.E.) > 0}\]