Option 1: Full Wave Rectifier
Step 1: Principle. A junction diode is a one-way device: it passes current only in forward bias. To obtain DC over the whole AC cycle, one diode must take over exactly when the other switches off. This handover is arranged with a centre-tapped transformer feeding two diodes.
Step 2: Components. Secondary of the transformer with a central tap C; upper end A goes to diode \(D_1\); lower end B goes to diode \(D_2\); the free ends of \(D_1\) and \(D_2\) are tied together and taken to the load \(R_L\), whose other end returns to C. Output is read across \(R_L\).
Step 3: Cycle-by-cycle account.
First half of input: A is positive, so \(D_1\) is ON and \(D_2\) is OFF. Path of current: A to \(D_1\) to \(R_L\) to C.
Second half of input: B is positive, so \(D_2\) is ON and \(D_1\) is OFF. Path of current: B to \(D_2\) to \(R_L\) to C.
In both halves the load current enters \(R_L\) from the same terminal, so its direction never reverses.
Step 4: Output waveform. The load therefore carries current during every half cycle in one fixed direction. The result is a unidirectional but pulsating voltage whose ripple frequency is \(2f\), double the mains frequency \(f\). A shunt capacitor filter can flatten this into nearly steady DC.
Step 5: Merit over half wave. Because no half of the input is thrown away, the rectifier efficiency (about \(81.2\%\)) is twice that of a half wave rectifier (about \(40.6\%\)), and the output is easier to smooth.
\[\boxed{D_1 \text{ and } D_2 \text{ conduct in turn} \Rightarrow \text{continuous DC through } R_L}\]
Option 2: Single Slit Diffraction
Step 1: Split the slit. Take a slit of width \(a\). To find the first dark fringe, imagine the slit divided into two equal halves, each of width \(a/2\). Pair every point in the upper half with the point exactly \(a/2\) below it in the lower half.
Step 2: First minimum. Each such pair cancels when its path difference is \(\lambda/2\): \((a/2)\sin\theta = \lambda/2\), which gives \(a\sin\theta = \lambda\). Dividing the slit into \(2n\) equal strips and pairing them the same way generalises this to the full minima condition
\[a\sin\theta = n\lambda, \quad n = 1, 2, 3, \dots\]
Step 3: Screen position. With slit-screen distance \(D\) and small \(\theta\), \(\sin\theta \approx x_n/D\), so
\[x_n = \frac{n\lambda D}{a}\]
The dark fringes are almost evenly spaced by \(\lambda D/a\).
Step 4: Central maximum. The bright centre is bounded by the two first-order minima (\(n = 1\)) at \(x = \pm\,\lambda D/a\). Hence its width is
\[W = \frac{2\lambda D}{a}\]
which is twice the spacing of the neighbouring fringes.
Step 5: Dependence. Since \(W \propto \lambda\), \(W \propto D\) and \(W \propto 1/a\), a longer wavelength, a larger screen distance, or a narrower slit each widens the central maximum. Conversely, if \(a\) is made very large, \(W \to 0\) and we recover the sharp geometrical shadow of ray optics.
\[\boxed{a\sin\theta = n\lambda, \quad W = \dfrac{2\lambda D}{a}}\]