Question

Explain full wave rectifying action of p-n junction diode with the help of circuit diagram.
OR
Draw the diffraction pattern of light due to single slit and write the expression for position of minima. On what factors does the width of central maxima depend?

Hint:

Option 1: in a centre-tapped full wave rectifier the two diodes conduct in alternate half cycles but push current through the load in the same direction, giving pulsating DC. Option 2: use \(a\sin\theta = n\lambda\) for the minima and \(W = 2\lambda D/a\) for the central maximum width.

Answer 1

Option 1: Full Wave Rectifier

Step 1: What a rectifier does. A rectifier is a circuit that converts alternating current (AC) into direct current (DC). A p-n junction diode conducts only when it is forward biased and blocks current when it is reverse biased, so it behaves like a one-way valve for current. In a full wave rectifier, both halves of the AC input are turned into DC output.

Step 2: Circuit arrangement. A centre-tapped step-down transformer is used. The central tap C of its secondary coil is connected to one end of the load resistance \(R_L\). The two outer ends A and B of the secondary are connected to two diodes \(D_1\) and \(D_2\). The remaining terminals of both diodes are joined together and connected to the other end of \(R_L\). The output DC voltage is taken across \(R_L\).

Step 3: Positive half cycle. During the positive half of the input, end A is positive and end B is negative with respect to the centre tap C. Now \(D_1\) is forward biased and conducts, while \(D_2\) is reverse biased and does not conduct. Current flows from A, through \(D_1\), through \(R_L\), and back to C. The current in \(R_L\) flows from top to bottom.

Step 4: Negative half cycle. During the negative half, end B becomes positive and end A negative. Now \(D_2\) is forward biased and conducts, while \(D_1\) is reverse biased. Current flows from B, through \(D_2\), through \(R_L\), and back to C. Importantly, the current in \(R_L\) again flows from top to bottom, i.e. in the same direction as in the positive half.

Step 5: Nature of output. Since current passes through \(R_L\) in the same direction during both halves of the input cycle, the output is a unidirectional (DC) but pulsating voltage. Both half cycles of the AC are used, so the output shows two humps per input cycle and the ripple frequency is twice the input frequency. This is why it is called a full wave rectifier, and it is more efficient than a half wave rectifier.

Circuit and waveform (described): Input AC is a complete sine wave; the output across \(R_L\) is a train of positive humps, all lying above the zero line, one hump from each half cycle. A filter capacitor across \(R_L\) can smooth this into steady DC.
\[\boxed{\text{Both half cycles drive current through } R_L \text{ in the same direction} \Rightarrow \text{DC output}}\]

Option 2: Single Slit Diffraction

Step 1: The pattern. When monochromatic light of wavelength \(\lambda\) passes through a narrow single slit of width \(a\) and falls on a screen, it produces a diffraction pattern: a bright central maximum at the centre, with alternate dark bands (minima) and dim bright bands (secondary maxima) on both sides. The central maximum is the brightest and the widest; the intensity of the secondary maxima falls off rapidly.

Step 2: Condition for minima. Destructive interference (a dark fringe) occurs when the path difference between the wavelets from the two edges of the slit equals a whole number of wavelengths: \[a\sin\theta = n\lambda, \quad n = 1, 2, 3, \dots\] Here \(\theta\) is the angle of diffraction and \(n\) is the order of the minimum.

Step 3: Position of the minima on the screen. Let \(D\) be the slit-to-screen distance and \(x_n\) the distance of the \(n\)th minimum from the centre. For small angles \(\sin\theta \approx \tan\theta = x_n/D\). Substituting in the condition above: \[\frac{a\,x_n}{D} = n\lambda \Rightarrow x_n = \frac{n\lambda D}{a}\]
Step 4: Width of the central maximum. The central maximum lies between the first minima on the two sides (\(n = 1\)), i.e. between \(x = +\lambda D/a\) and \(x = -\lambda D/a\). So its linear width is \[W = \frac{2\lambda D}{a}\] and its angular width is \(2\theta = 2\lambda/a\).

Step 5: Factors controlling the width. From \(W = 2\lambda D/a\):
(i) \(W\) is directly proportional to the wavelength \(\lambda\) of the light,
(ii) \(W\) is inversely proportional to the slit width \(a\) (a narrower slit gives a wider central maximum),
(iii) \(W\) is directly proportional to the distance \(D\) between the slit and the screen.
\[\boxed{x_n = \dfrac{n\lambda D}{a}, \qquad W_{\text{central}} = \dfrac{2\lambda D}{a}}\]