Question

Derive the formula \(\dfrac{n}{v}-\dfrac{1}{u}=\dfrac{n-1}{R}\) due to refraction of light on a spherical surface.
OR
Draw a circuit diagram of p-n junction full wave rectifier and describe its working. Also, give the graphical representation of input and output current.

Hint:

Use the exterior-angle geometry of triangles \(OAC\) and \(AIC\) with the paraxial Snell relation \(i=n\,r\), then apply the sign convention \(u<0,\ v>0,\ R>0\). For the rectifier, recall that a centre-tapped two-diode circuit conducts on both half cycles so the load current is unidirectional.

Answer 1

Option 1: Refraction at a single spherical surface

Setup and assumptions: Consider a convex spherical refracting surface of radius of curvature \(R\), pole \(P\) and centre of curvature \(C\). It separates a rarer medium of refractive index \(1\) (containing the object) from a denser medium of refractive index \(n\). A point object \(O\) lies on the principal axis. A paraxial ray \(OA\) meets the surface at \(A\); the normal at \(A\) is the line \(CA\) (a radius, since every normal to a sphere passes through its centre). The refracted ray \(AI\) bends towards this normal and cuts the axis at the image \(I\). We use the paraxial (small-angle) approximation and the Cartesian sign convention. Let \(i\) and \(r\) be the angles of incidence and refraction, and let \(\alpha=\angle AOP\), \(\beta=\angle AIP\), \(\gamma=\angle ACP\).

Step 1: Relate the angles by the exterior-angle theorem. In triangle \(OAC\), the exterior angle at \(A\) gives \[ i=\alpha+\gamma. \] In triangle \(AIC\), the exterior angle \(\gamma\) gives \(\gamma=r+\beta\), so \[ r=\gamma-\beta. \] Step 2: Apply Snell's law at \(A\): \(1\cdot\sin i=n\sin r\). For paraxial rays \(\sin i\approx i\) and \(\sin r\approx r\), so \[ i=n\,r\ \Rightarrow\ \alpha+\gamma=n(\gamma-\beta). \] Step 3: Express the angles through distances. Let \(M\) be the foot of the perpendicular from \(A\) to the axis; for paraxial rays \(M\) lies almost at the pole \(P\). Then (angle \(\approx\) tangent) \[ \alpha=\dfrac{AM}{PO},\quad \beta=\dfrac{AM}{PI},\quad \gamma=\dfrac{AM}{PC}. \] Substituting and cancelling the common height \(AM\): \[ \dfrac{1}{PO}+\dfrac{1}{PC}=n\left(\dfrac{1}{PC}-\dfrac{1}{PI}\right). \] Step 4: Apply the sign convention. Distances are measured from the pole \(P\), positive along the direction of incident light. Thus \(PO=-u\), \(PI=+v\), \(PC=+R\). Substituting, \[ \dfrac{1}{-u}+\dfrac{1}{R}=n\left(\dfrac{1}{R}-\dfrac{1}{v}\right), \] \[ -\dfrac{1}{u}+\dfrac{1}{R}=\dfrac{n}{R}-\dfrac{n}{v}. \] Rearranging the terms, \[ \dfrac{n}{v}-\dfrac{1}{u}=\dfrac{n}{R}-\dfrac{1}{R}=\dfrac{n-1}{R}. \] \[\boxed{\dfrac{n}{v}-\dfrac{1}{u}=\dfrac{n-1}{R}}\]

Option 2: p-n junction full wave rectifier (centre-tap type)

Circuit (described): The primary of a transformer is fed the AC input. Its secondary is centre-tapped, giving two ends \(A\) (top) and \(B\) (bottom) with the centre tap \(T\) in the middle. End \(A\) connects to the p-side (anode) of diode \(D_1\) and end \(B\) to the p-side of diode \(D_2\). The n-sides (cathodes) of \(D_1\) and \(D_2\) are joined together and taken to one terminal of the load resistance \(R_L\); the other terminal of \(R_L\) is connected to the centre tap \(T\). The output DC voltage is read across \(R_L\).

Working:
Positive half cycle: End \(A\) is positive with respect to \(T\), so \(D_1\) is forward biased and conducts, while \(D_2\) (with end \(B\) negative) is reverse biased and stays off. Current flows from \(A\) through \(D_1\), through \(R_L\), back to \(T\).
Negative half cycle: The secondary polarity reverses; now end \(B\) is positive, so \(D_2\) conducts and \(D_1\) is off. Current flows from \(B\) through \(D_2\), through \(R_L\), back to \(T\).
In both half cycles the current passes through \(R_L\) in the same direction, so both halves of the input are used. This is full wave rectification. The output is unidirectional pulsating DC whose ripple frequency is twice the input frequency.

Graphs: The input current/voltage is a complete sine wave with alternate positive and negative halves. The output current is a train of positive humps only, one for every half cycle (the negative halves are flipped up and also appear positive), i.e. continuous pulsating DC with no missing halves, unlike a half wave rectifier which loses the negative halves.
\[\boxed{\text{Both half cycles drive } R_L \text{ in one direction: full-wave pulsating DC}}\]