Option 1: Same result via Snell's law first, then geometry
Step 1: A convex surface of radius \(R\) (pole \(P\), centre \(C\)) separates air (\(n_1=1\), object side) from a denser medium (\(n_2=n\)). A paraxial ray from the axial object \(O\) strikes at \(A\) where the outward normal is the radius \(CA\); the refracted ray reaches the image \(I\) on the axis. Snell's law at \(A\) in the small-angle form is \(n_1\, i=n_2\, r\), that is \(i=n\,r\).
Step 2: Take the three small angles the ray lines make with the axis: \(\alpha=\angle AOP\) at the object, \(\beta=\angle AIP\) at the image, and \(\gamma=\angle ACP\) at the centre. Dropping a perpendicular of height \(h\) from \(A\) to the axis (its foot near \(P\) for paraxial rays), \(\alpha=\dfrac{h}{PO},\ \beta=\dfrac{h}{PI},\ \gamma=\dfrac{h}{PC}\).
Step 3: The incidence angle is the exterior angle of triangle \(OAC\): \(i=\alpha+\gamma\). The refraction angle is \(r=\gamma-\beta\) from triangle \(AIC\). Feed these into \(i=n\,r\): \(\alpha+\gamma=n(\gamma-\beta)\).
Step 4: Replace the angles by \(h/PO,\ h/PI,\ h/PC\) and cancel \(h\): \(\dfrac{1}{PO}+\dfrac{1}{PC}=n\!\left(\dfrac{1}{PC}-\dfrac{1}{PI}\right)\). Using the Cartesian signs \(PO=-u,\ PI=+v,\ PC=+R\) gives \(-\dfrac{1}{u}+\dfrac{1}{R}=\dfrac{n}{R}-\dfrac{n}{v}\), and collecting terms, \(\dfrac{n}{v}-\dfrac{1}{u}=\dfrac{n-1}{R}\).
\[\boxed{\dfrac{n}{v}-\dfrac{1}{u}=\dfrac{n-1}{R}}\]
Option 2: Full wave rectifier explained as a bridge circuit
Circuit (described): A full wave output can also be obtained with a bridge of four p-n junction diodes \(D_1,D_2,D_3,D_4\) and an ordinary (non centre-tapped) transformer secondary. The four diodes form the four arms of a bridge. The two input corners of the bridge receive the AC from the secondary; the other two corners feed the load \(R_L\), whose output is the DC.
Working:
Positive half cycle: Two diagonally opposite diodes (say \(D_1\) and \(D_3\)) become forward biased and conduct, while \(D_2\) and \(D_4\) are reverse biased. Current is routed through \(R_L\) in a fixed direction.
Negative half cycle: The input polarity reverses, so now \(D_2\) and \(D_4\) conduct and \(D_1,D_3\) are off. Crucially, the path is arranged so the current through \(R_L\) still flows in the same direction as before.
Because every half cycle sends current one way through \(R_L\), the whole input wave is rectified. The result is pulsating DC of ripple frequency twice the mains frequency, and no half of the input is wasted.
Graphs: Input is a full alternating sine wave; the output across \(R_L\) is a series of back-to-back positive humps (both input halves appear as upward pulses), i.e. gap-free pulsating DC, in contrast with the half wave case where alternate humps are missing.
\[\boxed{\text{Bridge (4-diode) rectifier gives full-wave unidirectional output}}\]