Q.Why is the base of a transistor made thin and is lightly doped?

[!TLDR] A thin base minimises electron-hole recombination so nearly all emitter-injected carriers reach the collector, while light doping (relative to the emitter) keeps the base current small -- together these give a transistor its large current gain. [!ANSWER] The base is made thin and lightly doped so that almost all carriers injected from the emitter pass through to the collector rather than recombining in or being supplied by the base, giving the transistor a large current amplification factor.

The base's thinness and light doping (compared to the heavily-doped emitter) work together to produce the transistor's essential amplifying action. Making the base very THIN means that electrons (in an n-p-n transistor) injected from the emitter have only a very short distance to diffuse across before reaching the collector-base depletion region, where the junction's field sweeps them into the collector -- this short transit distance leaves very little time or space for these electrons to recombine with the base's own holes along the way, so the vast majority (about 95%) survive the crossing and contribute to the collector current $I_C$. Making the base LIGHTLY doped (relative to the emitter, and even further, ten times more lightly than the collector) means the base itself has relatively few majority carriers (holes, for an n-p-n transistor) available to supply the small amount of recombination current that DOES occur, or to be injected 'backwards' into the emitter -- keeping the base current $I_B$ small (only a few percent of $I_E$). Together, a thin, lightly-doped base is exactly why $I_C\approx I_E$ (most of the emitter's supplied carriers reach the collector) while $I_B$ stays small -- and it is this large ratio $I_C/I_B=\beta_{DC}$ that gives the transistor its large current amplification (gain) when used in the common emitter configuration. If the base were instead thick and/or heavily doped, far more injected carriers would recombine within it, $I_B$ would be much larger relative to $I_C$, and the transistor's current gain would collapse. [!ANSWER] The base is made thin and lightly doped so that almost all carriers injected from the emitter pass through to the collector rather than recombining in or being supplied by the base, giving the transistor a large current amplification factor.

Trace how the base's geometry (thin) and doping (light, relative to the emitter) each independently reduce recombination/base current, and connect this directly to why $I_C\approx I_E$ and $\beta_{DC}=I_C/I_B$ comes out large.

Explaining only ONE of the two properties (thinness OR light doping) and missing that both act together -- thinness alone limits recombination distance, while light doping independently limits the majority carriers available in the base for that recombination and for reverse injection into the emitter.