Electronic – resistor R2 doing

It's not for protection, it's to form a voltage divider with the photocell.

For a typical photocell, the resistance may vary between say, 5 kΩ (light) and 50 kΩ (dark)
Note that the actual values may be quite different for your sensor (you'll need to check the datasheet for those)

If we leave the resistor out, the analog input will see 5 V either way (assuming an analog input of a high enough impedance not to affect things significantly)
This is because there is nothing to sink the current and drop voltage.

No Resistor

Let's assume the sensor is connected to an opamp with an input resistance of 1 MΩ(pretty low as opamps go, can be 100's of MΩ)

When there is no light shining on the photocell and it's resistance is at 50 kΩ we get:

$$ 5~\mathrm{V} \times \frac{1~\mathrm{M}\Omega}{1~\mathrm{M}\Omega + 50~\mathrm{k}\Omega} = 4.76~\mathrm{V} $$

When there is light shining on the photocell and it's resistance is at 5 kΩ, we get:

$$ 5~\mathrm{V} \times \frac{1~\mathrm{M}\Omega}{1~\mathrm{M}\Omega + 5~\mathrm{k}\Omega} = 4.98~\mathrm{V} $$

So you can see it's not much use like this - it only swings ~200 mV between light/dark. If the opamps input resistance was higher as it often will be, you could be talking a few µV.

With Resistor

Now if we add the other resistor to ground it changes things, say we use a 20 kΩ resistor. We are assuming any load resistance is high enough (and the source resistance low enough) not to make any significant difference so we don't include it in the calculations (if we did it would look like the bottom diagram in Russell's answer)

When there is no light shining on the photocell and it's resistance is at 50 kΩ, we get:

$$ 5~\mathrm{V} \times \frac{20~\mathrm{k}\Omega}{20~\mathrm{k}\Omega + 50~\mathrm{k}\Omega} = 1.429~\mathrm{V} $$

With there is light shining on the photocell and it's resistance is 5k we get:

$$ 5~\mathrm{V} \times \frac{20~\mathrm{k}\Omega}{20~\mathrm{k}\Omega + 5~\mathrm{k}\Omega} = 4.0~\mathrm{V} $$

So you can hopefully see why the resistor is needed in order to translate the change of resistance into a voltage.

With load resistance included

Just for thoroughness let's say you wanted to include the 1 MΩ load resistance in the calculations from the last example:

To make the formula easier to see, lets simplify things. The 20 kΩ resistor will now be in parallel with the load resistance, so we can combine these both into one effective resistance:

$$ \frac{20~\mathrm{k}\Omega \times 1000~\mathrm{k}\Omega}{20~\mathrm{k}\Omega + 1000~\mathrm{k}\Omega} \approx 19.6~\mathrm{k}\Omega $$

Now we simply replace the 20 kΩ in the previous example with this value.

Without light:

$$ 5~\mathrm{V} \times \frac{19.6~\mathrm{k}\Omega}{19.6~\mathrm{k}\Omega + 50~\mathrm{k}\Omega} = 1.408~\mathrm{V} $$

With light:

$$ 5~\mathrm{V} \times \frac{19.6~\mathrm{k}\Omega}{19.6~\mathrm{k}\Omega + 5~\mathrm{k}\Omega} = 3.98~\mathrm{V} $$

As expected, not much difference, but you can see how these things may need to be accounted for in certain situations (e.g. with a low load resistance - try running the calculation with a load of 10 kΩ to see a big difference)

Here is some insight. The + input of the op-amp is grounded, and the negative feedback action has the effect of keeping the - input of the op-amp at the same voltage as the + input. So the - input is a virtual ground.

You can solve this by pretending that the op-amp is not there, and then solving the remaining network of resistors and current source by finding the value of \$V_o\$ which is necessary for the junction between the \$3\Omega\$ and \$2\Omega\$ resistors to be at zero volts.

The easiest way to do this is to pretend that the junction is actually grounded (not simply at 0V potential). Determine how much current is dumped into that ground through the \$2\Omega\$ resistor. But then, recognize that there is no ground there and so the current flowing toward that node actually returns via the \$3\Omega\$ feedback resistor.

The current across the feedback resistor gives you a straightforward way to determine the voltage across that resistor, which directly determines \$V_o\$.