The ground connection on the left should not be be there. The whole point of the optoisolator is to provide galvanic isolation between the low voltage and mains sides.
It's also conventional to have signal flow from left to right, so normally the schematic should show the switch at the left and the output on the right. That's a style thing, it does not affect the actual circuit.
The input circuit has a problem- you will destroy the LED as soon as the switch is depressed since there is no series current-limiting resistor.
The output circuit has problems as well, the MOC3042 is not designed to switch a lamp load, it's main purpose is to switch a larger triac. If the incandescent bulb in your schematic is 100W and you have calculated the 580 ohms as the resistance, that is the resistance when it is hot. When you first apply power, the resistance will be much less, perhaps 1/10 to 1/20 of the resistance at operating temperature. That means that the current at switch-on could be in the 5-10A range, which could be enough to damage the device (the only hint is that the repetitive current is limited to 1A in the datasheet, and that will certainly be exceeded at times with such a load).
Even if it survives the initial turn-on, the power dissipation rating will be exceeded. See "Figure 2 On State Characteristics", and the maximum dissipation rating Pd of 150mW - 1.76mW/°C- which implies you should keep current to maybe 25mA to allow for moderately high ambient temperatures.
A small triac (8-16A range) will provide a lot more beefy ratings for switching a serious load, and the MOC3042 can be used to switch it. And don't forget the input resistor.
You could easily connect them all in parallel. or in any grid ie (3*10 or 6*5) and just control the input to the grid independently. As others have said incandescent bulbs is mostly a resistive load. If you go with ,say 5 in series and each of those 5 is in parallel then you would require 7.5V @ 1.8A. Choosing a configuration is simply a matter of what you have available.
Except for requiring more power than LEDs the incandescent bulbs can be pwm much easier. You might even be able to get away with a much lower switching frequency because the bulbs will tend to filter higher frequencies.
Series is easier to switch m but if one bulb do fails , the whole light fails. Going parallel you need to switch a higher current but its trivial to find a blown bulb. Using a grid like arrangement mixes these two features partially so you get lower switching current than when all is in parallel and its easier to find faults than when everything is in series.
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When Arduino is off , there is no current thru R1 and thus R2 will keep the FET turned off with low Vgs. There are no shared parts in this design without expecting some changes in performance depending on how the Optocoupler loads are enabled. We call this load regulation when a series string of Rs/(Rs+Rload) causes a change in output voltage due to variations in load. If you drive all Optos together then R2+R3 the load is reduced to 25% in series with R4.
With some effort you can simplify this to make some DC parts common, but then you will quickly realize it is far simpler to create a ZCS and software phase control with a triac for 4 ports and possibly even get away with non-isolated DC power if your interface to communicate to Arduino is isolated.
Misc info.
But in general, there are many reasons why PWM is a bad choice for tungsten bulbs.
PWM switches are efficient only when the switch impedance is relative low compared to the load. i.e. < 5% for 5% loss of load. This is not the case for cold start on tungsten and is why even triac dimmers surge on from off and have hystereis in the low range.
Tungsten resistance from cold to hot, will rise to 10x the cold value due to the ~2500'K rise in temp.
Tungsten coils are slightly inductive so rise time causes a phase shift.
This is why PWM is never used for tungsten bulbs and only use line frequency Triac phase controls.
Consider a 120W bulb @ 120V is 120Ω when hot, then the R_cold ~ 12 Ω and it's power dissipation at P=V²/R= (120V)²/12 = 1200 W or 10 times the steady state.
If using a FET bridge to drive PWM the conduction state depends mainly on the Vgs/Vgs(th) ratio and not the load resistance and if you set a low duty cycle where the filament does not heat up much like 10% of 120W or 12W, your bridge can be overheating.
Why ? because of the RdsOn/ load ratio when the bulb is relatively cool.
Why do triacs work better?
Triacs will not fire if the load resistance is too low or in other words current ratio of the load to trigger current is too high. this is due to the internal saturation or ESR of the Vbe junctions not being driven hard enough to latch.
( Triacs are basically two BJT's PNP & NPN with cross connections between CE to drive BE so they are trigger current and load sensitive with current gains dropping to <20% of hFE at saturation.)
So what happens is as the phase angle is increased, the bulb jerks on well above the minimum when dimming down. But this would not be the case for a voltage controlled FET bridge in PWM mode. The lamp would turn on and the bridge would consume almost as much as the bulb if you chose RdsOn to be 10% of the load.
But if you chose a bridge with much lower RdsOn like < 1% of load , OK, but this then becomes costly compared to Triacs.
Examine this design choice of FET
60W @ 230V , R_hot= V²/P = 230V²/60W = 882 Ω , R_cold= 88 Ω , RdsOn @ Vgs = 10 V = 0.40 Ω = 0.5% of load , so good choice but if you try a couple 100W bulbs , what out for hot device when slow rampup.