Q: Having trouble with light activated circuit. 
R1 = 10K 1/4W Resistor
R2 = Photoresistor (any type)
R3 = 2M2 1/4W Resistor
R4 = 1M 1/4W Resistor
C1= 10µF 25V Electrolytic Capacitor
C2 = 100nF 63V Polyester Capacitor
D1 = 1N4148 75V 150mA Diode
IC1, IC2 = 7555 or TS555CN CMos Timer ICs
BZ1 = Piezo sounder (incorporating 3KHz oscillator)
B1 = 3V Battery (2 x 1.5V AA, AAA or smaller type Cells in series)
When a beam of light enters from the opening, or the fridge lamp lights, the photo resistor lowers its resistance (<2K) stopping C1 charging current. Therefore IC1, wired as an astable multivibrator, starts oscillating at a very low frequency and after a period of about 24 sec. its output pin (#3) goes high, enabling IC2.
This chip is also wired as an astable multivibrator, driving the Piezo sounder intermittently at about 5 times per second. The alarm is activated for about 17 sec. then stopped for the same time period and the cycle repeats until the fridge door closes.
So I'm trying to build this circuit and have a question
- Does this circuit start sounding the alarm 24 seconds after the light enters or just after? I don't quite understand it
and also how should I calculate to get the values 24 seconds and 17 seconds.
Thank you in advance.

A: Understanding how the circuit works requires that you know how a photocell responds to light. 
I'm assuming that you are using a standard CdS photocell. These have high resistance in the dark and the resistance drops to a low value in the presence of light. 
When the photocell is dark, timing capacitor C1 is held High by R1.  When the photocell is exposed to light, the voltage across the photocell drops to a low value. D1 prevents C1 from being discharged by the photocell. 
Instead, C1 slowly discharges via the 1M resistor. When the voltage across C1 drops to about 1/3 the supply voltage, the output of the 555 timer goes Hi and turns on the tone generator. 
