All I got was two parallel bright fringes instead, like the ones you would get by shining a torch through two very thick slits.
It means the separation between slit is not close and the slits size is not small enough! Those two light beam must overlap to have interference. Small slit size is required to have large diffraction, the optimal slit size is certainly small than wavelength $\lesssim\lambda\approx0.5\mu m$ which gives you large diffraction. However, larger slits size is ok, but you have to (a) Make two slits as close as possible (b) move the setup far from the screen. You will know that it is enough when the light beam can overlap.
For the slit, you need better tools than a knife as well as a better material. First, you should use a shape cutter. Second, you need a material that can have a sharper edge such as film. I believe that film was used in the first few experiments of this kind. You have mentioned a hair is enough so $10\mu m$ should probably be ok, you just need to move the screen further away.
For the light source, you should always use a laser, since a high coherent light is required. Any laser out there is ok, it just cost 1 dollar and I can sure you can borrow a laser pointer near you. As I remember when I was doing Michelson Morley experiment, a tungsten light only gives interference pattern for $<0.1m$ with short coherent length, but a laser can have coherent length $>2m$. It means your life can be easier as you can use a 20 times larger slit with a laser!
Edit: Additional info on the methods Young used for this experiment.
The wiki about Young' interference experiment has quoted his paper on "On the nature of light and colours" (Also around page p.140 in the book Method and Appraisal in the Physical Sciences). The relevant excerpt is:
In order that the effects of two portions of light may be thus combined, it is necessary that they be derived from the same origin, and that they arrive at the same point by different paths, in directions not much deviating from each other. This deviation may be produced in one or both of the portions by diffraction, by reflection, by refraction, or by any of these effects combined; but the simplest case appears to be, when a beam of homogeneous light falls on a screen in which there are two very small holes or slits, which may be considered as centres of divergence, from whence the light is diffracted in every direction.
So, I guess the experiments were carried out as follow:
- Light source: In a room with all windows covered with thick curtains so that the inside is completely dark. Then let a small beam of sunlight go in.
- Monochromatic light: Use prism to split light into different color (This is known method back to Netwon). To get high quality single frequency light, a slit is required in front of prism to get a narrow sunlight beam.
- Point source of monochromatic light: Add another slit to get the required color (S1 in Fig. 1), the output monochromatic light is therefore from a single point source.
- Interference: Add another double slits (S2 in Fig. 1) so that the light can have two different path. Make sure that light from S1 falls on the slits S2. To ease observations, the screen should be far away.
Since his results cover all color, so it is very likely that he used sunlight rather than other light source such as candle (There was no light bulk at that time). Also, there is no diffraction grating, so it is likely that he was just using a simple prism.
For home experiments carried out these day, we can use LED as a monochromatic light source so that step 1 and 2 can be skipped. If you use a torch, you still need the step 2.
Best Answer
The resultant amplitude of two interfering waves is $$ {A_{net}}^2 = {A_1}^2 + {A_2}^2 +2A_1A_2\cos\theta $$ where $\theta$ is the phase difference between the waves.
Since intensity is proportional to the square root of the amplitude we have $$I_{net} = I_1 + I_2 +2\root\of{I_1}\root\of{I_2}\cos\theta$$ Normally in a double slit experiment the sources are same and coherent and that gives $$ I_1=I_2=I' (say)$$ and the formula for $I_{net}$ reduces to the one you mentioned.
But Since the source intensities are not same you can't apply that formula.
Instead just use $ I_1=\frac{I_2}{2}=I_0 $.
For the maximum intensity $\cos\theta = 1$ and for minimum intensity $\cos\theta = -1$
Find the $I_{net}$ for both these cases and take the ratio. You get
$$ (3+2\root\of2)^2$$