The figure shows a two-slit arrangement with a source that emits unpolarised light. $P$ is a polariser with an axis whose direction is not given. If $I_0$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

(N/A) In the absence of a polariser, the intensity of the principal maxima is $I_0 = 4I$, where $I$ is the intensity of light from each slit.
When a polariser is placed in one of the paths (say path $2$), the light passing through it becomes linearly polarised. Let the amplitude of the light from each slit be $a$. The intensity $I = a^2$.
For the slit without the polariser, the light remains unpolarised. For the slit with the polariser, the light becomes polarised.
When these two beams superpose, the unpolarised light can be considered as two incoherent components of intensity $I/2$ each, vibrating in mutually perpendicular directions.
The polarised light from the second slit has intensity $I' = I/2$ (after passing through the polariser).
At the principal maxima, the two beams are in phase. The intensity of the unpolarised beam is $I$ and the polarised beam is $I/2$. The resultant intensity is $I_{max} = I + I/2 + 2\sqrt{I \cdot I/2} \cdot \cos(0) = I + I/2 + \sqrt{2}I = I(1.5 + 1.414) \approx 2.914I$. Since $I_0 = 4I$, $I = I_0/4$. Thus, $I_{max} = (2.914/4)I_0 = 0.7285I_0$.
At the first minima, the phase difference is $\pi$. The intensity is $I_{min} = I + I/2 - 2\sqrt{I \cdot I/2} = I(1.5 - 1.414) = 0.086I = 0.086(I_0/4) = 0.0215I_0$.