1. In Young’s double slit experiment, if each slit alone produces intensity $I_0$, then for two coherent sources of equal intensity, the resultant intensity is

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$

Here, since both slits have equal individual intensity $I_0$,

$I = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos\phi = 2I_0(1+\cos\phi)$

where phase difference

$\phi = \frac{2\pi}{\lambda}\,\Delta$

with $\Delta$ = path difference.

2. For the first point, path difference is

$\Delta_1 = \frac{\lambda}{4}$

So phase difference is

$\phi_1 = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{4} = \frac{\pi}{2}$

Hence,

$I_1 = 2I_0(1+\cos\tfrac{\pi}{2}) = 2I_0(1+0) = 2I_0$

3. For the second point, path difference is

$\Delta_2 = \frac{\lambda}{3}$

So phase difference is

$\phi_2 = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{3} = \frac{2\pi}{3}$

Hence,

$I_2 = 2I_0(1+\cos\tfrac{2\pi}{3})$

Since

$\cos\frac{2\pi}{3} = -\frac{1}{2}$

therefore,

$I_2 = 2I_0\left(1-\frac{1}{2}\right)=2I_0\cdot \frac{1}{2}=I_0$

4. Now add the two intensities:

$I_1+I_2 = 2I_0 + I_0 = 3I_0$

Therefore,

$\frac{I_1+I_2}{I_0} = 3$

5. Comparison with stored answer:

Derived answer = $3$

Stored correct answer = $3$

They agree.