Justification of the relationship $s/S<\lambda/d$ for interference fringes to be seen

Question

[My book] (NCERT Class 12 Physics text; Chapter no.:10 - Wave Optics) (https://www.google.co.in/url?sa=t&source=web&rct=j&opi=89978449&url=https://ncert.nic.in/ncerts/l/leph202.pdf&ved=2ahUKEwjcz4PR1-6CAxXHQfUHHVlYCuoQFnoECAgQAQ&usg=AOvVaw301VEzXnurxZ3o6DPXUjP-)

states that the condition $$s/S < \lambda/d$$ (where $S$ is the distance of the source from the slitted plane and $s$ is the size of the source) must be satisfied for interference fringes to be seen.

It further says that when this condition is not satisfied interference patterns produced by different parts of the source overlap and no fringes are seen.

What I ultimately understood is that when the size of the source increases, interference patterns produced by different parts of the source overlap and no fringes are seen.

Please correct me if my understanding is flawed and provide a detailed intuitive explanation for the condition mentioned above.

Answer 1

Yes, your understanding is correct. Let's analyse it a little bit further, for a 1D source, parallel with the apparatus. Consider a fixed point P in the apparatus where the interference fringes are supposed to be seen. To calculate the difference in phase of two starting points of your extensive source, it's enough to know how would be the phase difference of two different points of the apparatus given a fixed initial point in the source. (If I have time, I will make a figure). Source and apparatus are basically symmetric with relation to slits. Now the phase difference will be $$ \Delta\phi = \frac{2\pi}{\lambda} d \frac{y}{S} $$ where $y$ is the distance between two points. Now if you allow unconstrained $y$, for any point $Q$ on the source with coordinate $y$, there will be another point $Q^{'}$ with coordinate $y^{'}=y+S/2\lambda d$ giving the exact opposite phase and resulting in a destructive interference. To avoid this situation, we should constraint the Source dimension. To achieve this, we impose that the phase difference between the highest point with the smallest one should not be greater than $2\pi$ $$ \frac{2\pi}{\lambda} d \frac{s}{S} < 2\pi\\ \Rightarrow \frac{s}{S} < \frac{\lambda}{d} $$