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## Homework Statement

Two slits (of width ##a=39 \mu m##) are lighted up with a monocromatic wave of ##\lambda=632,8 nm##. The distance between slits and the screen id ##D=4 m##. The distance between the slits id ##d=195 \mu m##.

In front of the slits there are a convergent lens with focal length ##f=60 cm##.

Find the width of central maximum of interference and diffraction.

## The Attempt at a Solution

I don't know what to do because I don't know what is the effect of the lens on the maximums and minimums.

Without lens the width of central maximum (of interference) is the distance between the first two minimum, so $$d\sin\theta=\frac{1}{2}\lambda \longrightarrow d\frac{y_{min}}{D}=\frac{1}{2}\lambda \longrightarrow \Delta y=2y=\frac{\lambda D}{d}$$

and for the diffraction it would be $$a\sin\theta=\lambda \longrightarrow a\frac{y_{min}}{D}=\lambda \longrightarrow \Delta y=2y=2\frac{\lambda D}{a}$$.

Can you give me a hint?