A double-slit interference experiment is done in a ripple tank. The slits are 4.90 cm apart, and a viewing screen is 1.56 m from the slits. The wave speed of the ripples in water is 0.027 m/s, and the frequency of the oscillator producing the ripples is 5.10 Hz. How far from the centerline of the screen will the first order maxima be found?

Question

anonymous · Answer

Let's call the position on the screen away from the centreline $y$, the distance from the slits to the screen $D$ = 1.56 m, the separation of the slits $d$ = 4.9 cm, and the angle of diffraction $\theta$. 

The wavelength $\lambda$ is the wavespeed divided (0.027 m/s) by the frequency (5.1 Hz).

The difference in path length from each slit to a point $y$ on the screen is $d\sin\theta$ where $y$ is related to $theta$ by

$$\tan\theta=\frac{y}{D}$$

The condition for the first order maximum is that the difference in path length must be 1 wavelength.

$$d\sin\theta=\lambda$$

$y$ may be worked out thus:

$$y=D\tan\sin^{-1}\Big(\frac{\lambda}{d}\Big)$$

or you could use the small-angle approximation

$$y = \frac{\lambda D}{d}$$

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