Answer to Question #92599 in Optics for Sridhar

Question

An electromagnetic wave emitted by a source travels 21km to arrive at a receiver. while travelling in another path is reflected from a surface at 19km away and travels further 12 km to reach the same receiver. If destructive interference occurs at the receiving end the maximum wavelength of the wave is 1) 5km 2) 1km 3) 5km 4) 10 k m

Answer 1

The path difference between two paths are:

$$\Delta L=L_1-L_2=(19+12)-21=10km$$

The condition of destructive interference is:

$$\lambda (k+\frac{1}{2})=\Delta L$$

But, as in our case one of the waves has been reflected once, its wave is shifted on half of the wavelength.

So, now, our destructive interference condition is:

$$\lambda k= \Delta L$$

And so, the maximum possible wavelength is:

$$\lambda_{max}=\Delta L = 10 km$$