Answer to Question #92459 in Optics for kaveendri

Question

A microwave detector is located at the shore of a lake at a height d above the water level. As a radio star emitting monochromatic microwaves of wavelength \lambda rises slowly above the horizon, the detector indicates successive maxima and minima of signal intensity.

(a) What is the difference in phase between Ray 1 and Ray 2 arriving at the detector (in terms of d, \lambda , and \theta )?

(b) At what angle \theta above the horizon is the radio star when the first maximum is received (in terms of d and \lambda )?

Answer 1

(a)   In the case of constructive interference, the difference in phase between Ray 1 and Ray 2 is given by formula

$$Δ=\frac {2 \pi}{\lambda}× Δr (1)$$

where

$$Δr=d × \tan{\theta} (2)$$

We put (2) in (1)

$$Δ=\frac {2 \pi d}{\lambda}×\tan{\theta} (3)$$

 

(b) The first term is given by the phase difference between two parallel rays. In this case, we have

$$Δ=\pi (4)$$

We put (4) in (3)

$$\pi =\frac {2 \pi d}{\lambda}×\tan{\theta} (5)$$

Using (5) we get

$$\tan{\theta}=\frac {\lambda}{2 d} (6)$$

$$\theta =\arctg{\frac {\lambda}{2 d}}$$

Answer

(a)

$$Δ=\frac {2 \pi d}{\lambda}×\tan{\theta}$$

(b)

$$\theta =\arctg{\frac {\lambda}{2 d}}$$