Question

Optical methods of measurement often rely on the interference of two light waves. If two light waves, identical except for a phase lag, are mixed together, the resulting intensity, or irradiance, is given by $I_1=4A^2\frac{(csc\theta -1)(sec\theta +\tan \theta )}{csc\theta sec\theta }$

Show that this is equivalent to$I_1={(2A\cos \theta )}^2$

Short answer

The given expressions are equivalent.

Step-by-step solution

Step 1. Given Information

The given expressions are$I_1=4A^2\frac{(csc\theta -1)(sec\theta +\tan \theta )}{csc\theta sec\theta }and$$I_1={(2A\cos \theta )}^2$

Step 2. Calculation 

Consider one expression and prove it is equivalent to the other given expression.

$$\begin{gathered} 4A^2\frac{(csc\theta -1)(sec\theta +\tan \theta )}{csc\theta sec\theta }=4A^2\frac{({\displaystyle \frac{1}{\sin \theta }}-1)({\displaystyle \frac{1}{\cos \theta }}+{\displaystyle \frac{\sin \theta }{\cos \theta }})}{{\displaystyle \frac{1}{\sin \theta }}\times {\displaystyle \frac{1}{\cos \theta }}} \\ =4A^2\frac{\left({\displaystyle \frac{1-\sin \theta }{\sin \theta }}\right)\left({\displaystyle \frac{1+\sin \theta }{\cos \theta }}\right)}{{\displaystyle \frac{1}{\sin \theta \cos \theta }}} \\ =4A^2\frac{{\displaystyle \frac{(1-\sin \theta )(1+\sin \theta )}{\sin \theta \cos \theta }}}{{\displaystyle \frac{1}{\sin \theta \cos \theta }}} \\ =4A^2\frac{(1-\sin \theta )(1+\sin \theta )}{1} \\ =4A^2(1-{\sin }^2\theta ) \\ =4A^2\left({\cos }^2\theta \right) \\ ={\left(2A\cos \theta \right)}^2 \end{gathered}$$

Thus, we can conclude that the given expressions are equivalent.