Chapter 7.4, Problem 106AYU

Optical Measurement 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 $I_t=4A^2\frac{\left(\csc \theta -1\right)\left(\sec \theta +\tan \theta \right)}{\csc \theta \sec \theta }$.

Show that this is equivalent to $I_t={\left(2A\cos \theta \right)}^2$.

To determine

To find: 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,

$I_t=4A^2\frac{\left(\csc \theta -1\right)\left(\sec \theta +\tan \theta \right)}{\csc \theta \sec \theta }$

Show that this is equivalent to $I_t=4A^2\frac{(\csc \theta -1)(\sec \theta +\tan \theta )}{\csc \theta \sec \theta }\text{ and }{(2A\cos \theta )}^2$.

Answer to Problem 106AYU

Solution:

$I_t=4A^2\frac{(\csc \theta -1)(\sec \theta +\tan \theta )}{\csc \theta \sec \theta }\text{ and }{(2A\cos \theta )}^2$ are equal.

Explanation of Solution

Given:

$I_t=4A^2\frac{\left(\csc \theta -1\right)\left(\sec \theta +\tan \theta \right)}{\csc \theta \sec \theta }$

Calculation:

$I_t=4A^2\frac{(\csc \theta -1)(\sec \theta +\tan \theta )}{\csc \theta \sec \theta }\text{ and }{(2A\cos \theta )}^2$

By reciprocal identity and quotient identity.

$I_t=4A^2\frac{\left(\frac{1}{\sin \theta }-1\right)(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta })}{\frac{1}{\sin \theta }\frac{1}{cos\theta }}\text{ and }{(2A\cos \theta )}^2$

Adding up the quotients.

$I_t=4A^2\frac{\left(\frac{1-\sin \theta }{\sin \theta }\right)( \frac{1+\sin \theta }{\cos \theta })}{\frac{1}{\sin \theta }\frac{1}{cos\theta }}\text{ and }{(2A\cos \theta )}^2$

Cancelling the common factor $\sin \theta \cos \theta$ and by Pythagorean identity.

$I_t=4A^2\left(1-{\sin }^2\theta \right)\text{ and }{(2A\cos \theta )}^2$.

Taking inside the brackets.

$I_t={\left(2A\text{cos}\theta \right)}^2\text{ and }{(2A\cos \theta )}^2$.

Both are equivalent.