Question

The path difference between two wavefronts generated by coherent source is 2.1 micron.If the phase difference between these wavefronts at a point is 7.692π,the wavelength of light emitted from the source will be: a)5386Å b)5400Å c)5460Å d)5892Å.

Answer 1

Answer on Question #38006 – Physics – Other

Question: the path difference between two wave fronts generated by coherent source is 2.1 micron. If the phase difference between these wave fronts at a point is $7.692\pi$, the wavelength of light emitted from the source will be: $a)5386\,\text{\AA}$; $b)5400\,\text{\AA}$; $c)5460\,\text{\AA}$; $d)5892\,\text{\AA}$.

Solution: the phase of the wave is determined by the equation

$$\phi = k x - \omega t.$$

$k$ is the wave number, $k = \frac{2\pi}{\lambda}$. The phase difference between two wave fronts at the time $t$ is

$$\phi_2 - \phi_1 = k (x_2 - x_1) = k \Delta x = \frac{2\pi\Delta x}{\lambda}$$

Now we can calculate the value of the wavelength

$$\lambda = \frac{2\pi\Delta x}{\Delta\phi} = \frac{2\pi \cdot 2.1 \cdot 10^{-6}}{7.692\pi} = 5460\,\text{\AA}.$$

Answer: $c)5460\,\text{\AA}$.