Question 288: Wave Optics

Question

Two light rays having the same wavelength ' $\lambda$ ' in vacuum are in phase initially. Then, the first ray travels a path ' $\mathrm{L}_1$ ' through a medium of refractive index ' $\mu_1$ ' while the second ray travels a path of length ' $L_2$ ' through a medium of refractive index ' $\mu_2$ '. The two waves are then combined to observe interference. The phase difference between the two waves is

Accepted Answer

To understand how the phase difference between two light waves is determined, consider the following:
Initial Conditions
: Two light rays with the same wavelength, denoted as 'ÃÂ»', are initially in phase.
Traveling through Different Media
:
The first ray travels a distance 'LÃ¢ÂÂ' through a medium with a refractive index 'ÃÂ¼Ã¢ÂÂ'.
The second ray travels a distance 'LÃ¢ÂÂ' through a medium with a refractive index 'ÃÂ¼Ã¢ÂÂ'.
Optical Path Difference
: When light travels through a medium, the distance is effectively multiplied by the medium's refractive index to obtain the optical path length. Hence, the optical path for the first ray is 'ÃÂ¼Ã¢ÂÂLÃ¢ÂÂ', and for the second ray, it is 'ÃÂ¼Ã¢ÂÂLÃ¢ÂÂ'.
Phase Difference Formula
:
$    	ext{Optical Path Difference} = \mu_1 L_1 - \mu_2 L_2    $
The phase difference (ÃÂÃÂ) between the two waves is directly related to their optical path difference and is given by:
$    \Delta \phi = \frac{2 \pi}{\lambda} 	imes (	ext{Optical Path Difference})    $
Simplifying, we have:
$    \Delta \phi = \frac{2 \pi}{\lambda} \left[\mu_1 L_1 - \mu_2 L_2ight]    $
This formula helps determine how the difference in paths affects the phase relationship of the two waves when they recombine, thus influencing interference patterns.

Answer

$\frac{2 \pi}{\lambda}\left(\mu_1 L_1-\mu_2 L_2\right)$

Answer

$\frac{2 \pi}{\lambda}\left(L_2-L_1\right)$

Answer

$\frac{2 \pi}{\lambda}\left(\frac{\mathrm{~L}_1}{\mu_1}-\frac{\mathrm{L}_2}{\mu_2}\right)$

Answer

$\frac{2 \pi}{\lambda}\left(\mu_2 \mathrm{~L}_1-\mu_1 \mathrm{~L}_2\right)$

Wave Optics

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