Question

What will be the change in frienge pattern in young double slit experiment even if two micafoils having same refracting indecs and thickness L1 and L2 (L1>L2) are placed in the path of interfaring raise

Answer 1

Answer on Question #61774, Physics / Optics

What will be the change in fringe pattern in young double slit experiment even if two mica foils having same refracting indices and thickness L1 and L2 (L1>L2) are placed in the path of interfering raise

Solution:

Consider the light rays from the two coherent point sources made from infinitesimal slits a distance $d$ apart. We assume that the sources are emitting monochromatic light of wavelength $\lambda$ .

The rays are emitted in all forward directions, but let us concentrate on only the rays that are emitted in a direction $\theta$ toward a distant screen ( $\theta$ measured from the normal to the screen, diagram below). One of these rays has further to travel to reach the screen, and the path difference is given by $d\sin \theta$ . If this path difference is exactly one wavelength $\lambda$ or an integer number of wavelengths, then the two waves arrive at the screen in phase and there is constructive interference, resulting in a bright area on the screen.

$$y _ {b r i g h t} = \frac {\lambda L}{d} m$$

Our case equal to situation, when a transparent foil of thickness (L1-L2) and refractive index $n$ is placed in one of the incoming wave path, due to the increase of the path by (n-1)(L1-L2), the interference pattern undergoes a shift $s$ .

Fig. Equal effective path lengths without (left) and with (right) mica foil.

If the foil has an effective thickness (L1-L2), then there are $\frac{L_1 - L_2}{\lambda / n}$ complete wavelengths that travel through it, while there are $\frac{L_1 - L_2}{\lambda / 1}$ wavelengths that travel through the same thickness of air.

The number of fringes shifted is

$$m = \left| \frac {L _ {1} - L _ {2}}{\lambda / n} - \frac {L _ {1} - L _ {2}}{\lambda} \right| = \frac {L _ {1} - L _ {2}}{\lambda} (n - 1)$$

Shift of pattern $s = y_{bright} = \frac{\lambda L}{d} m = \frac{\lambda L}{d}\frac{L_1 - L_2}{\lambda} (n - 1) = \frac{L}{d} (n - 1)(L_1 - L_2)$ .

Answer: Due to the increase of the path by $(n - 1)(L1 - L2)$ , the fringe system shifts on the side of the thicker (L1) mica foil.

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