Does the spacing between fringes in a two-slit interference pattern increase, decrease, or stay the same if

(a) the slit separation is increased,

(b) the color of the light is switched from red to blue, and

(c) the whole apparatus is submerged in cooking sherry?

(d) If the slits are illuminated with white light, then at any side maximum, does the blue component or the red component peak closer to the central maximum?

Interference from a pair of slits.

The interference fringefor slit separation $\mathit{d}$, screen distance$\mathit{D}$ ,and wavelength$\mathit{\lambda }$ is

$\mathbf{∆}\mathit{x}\mathbf{=}\frac{\mathbf{\lambda }\mathbf{D}}{\mathbf{d}}$….. (1)

From equation (1) the fringe width is inversely proportional to the separation of the two slits. Thus, if slit separation is increased the fringe width will decrease.

From equation (1) the fringe width is directly proportional to the wavelength of incident light. The wavelength of red light is greater than the wavelength of blue light. When light is switched from red to blue the wavelength decrease, thus the fringe width also decreases.

From equation (1) the fringe width is directly proportional to the wavelength of incident light. Cooking sherry has a refractive index greater than air. Thus,the wavelength of light in cooking sherry will be lower than that in air. Thus, the fringe width decreases.

It has previously been shown that the fringe width for blue light is smaller than the fringe width for a red light. Thus, when the incident light is white containing all the colors, the non-zero order maxima from blue light will be closer to the central maxima than the non-zero order maxima from red.