In double slit experiment, 0.10mm wide slits are separated by 0.5 mm and are ill

Question

In double slit experiment, 0.10mm wide slits are separated by 0.5 mm and are illuminated by light of wavelength 600 nm. The slits are placed 2.0 m from a screen.

a) Consider a point on the screen located at 12 mm above the symmetry axis of the slits, is this point bright or dark? show calculation supporting you answer

b) How many interference fringes are visible in the central diffraction maximum?

c) In order to have more fringes within the central maxima, we want to change the wavelength, or spacing or width. How should each one change ( mention increasing or decreasing, no change).

d) Explain any changes in intensity pattern if three similar slits are added between the original two slits.

Explanation / Answer

b). Relevant equations

d*sin(?) = m*? for interference maxima with slit separation d, maxima are at m = 0, 1, 2, 3 ...

a*sin(?) = m*? for diffraction minima with slit width a, minima are at m = 1, 2, 3 ...

The number of interference fringes visible in the central diffraction peak: 2(d/a) - 1

slit separation d=0.5mm

slit width a=0.10mm
Using this, We get: 2(0.5/0.10) - 1 = 9, So 9 peaks.

c). Wavelength decreased --> the fringes will be more closely placed and so number of fringes increases. sin(theta)=(m*wavelength)/wavelength

space between slit increased: fringes will be more closely spaced and hence the number of fringes willl be increased.

distance to the screen decreased: fringes more closely packed and so the number of fringes will increase in central maxima spacing.

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