Question

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 39.0 cm away from the slit, when light of wavelength 550 nm is used. Find the slit width.

Answer 1

For minima:

y = mD $\lambda$ /a : where a =slit width ,m= order of minima, $\lambda$ = wavelength,D= slit-screen distance

And here y₅ - y₁ = 5*(D $\lambda$ /a) - (D $\lambda$ /a) = 4*(D $\lambda$ /a) = 0.5 mm = 4*(39cm*550nm/a)

=> a = 4*0.39*2000*550nm = 1716000 nm = 1.716 mm

So, slit width = 1.716 mm

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