 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2011/paper-1/section-a/3/solution

For source and screen distances

s, r

u

u^{2} (1/ s + 1/ r) /2

. The outer boundary of Fresnel half-period zone

n

has excess

nλ /2

, hence

u_{n}^{2} = \frac{nλ sr}{s + r}, u_{2} = \frac{2 ( 500 \times 1 0 ^{- 9} ) ( 0.5 ) ^{2}}{1.0} = 0.500 mm .

This is the outer radius of the second zone, measured from the axis.

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