Question

The first dark ring forms the boundary for the bright Airy disk at the center of the diffraction pattern. What is the area of the Airy disk on the screen from Part B?
Express your answer in , to three significant figures.
=




Diffraction due to a circular aperture is important in astronomy. Since a telescope has a circular aperture of finite size, stars are not imaged as points, but rather as diffraction patterns. Two distinct points are said to be just resolved (i.e., have the smallest separation for which you can confidently tell that there are two points instead of just one) when the center of one point's diffraction pattern is found in the first dark ring of the other point's diffraction pattern. This is called Rayleigh's criterion for resolvability.

Consider a telescope with an aperture of diameter 0.970 mm^2

Answer 1

(A) sin? = 1.22 ? /d
? = sin-1 (1.22 ? /d ) .................(1)
Here d = diametre of the pin hole = 0.1*10-3 m
wave length ? = 632.8*10-9 m
? = 0.4420
(B) when ? is small
tan? = sin? = y / L
Radius of the first dark fringe y = Ltan? = ( 3m )( tan 0.4420 ) = 0.023.2 m = 23.2 mm
(C) Area = py2 = 3.14 ( 0.023.2 m ) 2 = 1690.0736( mm)2

Answer 2

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(A) sinλ = 1.22 λ /d
λ = sin-1 (1.22 ? /d ) .................(1)
Here d = diametre of the pin hole = 0.1*10-3 m
wave length λ = 632.8*10-9 m
λ = 0.4420m

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(B) when λ is small
tanλ = sin? = y / L
Radius of the first dark fringe y = Ltan? = ( 3m )( tan 0.4420 ) = 0.023.2 m = 23.2 mm

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(C) Area = pi2 = 3.14 ( 0.023.2 m ) 2 = 1690.0736( mm)2