Question

Part A: At what angle θ₁  to the normal would the first dark ring be observed?

answer is .442^o

Part B: Suppose that the light from the pinhole projects onto a screen 3 meters away. What is the radius of the first dark ring on that screen? Notice that the angle from Part A is small enough that sinθ≈tanθ .

Answer is 23.2mm

Part C: The first dark ring forms the boundary for the bright Airy disk at the center of the diffraction pattern. What is the area A of the Airy disk on the screen from Part B? answer must be in mm²

Answer 1

(A)   sinθ =   1.22 λ /d   

                    θ = sin^-1 (1.22 λ /d ) .................(1)

Here d = diametre of the pin hole = 0.1*10^-3 m

wave length λ =   632.8*10^-9 m  

             θ = 0.442⁰

(B) when θ is small   

             tanθ   = sinθ = y / L

Radius of the first dark fringe y   = Ltanθ   =   ( 3m )( tan   0.442⁰ )   = 0.023.2 m =   23.2 mm

(C) Area     = πy²   = 3.14 ( 0.023.2 m ) ² = 1690.0736( mm)²