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IPhO-37_Th1-solve

# IPhO-37_Th1-solve - OUT 1 when the optical path lengths of...

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SOLUTIONS to Theory Question 1 Geometry Each side of the diamond has length L = a cos θ and the dis- tance between parallel sides is D = a cos θ sin(2 θ ) = 2 a sin θ . The area is the product thereof, A = LD , giving 1.1 A = 2 a 2 tan θ . The height H by which a tilt of φ lifts OUT 1 above IN is H = D sin φ or 1.2 H = 2 a sin θ sin φ . Optical path length Only the two parallel lines for IN and OUT 1 matter, each having length L . With the de Broglie wavelength λ 0 on the IN side and λ 1 on the OUT 1 side, we have Δ N opt = L λ 0 - L λ 1 = a λ 0 cos θ 1 - λ 0 λ 1 . The momentum is h/λ 0 or h/λ 1 , respectively, and the statement of energy conservation reads 1 2 M h λ 0 2 = 1 2 M h λ 1 2 + MgH , which implies λ 0 λ 1 = 1 - 2 gM 2 h 2 λ 2 0 H . Upon recognizing that ( gM 2 /h 2 ) λ 2 0 H is of the order of 10 - 7 , this simplifies to λ 0 λ 1 = 1 - gM 2 h 2 λ 2 0 H , and we get Δ N opt = a λ 0 cos θ gM 2 h 2 λ 2 0 H or 1

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1.3 Δ N opt = 2 gM 2 h 2 a 2 λ 0 tan θ sin φ . A more compact way of writing this is 1.4 Δ N opt = λ 0 A V sin φ , where 1.4 V = 0 . 1597 × 10 - 13 m 3 = 0 . 1597 nm cm 2 is the numerical value for the volume parameter V . There is constructive interference (high intensity in
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Unformatted text preview: OUT 1 ) when the optical path lengths of the two paths diﬀer by an integer, Δ N opt = 0 , ± 1 , ± 2 ,... , and we have destructive interference (low intensity in OUT 1 ) when they diﬀer by an integer plus half, Δ N opt = ± 1 2 , ± 3 2 , ± 5 2 ,... . Changing φ from φ =-90 ◦ to φ = 90 ◦ gives Δ N opt ± ± ± ± φ =90 ◦ φ =-90 ◦ = 2 λ A V , which tell us that 1.5 ] of cycles = 2 λ A V . Experimental data For a = 3 . 6 cm and θ = 22 . 1 ◦ we have A = 10 . 53 cm 2 , so that 1.6 λ = 19 × . 1597 2 × 10 . 53 nm = 0 . 1441 nm . And 30 full cycles for λ = 0 . 2 nm correspond to an area 1.7 A = 30 × . 1597 2 × . 2 cm 2 = 11 . 98 cm 2 . 2...
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IPhO-37_Th1-solve - OUT 1 when the optical path lengths of...

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