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HW 9 Solution PDF

HW 9 Solution PDF - 7.9 Soleil Compensator Consider a...

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Unformatted text preview: 7.9 Soleil Compensator: Consider a Soleil compensator as in Figure 7Q9 that uses a quartz crystal. Given a light wave with a wavelength λ ≈ 600 nm, a lower plate thickness of 5 mm, calculate the range of d values in Figure 7Q9 that provide a retardation from 0 to π (half- wavelength). Figure 7Q9 Solution The phase difference φ between the two polarizations passing through the Soleil-Babinet compensator φ = 2 π λ ( n e- n o )( D- d ) ∴ d = D- λφ 2 π ( n e- n o ) When φ changes by Δ φ , then the change in d is Δ d , Δ d = λ Δ φ 2 π ( n e- n o ) Δ φ = π , Δ d = λ Δ φ 2 π ( n e- n o ) = (600 × 10- 6 mm)( π ) 2 π (1.5533- 1.5442) = 0.033 mm = 33 microns 7.15 Transverse Pockels cell with LiNbO 3 Suppose that instead of the configuration in Figure 7.20, the field is applied along the z-axis of the crystal, the light propagates along the y- axis. The z-axis is the polarization of the ordinary wave and x-axis that of the extraordinary wave. Light propagates through as o- and e-waves. Given that E a = V/d , where d the crystal length along z , the indices are ′ n o ≈ n o + 1 2 n o 3 r 13 E a and ′ n e ≈ n e + 1 2 n e 3 r 33 E a Optic axis Optic axis d D Wedges can slide Plate E 1 E 2 Soleil-Babinet Compensator Show that the phase difference between the o- and e-waves emerging from the crystal is, Δ φ = φ e- φ o = 2 π L λ n e- n o ( ) + 2 π L λ 1 2 n e 3 r 33- n o 3 r 13 ( ) V d where L is the crystal length along the y- axis....
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HW 9 Solution PDF - 7.9 Soleil Compensator Consider a...

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