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MIT2_71S09_practice2_sol

# MIT2_71S09_practice2_sol - 2.71/2.710 Optics Practice Exam...

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2.71/2.710 Optics Practice Exam 2 - Solutions Spring ‘09 1. What is the Fraunhofer diffraction pattern of a 1-D slit of size a ? x Slit description (1D): f ( x ) = rect a Fourier transform of slit: F ( u ) = a sinc( au ) x 2 + y 2 x λz Diffracted far field: g ( x ) = e × F λz ax Fraunhofer diffraction pattern (intensity): | g ( x ) | 2 = a 2 sinc 2 λz 2. What is the Fraunhofer diffraction pattern of this sinusoidal amplitude grating, where Λ is the grating period? 1

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�� Solution: 1 x f ( x ) = 1 + cos 2 π 2 Λ = DC term (0th order) + diffracted orders 1 1 i 2 π x 1 e i 2 π x Λ Λ = + e + 2 4 4 ���� � �� � �� plane wave, plane wave, plane wave, u = 1 1 1 1 1 F ( u ) = 2 δ ( u ) + 4 δ u Λ + 4 δ u + Λ �� x 2 + y 2 1 x 1 x 1 1 x 1 λz g ( x ) = e 2 δ λz + 4 δ λz Λ + 4 δ λz + Λ 1 Λ Note: without being too rigorous mathematically, we treat the intensity corresponding to the δ -function field as a “very bright and sharp” spot. u =0 1 Λ u = 3. How does the result of problem 2 change if the illumination is a plane wave incident at angle θ 0 with respect to the optical axis? ( θ 0 << 1) Solution: Let u 0 = sin θ 0 , so the plane wave is e i 2 πu 0 x (at z = 0). λ 2
1 �� 1 x 1 1 1 1 F e i 2 πu 0 x × 2 1 + cos 2 π Λ = 2 δ ( u u 0 )+ 4 δ u u 0 Λ + 4 δ u u 0 + Λ 4. What is the Fraunhofer pattern of this truncated sinusoidal amplitude grating? Assume that a >> Λ. �� 1 x x f ( x ) = 1 + cos 2 π rect 2 Λ a Solution: �� 1 x 1 1 1 1 1 f 1 ( x ) = 1 + cos 2 π F 1 ( u ) = δ ( u ) + δ u + δ u + 2 Λ 2 4 Λ 4 Λ x f 2 ( x ) = rect F 2 ( u ) = a sinc( au ) a According to the convolution theorem, F{ f 1 ( x ) · f 2 ( x ) } = F 1 ( u ) ⊗ F 2 ( u ). Recall that: δ ( u u 0 ) A ( u ) = δ ( u u 0 ) A ( u u ) du = A ( u u 0 ) −∞ �� 1 1 1 1 1 F 1 ( u ) ⊗ F 2 ( u ) = 2 δ ( u ) + 4 δ u Λ + 4 δ u + Λ a sinc( au ) �� �� 1 1 1 1 1 = a sinc( au ) + a sinc a u + a sinc a u + 2 4 Λ 4 Λ Note: When you take || 2 , cross-terms can be ignored. Why?

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