Fundamental of Optical Engineering Lecture 8

 A linearly polarized plane wave with Ē vector described by is incident on an optical element under test. Describe the state of polarization of the output wave (linear, elliptrical, or circular) if the optical element is:

 (a) A linear polarizer oriented to transmit light polarized in the e x direction.

 (b) A half-wave plate with birefringence axes oriented to coincide with e x and e y.

 (c) A half-wave plate with birefringence axes oriented at 45º relative to e x and e y.

 (d) A quarter-wave plate with birefringence axes oriented to coincide with e x and e y.

 (e) A quarter-wave plate with birefringence axes oriented at 45º relative to e x and e y.

 (f) A half-wave plate with birefringence axes oriented at 25º relative to e x and e y.

 (g) A quarter-wave plate with birefringence axes oriented at 25º relative to e x and e y.

 A linearly polarized light propagating in the z-direction with polarization vector in the x- direction is incident on a birefringent crystal. What is the state of polarization of the light after passing through the crystal if: (a) the crystal is a quarter-wave plate with optic axis in the xy plane oriented at 30º relative to the y- axis?

◦ (b) the crystal is a half-wave plate with optic axis in the y-direction?

◦ (c) the crystal is a half-wave plate with optic axis in the xy plane oriented at 11º relative to the x-axis?

◦ (d) the crystal is a quarter-wave plate with optic axis in the xy plane oriented at 45º relative to the y-axis?

◦ (e) the crystal is a quarter-wave plate with optic axis in the z-direction?

 For a birefringent median with n0 = and nE = as shown in the figure. Find the length L that makes it be (a) a full wave plate (b) a half wave plate (c) a ¼ -wave plate if the wavelength is 656 nm.

 This makes use of electrooptic effect (applied electric fields used to change the optical properties).  There are 2 kinds of electrooptic effect: linear and quadratic.

 The linear electrooptic effect is called “Pockels effect”.  This refers to the change in the indices of the ordinary and extranordinary rays proportional to applied electric field.  This effect exists only in crystals without an inversion symmetry such as LiNbO 3.

 For a crystal with an inversion symmetry, the linear electrooptic effect can not exist, while the quadratic electrooptic effect known as “Kerr effect” is observed.  This is where the induced index change is proportional to the square of applied electric field.

 V  (pi-voltage) or half-wave voltage is the applied voltage that makes the relative phase shift be  in a cube of material.  In general, = refractive index changes produced by applied voltage.

 It is preferable to design L >> h to have a low applied voltage V.  After applying a voltage, indices are changed as

 V  for the material is 2,700 V, L = 2 cm, and h = 0.5 mm. Find applied voltage V to have Δ  =  (complete extinction).

 An electrooptic crystal has dimensions of 2x2x3 along the x,y, and z axes with n E = and n O = An input wave propagating in the z-direction at λ = 0.63 μ m is linearly polarized at a 45 º angle relative to the x- and y- axes. A voltage V applied across the crystal in the x-direction. The voltage is increased from V = 0 until, when V = 245 V, the output polarization from the crystal is the same as that observed for V = 0. Assume that optic axis is along the y- axis. ◦ (a) What is the total phase retardation, in rad, for V = 0? ◦ (b) What is pi-voltage for the material? ◦ (c) What is the refractive index change Δ n x produced by the applied voltage of 245 V, assuming Δ n y =0?

 Recall: interference eq.

 Assume that

 Reflectance  We can consider R into 3 cases: ◦ n 1 = n 3. ◦ n 1 n 3, n 1 ≠ n 3. ◦ n 1 < n 2 < n 3.

 Case 1: n 1 = n 3  From a definition: A ij = A ji

 Max in R for  Min in R for

 n 1 = 1.5, n 2 = 1.6, and λ = 0.63 μ m. Find t 2 for R max and R min.

 Case 2: n 1 n 3, n 1 ≠ n 3.

 Max in R for  Min in R for

 n 1 = 1.5, n 2 = 1.6, n 3 = 1.4, λ = 0.63 μ m. Find t 2 for R max and R min.

 Case 3: n 1 < n 2 < n 3

 Max in R for  Min in R for

 n 1 = 1.5, n 2 = 1.6, n 3 = 1.7, λ = 0.63 μ m. Find t 2 for R max and R min.