= n. 1 n. Next apply Snell s law at the cylindrical boundary between the core and cladding for the marginal ray. = n 2. sin π 2.

Transcription

1 EXERCISE.-9 First, re-derive equation.-5 considering the possibility that the index at the entrance to the fiber bay be different than one. Snell s law at the entrance to the fiber is where all symbols are as shown in Figure.-8 except now n is the index at the entrance to the fiber. Use trigonometric relations to write the sine of the acceptance angle in terms of the sine of the critical angle cosθ c sin θ c ext apply Snell s law at the cylindrical boundary between the core and cladding for the marginal ray. Solving for sine of the critical angle n sin π n Substituting into the expression for the acceptance angle we obtain the desired result. n n n n n n The numerical aperture is defined in terms of the acceptance angle. According to this section of Saleh and Teich A ST but, according to Wikipedia the usual definition is A W You may use either definition for full credit. For the bare core in air, we have n.46 this leads to a value greater than one in the expression for the sine of the acceptance angle, which has no solution for the angle. This means there is no marginal ray for the full range of possible angles, to π/. All angles in the possible range are accepted, so the acceptance angel is π/. θ a π 9 degrees Both definitions for the numerical aperture give the same answer when the index at the entrance is one. A ST A W In water we have.33 n.46 the expression above gives a value less than one and there is a marginal ray. θ a degrees A ST.453 A ST.6

2 EXERCISE.4-8 From equation.4-3, the transfer function for one layer is t q ( x, y) exp( jn q k d q ) The overall transfer function is the product of the individual transfer functions of the layers ( ) t ( x, y) t q ( x, y) exp jn q k d q q q A product of exponentials can be evaluated by summing the exponents t ( x, y) exp( jn q k d q ) exp jk n q d q q q Equating this expression to a single transfer function through some distance, d, of free space exp jk n q d q q exp( jk d) Solving for d d Compare with the definition of optical path length Given our function of index where n( r) OPL q B A n q d q ( ) n r ds n < n < n 3 < 3 n < q + d i i Integrating across the layers in the direction we find B q q n( r) ds n( r) d n q d n q d A q q q q which is the same as the expression we found for d. q q n q d q

3 EXERCISE 3.-3 The condition f says that the limit of ray optics applies to the first lens. We can see this from equations 3.-6 and First we note that this condition means that and thus r f M M r + r M f f r f f Then applying 3.-6 to the first lens we find f f f ( f ) f f where all the notation is as in figure Solving for we find f + f f + f f f f f f ( ) f which gives the familiar expression from ray optics f + f The other condition says that the waist of the input beam is far away from the input lens f Taking the inverse f which says that the second term in the ray optic expression may be neglected and thus and f f In other words the input light comes in nearly parallel so the first lens focuses the light at the focal length. From solving problem 3.- we know that to collimate the output we need to have f + To eliminate the intermediate Rayleigh range, we use equation 3.-7 M Substituting the expression for M from above f f f where we have used the fact that f is small compared to. We can further simplify by noting that

4 f + f f f + f f The first term in the right hand parenthesis is greater than one, since we are given that the first lens has a short focal length and the second lens has a long focal length. The second term is a product of two factors, both of which are much smaller than one. (Since -f is much greater than, is also much greater than.) So we have f f f f Finally we find d + f + f The total magnification is the product of the individual magnification M M M From above, M is given by M f The parameter, r, simplifies due to the collimation condition r f Then for M we have M M r + r M r Again using the collimation condition f f M f f f To eliminate the intermediate Rayleigh range we use the expression derived above M f f f f f Combining the two magnifications M f f f f f

5 PROVE A PLAE WAVE IS TRASVERSE Use Maxwell s source free equation D For linear isotropic media D εe Substituting into the above equation ( εe) For homogenous media the permittivity is independent of the coordinates so may be pulled out of the divergence operator E E x x + E y y + E where the operator has been written out in Cartesian coordinates. Our plane wave, also written out in Cartesian coordinates E E e jωt e jk r ( E x ˆx + E y ŷ + E ẑ)e jωt exp y + k ) Compute the partial derivatives E x x E e jωt x ( jk x )exp y + k ) E y y E e jωt y ( jk y )exp y + k ) E E e jωt ( jk )exp y + k ) The sum of these derivatives must be ero as shown above E x x + E y y + E j ( E k + E k + E k x x y y )e jωt exp y + k ) includes a term which is the dot product of the electric field complex envelope E x x + E y y + E je ke jωt exp y + k ) Since the exponentials are not in general ero, this dot product must be E k If the electric field and the wave vector point in real directions, then they are perpendicular to each other. E k

6 EXERCISE 6.-8 The total Jones matrix is the product of the three component matrices T T c T a The a component is a π/ wave retarder with fast y-axis T a e jγ j The b component is a π wave retarder with fast axis 45 degrees to the x axis. cosθ sinθ cos( θ ) sin( θ ) osθ e jγ sin( θ ) cos( θ ) Letting Γ π and θ π/4 Multiplying the matrices The c component is a π/ wave retarder with fast x-axis T c e jγ j The total matrix j T j Multiplying the matrices j j T T This matrix swaps the components of the Jones vector and negates one component, exactly what is required to give the orthogonal state for any Jones vector where the two components have the same phase. That is to say that this matrix gives the orthogonal state only for linear polariations. Check it with a general linearly polaried input Jones vector

7 The output Jones vector Check the inner product J TJ Reversing the order of the retarders T J cosθ sinθ cosθ sinθ sinθ cosθ J J cosθ sinθ + sinθ ( cosθ) T j j j T This device gives the other orthogonal state, it rotates 9 degrees in the opposite direction. j