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PHY227 Homework 1, answers Question 6 Consider electric field oscillations of the electromagnetic wave propagating towards you in a fixed plane (i.e. only the temporal evolution will be taken into account). Any polarisation state will be represented by two linearly polarised components polarised along the fast (y) and slow (x) axes. The right-circularly polarised light is then given by E x E0x sin ωt E y E0y sin(ωt ) Here =/2+2m (m – integer). After passing through the QWP an additional phase shift of /2 will be acquired by the y-component so that E x E0x sin ωt E y E0y sin(ωt 2πm π) This is linearly polarised light at an angle =-45 degrees with respect Ox. The relation between E0x and E0y is obtained from E0x E0 | cos | E0y E0 | sin | E0x / E0y | cos | / | sin | The linear polariser (LP) blocking this light has the transmission axis at =+45 degrees with respect Ox. For the elliptically polarised light, again linearly polarised light is produced by the QWP, as the light can be completely blocked by the LP. Now its axis is rotated by 30 degrees clockwise, i.e. forms =-75 degrees with respect Ox. Following the expression above E0x / E0y | cos(75) | / | sin(75) | 0.268 Linearly polarised light produced by the QWP can be described as before as E x E0x sin ωt E y E0y sin(ωt 2πm) In a similar way, an additional phase shift of /2 has been acquired by the y-component so that the original waves before the QWP are: E x E0x sin ωt E y E0y sin(ωt 2πm π/2) Thus the answer is an elliptically polarised light with the electric field vector rotating clockwise and the ratio of the horizontal and vertical components of 0.268.