A Kerr cell is positioned between two crossed Nicol prisms so that the direction of electric field `E` in the capacitor forms an angle of `45^(@)` with the principle directions of the prisms. The capacitor has the length `l = 10 cm` and is falled up with nitrobenzence. Light of wavelength `lambda = 0.50 mu m` passes through the system. Taking into account that in this case the kerr constant is equal to `B = 2.2.10^(-10)cm//V^(2)`, find: (a) the minimum strength of electric field `E` in the capacitor at which the intensity of light that passes through this syetm is independent of rotation of the rear prism, (b) how many times per second light will be interrupted when a sinusoidal voltage of frequecny `v = 10 MHz` and strength amplitude `E_(m) = 50kV//cm` is applied to the capacitor. Note. The Kerr constant is the coefficient `B` in the equation `n_(e) = n_(0) = Blambda E^(2)`.

A Kerr cell is positioned between two crossed Nicol prisms so that the

1.	A Kerr cell is positioned between two crossed Nicol prisms so that the direction of electric field `E` in the capacitor forms an angle of `45^(@)` with the principle directions of the prisms. The capacitor has the length `l = 10 cm` and is falled up with nitrobenzence. Light of wavelength `lambda = 0.50 mu m` passes through the system. Taking into account that in this case the kerr constant is equal to `B = 2.2.10^(-10)cm//V^(2)`, find: (a) the minimum strength of electric field `E` in the capacitor at which the intensity of light that passes through this syetm is independent of rotation of the rear prism, (b) how many times per second light will be interrupted when a sinusoidal voltage of frequecny `v = 10 MHz` and strength amplitude `E_(m) = 50kV//cm` is applied to the capacitor. Note. The Kerr constant is the coefficient `B` in the equation `n_(e) = n_(0) = Blambda E^(2)`.
Answer» In passing through the Kerr cell the two perpendicualr components of the electirc field will acquire a phase difference. When its phase difference quals `90^(@)` the emergent light will be circularly polarized because the two perpendicualr components `O & E` have the same magnitude since it is given that the direction of electric field `E` in the capacitor forms an angle of `45^(@)` with the principle directions of then icols. In this case the intensity of light that emerges from this system will be independent of the rotation of the analyser prism. Now the phase difference introduced is given by `delta = (2pi)/(lambda) (n_(e) - n_(0)) l` In the present case `delta = (pi)/(2)` (for minimum electric field) `n_(e) - n_(0) = (lambda)/(4l)` Now `n_(e) - n_(0) = B lambda E^(2)` so `E_(min) sqrt((1)/(4Bl)). = 10^(5)// sqrt(88) = 10.66 kV// cm`. (b) If the applied electric fiel is `E = E_(m) sin omegat, omega = 2pi v` then the Kerr cell introduces a time varying phase difference `delta = 2pi B\|E_(m)^(2) sin^(2) omegat` `= 2pi xx 2.2 xx 10^(-10) xx 10 xx (50 xx 10^(3))^(2) sin^(2)omegat` `= 11pi sin^(2) omega t` In one half-cycle (i.e. in time `(pi)/(omega) = T//2 = (1)/(2v)`) this reaches the value `2kpi` when `sin^(2) omegat = 0, (2)/(11), (4)/(11), (6)/(11), (8)/(11), (10)/(11)` `(2)/(11), (4)/(11), (6)/(11), (8)/(11), (10)/(11)` i.e. `11` times. On each be interrupted when the Kerr cell (placed between crossed Nicols) introduced a phase difference of `2 k pi` and in no other case.)

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