Answer Happy

8 Vertically polarized light of irradiance I, is incident on a series of N successive linear polarizers, each with its trans- mission axis offset from the previous one by a small angle. With the help of the Law of Malus, determine the value of N such that the final transmitted irradiance is Ix = 0.976 when the small angle offsets sum to 90°, that is when the initial vertical polarization is rotated to a horizontal polar- ization. Solution: Given that a vertically polarized light of intensity leis incident on a polarizer whose axis is at an angle According to Malus' law the intensity of light after coming out of the polarizer will be 1 =1,coso Also given that there are such polarizer's whose axes are at an angle with the adjacent ones. When light passes through these polarizers, the intensity of light transmitted will be 1, 1. (cos' O)(cos' )(cos?)... times - 1. (cos?o)" = 1. (cos***) Given that The final transmittance should be equal to 1,091, The sum of the angles is 90 Therefore Ne-90 90 Or On substituting the numerical values in equation (1) 1,-1.co 1.com -0.91 (1) -0.9 Solving this equation numerically, we can get lies between 23 and 24. If we consider N-24, we get -1.03.75")" - 1 (0,99785892323860380673806979127278)" -1.0.90223355446426404962676315003607) 0.91 Therefore the numbers of polarizers are N = 24