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Questions 124 and 125 are based on paragraph given below: Unpolarised light of intensity 32 W m^(-2) passes through three polarisers such that transmission axis of last polariser is crossed with first, the intensity of emergent light is 3 Wm^(-2). 124. The nagle between transmission axis of first two polarisers is |
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Answer» `10^(@)` `theta_(2)` = angle between second and third polariser I = initial intensity of unpolarised light and `I_(1), I_(2) and I_(3)` = intensities transmitted through first, second and third polarisers, then `I_(1) = I_(2)`and `I_(2) = I_(1) cos^(2)THETA, I_(3) = i_(2) cos^(2)theta_(2)` Now `theta_(1) + theta_(2) = (pi)/(2)` `theta_(2) = (pi)/(2) - theta_(1)` Then `I_(3)= I_(2)cos^(2)((pi)/(2) - theta_(1)) = I_(2) sin^(2) theta_(1)` `I_(3) = I_(1) cos^(2)theta_(1). sin^(2)theta_(1) = (I)/(2) cos^(2)theta_(1). sin^(2)theta_(1)` ` = I/2((sin 2theta_(1))/(2))^(2) = I/8 sin^(2)2theta_(1)` But ` i = 32 Wm^(-2)` and `I_(3) = 2Wm^(-2)` Then `theta_(1) = 30^(@)`. |
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