InterviewSolution
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InterviewSolution
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A plane monochromatic light wave with intensity `I_(0)` falls normally on the surfaces of the opaque screens shown in Fig. Find the intensity of light `I` at a point `P` (a) located behind the corner ponts of screens `1-3` and behind the edge of half-plane `4`, (b) for which the rounded-off edge of screens `5-8` coincides with the boundary of the first formula describing the result obtained for screens `1-4`, the same, for screens `5-8`. |
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Answer» When the screen is fully transparent, the amplitude of vibrations is `(1)/(2)A_(1)` (with intensity `I_(0) = (1)/(4)A_(1)^(2))`. (a) `(1)` In this case `A = (3)/(4) ((1)/(2)A_(1))` so squaring `I = (9)/(16)I_(0)` `(2)` In this case `(1)/(2)` of the plane s blacket out so `A = (1)/(2) ((1)/(2)A_(1))` and `I = (1)/(4)I_(0)` `(3)` In this case `A = (1)/(4) (A_(1)//2)` and `I = (1)/(16)I_(0)`. `(4)` In this case `A = (1)/(2) ((1)/(2)A_(1))` again and `I = (1)/(4)I_(0)` so `I_(4) = (I)/(2)` In general we get `I(varphi) = I_(0) ((2varphi)/(2pi)))^(2)` where `varphi` is the total angle blocket out by the screen. `(b) (5)` Here `A= (3)/(4) ((1)/(2)A_(1)) + (1)/(4)A_(1)` `A_(1)` being the contribution of the first Fresnel zone. Thus `A = (5)/(8)A_(1)` and `I = (25)/(16)I_(0)` `(6) A = (1)/(2) ((1)/(2)A_(1)) + (1)/(2)A_(1) = (3)/(4)A_(1)` and `I = (9)/(4)I_(0)` `(7) A = (1)/(4) ((1)/(2)A_(1)) + (3)/(4)A_(1) = (7)/(8)A_(1)` and `I = (49)/(16)I_(0)` `(8) A = (1)/(2) ((1)/(2)A_(1)) + (1)/(2)A_(1) = (3)/(4)A_(1)` and `I = (9)/(4)I_(0) (I_(8) - I_(6))` In `5` to `8` the first term in the expression for the amplitude is the contribution of the plane part and the second term gives the expression for the Fresnel zone part. in general in `(5)` to `(8) I = I_(0) (1+ ((varphi)/(2pi))^(2))` when `varphi` is the angle convered by the screen. |
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