InterviewSolution
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InterviewSolution
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A plane monochromatic light wave with intensity `I_(0)` falls normally on an opaque disc closing the first Fresnel zone for the observation point `P`. What did the intensity of light `I` at the point `P` become equal to after (a) half of the disc (along the diameter) was removed, (b) half of the external half of the first Fresnal zone removed (along the diameter)? |
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Answer» Suppose the disc does not abstruct light at all. Then `A_(disc) + A_("remainder") = (1)/(2)A_(disc)` (because the disc covers the first Fresnel zone only). So `A_("remainder") =- (1)/(2)A_(disc)` Hence the amplitude when half of the disc is removed along a diameter `= (1)/(2)A_(disc) + A_("remainder") = (1)/(2)A_(disc) - (1)/(2)A_(disc) ~~0` Hence `I = 0`. (b) In this case `A = (1)/(2)A_("external") +A_("remainder")` `= (1)/(2)A_("external") - (1)/(2)A_(disc)` We write `A_(disc) = A_(in) + iA_(ou)` where `A_(in) (A_(out))` stands for `A_("internal") (A_("external"))`. The factor `i` takes account of the `(pi)/(2)` pahse difference between two halves of the first Fresnel zone. Thus `A =-(1)/(2)A_("in")` and `I = (1)/(4)A_("in")^(2)` On the other hand `I_(0) = (1)/(4) (A_("in")^(2) + A_(out)^(2)) = (1)/(2) A_("in")^(2)` so `I = (1)/(2)I_(0)`. |
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