Visibility in Optics: Integration help

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The discussion focuses on integrating the visibility of a non-point source interference pattern using the equation dI = (C/w)(cos^2[(ud/gL)(y-y_0)]).dy_0, specifically between w/2 and -w/2. Participants highlight the need to evaluate the integral of A*cos^2(B(y-y_0)) for constants A and B. A substitution is suggested to simplify the integration, utilizing the identity cos^2{x} = 1/2(cos{2x} + 1). Step-by-step guidance is requested to clarify the integration process.
Master J
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In calculating the visibility of an non point source interfernce pattern, you integrate the following:

dI = (C/w)(cos^2[(ud/gL)(y-y_0)]).dy_0.

between w/2 and -w/2. C, u, d, g, l , y constants.

I'm finding this pretty tricky to integrate. Could someone help? Step by step would be really helpful.
 
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Hi Master J,
Master J said:
In calculating the visibility of an non point source interfernce pattern, you integrate the following:

dI = (C/w)(cos^2[(ud/gL)(y-y_0)]).dy_0.

between w/2 and -w/2. C, u, d, g, l , y constants.

I'm finding this pretty tricky to integrate. Could someone help? Step by step would be really helpful.
As you have it written (and assuming you mean to say that "L" is a constant), this just amounts to evaluating \int_{-w/2}^{w/2} A\cos^2{\left(B(y-y_0)\right)} \, dy_0 for constants y, A and B. Then, using a substitution, this amounts to integrating \cos^2{x}. To do that, use the identity cos^2{x} = \frac{1}{2}(\cos{2x} +1). Show your work so we know where abouts you're getting stuck.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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