Solve Width of Aperture: Find Width Now!

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Width of Aperture...please help!

Homework Statement


A point source of light illuminates an aperture 2.3 m away. A 10.2-cm-wide bright patch of light appears on a screen 1.02 m behind the aperture. How wide is the aperture?

Homework Equations





The Attempt at a Solution


I can't figure out how to do this?
 
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woops this was supposed to go under physics...please remove! thanks!
 
… just geometry …

BuBbLeS01 said:

Homework Statement


A point source of light illuminates an aperture 2.3 m away. A 10.2-cm-wide bright patch of light appears on a screen 1.02 m behind the aperture. How wide is the aperture?

Actually, this is geometry, not physics.

Just draw a diagram, apply ordinary geometry of similar triangles, and you should get the answer! :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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