Trig problems (Interference/diffraction and slits))

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Interference and diffraction problems involving slits require a solid understanding of geometry and trigonometry, particularly in deriving key equations like d sin θ = m λ and y = (mλL)/d. For small angles, sin θ and tan θ are approximately equal because the adjacent side of a right triangle is nearly the same length as the hypotenuse, which can also be understood through Taylor series expansion. The equation y = (mλL)/d is indeed used to determine the distance between bright and dark fringes in both single and double slit diffraction, with 'd' representing the distance between slits in double slit setups and the width of the slit in single slit diffraction. Developing the ability to make these connections often comes down to practice with trigonometric problems. Understanding approximations like sin x ≈ x for small x is crucial in these discussions.
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Interference and diffraction problems involving slits includes a lot of geometry and trig right? I'm having a hard time deriving the equtions: d sin \theta=m \lambda and y=\frac{m\lambda L}{d} because there are some many geometry stuff you need to do to connect the angles.

Can someone explain to me why tan\theta=sin\theta if theta is small? thanks
 
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Can someone explain to me why sin \theta = tan \theta if theta is small? thanks



Look at a right triangle. Now consider the definition of Sin and Tan.

Sin \theta = \frac o h

Tan \theta = \frac o a

if \theta is small the adjacent side is very nearly the same length as the hypotenuse. So Sin and Tan are very nearly equal. Another way to see this is to look at the Taylor series expansion of Tan and Sin both have the same first term \theta when you drop all but the linear term, you have sin \theta = tan \theta = \theta
 
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Okay I get it now, thanks. How does one develop the ability to see these kinds of connections without drawing triangles etc.? Is it just practice with these type of trig problems?

Anyways, is this equation used to determine the distance bewteen bright/dark fringes in single slit diffractions of light? y=\frac{m\lambda L}{d} Except that (d) is replaced with (w), the width of the slit right? This equation applies to both double slit and single slit? So the only difference is that (d) is the distance between the slits in the two slit diffractions while (d) is the width of the slit in single slit diffractions? Thanks.
 
Okay I get it now, thanks. How does one develop the ability to see these kinds of connections without drawing triangles etc.? Is it just practice with these type of trig problems?

sinx=x for small x is a pretty standard approximation that crops up time and again... and of course, tanx = sinx/cosx, and cosx is about 1 for small x
 
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