Trig problems (Interference/diffraction and slits))

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In summary, understanding interference and diffraction problems involving slits requires a lot of geometry and trigonometry. The equations d sin \theta=m \lambda and y=\frac{m\lambda L}{d} are used to calculate angles and distances in these problems. For small angles, sin \theta and tan \theta are approximately equal, which can be seen by looking at a right triangle and considering the definitions of sin and tan. This approximation is also shown in the Taylor series expansion of sinx and tanx. With practice, one can develop the ability to see these connections without drawing triangles. The equation y=\frac{m\lambda L}{d} is used to determine the distance between bright/dark fringes in both
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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: [tex]d sin \theta=m \lambda[/tex] and [tex]y=\frac{m\lambda L}{d}[/tex] because there are some many geometry stuff you need to do to connect the angles.

Can someone explain to me why [tex] tan\theta=sin\theta[/tex] if theta is small? thanks
 
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Can someone explain to me why [itex] sin \theta = tan \theta [/itex] if theta is small? thanks



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

[tex] Sin \theta = \frac o h [/tex]

[tex] Tan \theta = \frac o a [/tex]

if [itex] \theta [/itex] 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 [itex] \theta [/itex] when you drop all but the linear term, you have [itex] sin \theta = tan \theta = \theta [/itex]
 
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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? [tex]y=\frac{m\lambda L}{d}[/tex] 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.
 
  • #4
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
 

1. What is the difference between interference and diffraction in trig problems?

Interference and diffraction are both phenomena that occur when waves interact with each other. Interference is the result of two or more waves combining to form a new wave, while diffraction is the bending of waves around obstacles or through narrow openings.

2. How do slits affect interference and diffraction in trig problems?

Slits are commonly used in experiments involving interference and diffraction. They act as narrow openings through which waves can pass, causing them to diffract and interfere with each other.

3. What is the role of trigonometry in solving problems related to interference and diffraction?

Trigonometry is used to model and analyze the behavior of waves in interference and diffraction problems. It helps determine the angles and distances between waves, as well as the intensity of the resulting interference patterns.

4. Can trigonometry be used to calculate the position of interference fringes in a double-slit experiment?

Yes, trigonometry can be used to calculate the position of interference fringes in a double-slit experiment. The equation for the position of a fringe is given by d sinθ = mλ, where d is the distance between the slits, θ is the angle of the fringe, m is the order of the fringe, and λ is the wavelength of the wave.

5. How can trigonometry be applied to real-world situations involving interference and diffraction?

Trigonometry is used in various industries, such as telecommunications and optics, to design and analyze systems that utilize interference and diffraction. It is also used in the study of sound and light waves, as well as in medical imaging techniques.

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