Small Angle Approximation in Single Slit Interference

[lulzury]

Homework Statement

A monochromatic light source is used with a double slit to create an interference pattern on a screen that is 2.00 meters away. If the 2nd bright spot is observed 8.73 mm above the central maximum, can the small angle approximation be used? Show and/or explain your reasoning

Homework Equations

$d\sinθ=mλ$

The Attempt at a Solution

I might be over-thinking this. I know you can have a small angle approximation if D >> d, but in this case I don't know d yet, so I first wanted to relate d to y without first assuming a small angle approximation (sinθ ≈ tanθ ≈ θ), but I get stuck:

I get the following relation:
$D=\frac{d}{2\tan(γ)}$
$D= (y+\frac{d}{2}) * \frac{1}{ \tan(θ)}$

$\frac{d}{2}\cot(γ) = y\cot(θ) + \frac{d}{2}\cot(θ)$

I'm not sure how to solve for d here so that I can show that D >> d.

Thanks in advance!

 

[mfb]

D>>d/2 is necessary for other approximations (all your formulas wouldn't work without that condition).

The small angle approximation that is relevant here is a small θ. You can calculate θ with D and y alone in a single step.

 

[lulzury]

mfb, that makes sense thank you!

So in this case tan(θ) = y/D

$θ = \arctan(\frac{y}{D})$
$θ = \arctan(\frac{8.73*10e-3}{2})$

θ = 0.00436 rad ~ 0.25 degrees
That is a very small angle, so I can use a small angle approximation here!

 

[mfb]

Right.