Why Is the Small Angle Approximation Used in Optics Problems?

[Gauss M.D.]

Homework Statement

Had the same problem as this threadstarter:

https://www.physicsforums.com/showthread.php?t=109059

Homework Equations

The Attempt at a Solution

I managed to find a ratio of tangents for the two angles. From there, it seems you're supposed to go "well tan(x) ≈ x for small angles so let's magically assume this is a small angle and go grab a donut".

Why is the small angle approximation appropriate for this problem and how do I avoid getting stuck on similar problems in the future?

 

[NascentOxygen]

You could try a range of small angles and judge for yourself, e.g.,

x=1°: x=... radians, sin x=..., tan x=...

x=2°: x=... radians, sin x=..., tan x=...

x=3°:


By working this out for yourself, you'll be left with a better appreciation of the result.

▣ Remember, the trig approximations expect x to be in radians.

 

[technician]

Do what nascent oxygen recommends...you will be surprised how 'BIG' the angle can be yet still be considered 'SMALL'

 

[Integral]

Of course the small angle approximation only works if you use radians.

 

[technician]

Of course the small angle approximation only works if you use radians.

As recommended !

 

[Gauss M.D.]

No, I get the small angle approximation, I just don't get how I am supposed to know that it is applicable here. I mean, we're not given any angles. We're supposed to figure it out through trig/geometry trickery.Theoretically, the angles could be pi/2 for all I know.