Interference Maxima and Newton's Rings (Please help with trig)

[mm2424]

Homework Statement

A lens with radius of curvature R sits on a flat glass plate and is illuminated from above by light with wavelength λ (see picture below). Circular interference patterns, Newton's Rings, are seen when viewed from above. They are associated with variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assuming r/R <<1.

Homework Equations

2L = (m + 1/2) λ

The Attempt at a Solution

I understand that we will use 2L = (m + 1/2) λ here. However, I can't figure out how we relate d to R or r. I have my professor's answer key, and he defines θ as the top angle in the picture (formed by R and the normal line to the glass surfaces). He then says that r/R = sinθ which equals θ. Then, he says d = R(1-cosθ) and uses the expansion of cosθ = 1 - θ^2/2 + θ^4/4!, etc.

I can't wrap my head around how he found d = R(1-cosθ). If someone can help me see it, I would be greatly appreciative.

Thanks!

 

[tiny-tim]

hi mm2424!

I can't wrap my head around how he found d = R(1-cosθ).

shift that little arrow to the middle of the picture …

it's the gap between the arc (of the circle) and the triangle …

= R - Rcosθ ​

 

[mm2424]

I'm still missing something, haha. I'm not sure what you mean by shift the arrow to the middle of the picture, and I'm not clear on where Rcosθ comes from. Is it some type of trig relationship involving arcs?

 

[tiny-tim]

… I'm not clear on where Rcosθ comes from.

you have a right-angled triangle with hypotenuse R and top angle θ …

so the vertical side has length Rcosθ ​

 

[asteriatic]

It seems that your professor is using the small angle approximation for trigonometric functions, where for small angles θ, sinθ ≈ θ and cosθ ≈ 1. This is commonly used in physics and engineering problems to simplify calculations.

In this case, since r/R << 1, we can assume that θ is a small angle. Using the definition of θ as the top angle in the picture, we can see that θ is also equal to the angle between the normal line and the line connecting the center of the lens and the center of the interference pattern.

Now, for a small angle θ, the length of the chord connecting the two points on the circle is approximately equal to the arc length between those points. In this case, the arc length is d, and the chord length is R(1-cosθ). Therefore, we can write d = R(1-cosθ).

I hope this helps you understand how your professor arrived at this equation. Remember, the small angle approximation is only valid for small angles, so make sure to check if it is applicable in other problems.