Lens Calculations using Refractive Index

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nao113
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Homework Statement
Consider a lens made of glass with refractive index 𝑛𝑛=1.5 is placed in the air (refractive index =1.0), as shown in the figure. The first surface is plane and the second surface is spherical convex shape whose radius of curvature is 𝑟𝑟=100 [mm]. z-axis is called optical axis. The first surface perpendicularly intersects the optical axis at O, and second surface intersects the optical axis at Q. The thickness of the lens OQ = 𝑑𝑑 = 20 [mm]. A light beam parallel to the optical axis is incident to the lens, where the distance (height) of the beam from the optical axis is h. The refracted beam intersects the optical axis at X. Derive the distance between the lens and the intersection X, Q
Relevant Equations
n1theta1 = n2 theta2
Ray diagram of a plano-convex glass lens: a horizontal ray at height h enters the flat first surface, refracts at spherical surface S with centre C and radius r, and meets the optical axis at X beyond the lens, with labels O, Q and n.

Ray diagram of a plano-convex lens with a horizontal axis z, incoming ray at height h refracting at point S on the curved surface, marking angles theta 1, theta 2, theta 3, distances d and Delta, point Q on the axis, and point X; with four questions listed in red below.


Answer:
I already found the answer for Sin theta 2 like the pic below
Handwritten working on a tablet over printed text: sin θ1 = h/r = 2θ/(1 m) = 1/5 = 0.2; sin θ1 = (n2/n1) sin θ2; 0.2 = (1.0/1.5) sin θ2; sin θ2 = 0.3, with a red ray meeting a line marked X and Z.
But, I am still not sure about how to derive delta here. Can anyone show me they way? thank you
 
Can you find ##\theta_3## in terms of ##\theta_1## and ##\theta_2##? Then use Snell's law to eliminate ##\theta_2##.
 
kuruman said:
Can you find ##\theta_3## in terms of ##\theta_1## and ##\theta_2##? Then use Snell's law to eliminate ##\theta_2##.
Handwritten worked calculation on lined paper: n1 = 1.5, n2 = 1.0, for h = 1 mm, sin θ1 = h/r = 1 mm / 100 mm = 10^-2, Snell's law steps to sin θ2 = 15 x 10^-3, a triangle diagram with sides 20 mm, 1 mm and ΔQX, and a Pythagoras calculation giving Δ = 10.9 mm.
Handwritten calculations on lined paper for h = 10 and h = 20 mm, with sin θ values, refractive index steps, and Pythagoras results for ΔX and Δ, ending in boxed values of 199.75 mm and 199.9975 mm.
Handwritten geometry sketch of a triangle with vertices labelled S, O and X, sides marked 200 and 28, and an unknown side marked ?, beside a worked calculation: OX = 202 using Pythagoras, OX = OQ + QX, 202 = 20 + (190.9975 - Δ), Δ = 16.9975 mm.


I put it like this, how is it?
 
nao113 said:
It doesn't look good. How about finding a formula for Δ in terms of ##h##, ##r## and ##d##? Then instead of doing three separate calculations, you can substitute three separate values in the one formula. Even better, load the formula on a spreadsheet. That way it is less likely that you will get an inconsistent answer, you and we can troubleshoot your work more easily.

I see one glaring error that makes your numerical answers incorrect. That is your assertion that $$\sin\theta_3=\sin\theta_2-\sin\theta_1.~~\leftarrow~~\text{This is not correct}$$Why do you think that the sines are related this way? The way to do this problem is to answer the four questions in the order (2), (3), (4) and (1). So let's start with (2), "What is the relation between ##\theta_1##, ##\theta_2## and ##\theta_3##?"
 
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kuruman said:
How about finding a formula for Δ in terms of h, r and d?
Hint: This relationship is a standard piece of lens/mirror plane geometry known as the sagitta theorem. Learn it.
 
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hutchphd said:
Hint: This relationship is a standard piece of lens/mirror plane geometry known as the sagitta theorem. Learn it.
I didn't know it had a theorem name. I always (re)derived it as needed using the Pythagorean theorem.
 
I remember it because I ground a telescope mirror in Jr High school.
Now I am lucky to remember to put on pants...
 
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