2.5.13 · D5Optics
Question bank — Newton's rings — derivation
True or false — justify
Each line is a claim. Decide true/false and give the reason before revealing.
The centre of the pattern in reflected light is dark.
True — at contact the geometric path is zero, but exactly one reflection (air→glass) adds a flip, so → destructive. See Stokes' relations & phase change on reflection.
If you view the same setup in transmitted light, the centre is still dark.
False — transmitted light has no odd flip (both transmissions and the internal path add no net half-wave), so the centre is bright and the whole pattern is the complement of the reflected one.
The rings are equally spaced, like marks on a ruler.
False — , so radius grows as ; consecutive rings crowd closer together as you move outward.
Doubling the lens radius of curvature makes the rings larger.
True — , so a flatter (larger ) lens spreads the air wedge more gently and the rings expand.
Filling the air gap with a liquid of makes the rings expand.
False — , so larger shrinks every ring by ; this shrinkage is how the method measures a liquid's index (see Refractive index measurement).
Using blue light instead of red makes the rings smaller.
True — , and blue has the shorter , so all rings contract toward the centre.
The bright and dark rings would swap places if the flip did not occur.
True — the flip is exactly what converts the "usual" (normally bright) into the dark condition; remove it and centre becomes bright, orders invert.
The order number of a dark ring counts how many half-wavelengths of path difference it holds.
False for the centre, roughly true elsewhere — for dark rings , so counts whole wavelengths of the term, with the flip already folded into the condition.
You can determine from a single ring's diameter if you know .
Technically true from , but practically false — the uncertain contact point poisons any single ; only diameter differences give trustworthy results.
Spot the error
Each line states a piece of reasoning with a flaw. Name the flaw before revealing.
"The path difference is because the film has one thickness ."
Missing the factor 2 — ray B goes down and up through the film, so the geometric path is ; dropping the 2 halves any you extract.
"At the path difference is zero, so the waves are in phase and the centre is bright."
Ignores the Stokes flip — one reflection adds , so the total at contact, giving a dark centre.
"Rings crowd outward because the wavelength gets shorter far from the centre."
Wrong cause — is fixed by the source; the crowding comes from geometry, , making thickness (and thus order) rise as .
" simplifies to by neglecting compared to always, exactly."
The step is an approximation, valid only because microns while metres; it is not an identity, and for a thick gap or tiny it fails.
"To get , measure one diameter carefully and use ."
Uses an absolute diameter — the fuzzy contact centre adds an unknown constant to every ; the difference method is required to cancel it.
"Both reflected rays flip by , so the flips cancel and don't matter."
Only the air→glass reflection (ray B, going into denser medium) flips; the glass→air reflection (ray A, into rarer medium) does not — one flip, not two, so it does not cancel.
"Rings are fringes of equal path length."
They are fringes of equal thickness (hence circles of constant ); equal thickness gives equal path only because maps to path via — the primary invariant on each ring is .
Why questions
Give the mechanism, not just the fact.
Why are the fringes circles rather than straight lines?
The air gap depends only on distance from the contact point, so a locus of constant is a circle — fringes of equal thickness become concentric rings.
Why do we measure diameters instead of radii?
The contact point is deformed and its exact location is uncertain; a diameter spans two points on the same ring so any centre offset cancels, unlike a radius drawn from the doubtful centre.
Why does subtracting remove two nuisances at once?
A shifted centre adds the same constant to every , and the bookkeeping term is also a constant offset; subtracting two squared diameters kills both, leaving the clean linear .
Why does only one of the two reflecting rays get the flip?
Ray A reflects glass→air (dense→rare, no flip); ray B reflects air→glass at the plate (rare→dense, flip) — reflection flips phase only when entering a higher-index medium (Stokes' relations).
Why does the gap grow as and not as ?
The lens is a sphere; near the axis its surface dips below the rim by the sagitta, and circle geometry () forces that dip to scale with the square of the off-axis distance.
Why must the light be monochromatic (or filtered) for sharp rings?
Different wavelengths satisfy at different , so their ring sets overlap and smear; a single keeps every ring crisp (coherence discussed in Interference — path difference & coherence).
Why is Newton's rings a "fringes of equal thickness" phenomenon like the air wedge?
Both trap a film whose thickness varies with position, and each fringe marks a contour of constant — the ring case just has circular contours instead of the straight ones of Wedge fringes / air wedge.
Edge cases
Boundary and degenerate scenarios — reason through each.
What happens to the pattern if the lens and plate have identical refractive index but the gap is filled with a medium of higher index than both?
Now the plate reflection is dense→rare (no flip) while the lens-bottom reflection becomes rare→dense (flip) — the single flip still exists but on the other ray, so the centre stays dark; what matters is that exactly one flip occurs.
What if the gap medium's index exactly equals the glass on both sides (index-matched)?
With no index step at either surface there is essentially no reflection and no contrast — the rings vanish because there is nothing to reflect and interfere.
As very large (far from centre), why do the rings eventually become invisible?
Ring spacing shrinks as so they crowd below resolvable width, and the growing thickness eventually exceeds the coherence length, washing out contrast.
At the exact geometric contact point, is the thickness truly zero?
Ideally yes, but real glass deforms under weight and dust intrudes, so the "point" is a small flattened patch — this is precisely why absolute radii are untrustworthy and diameter differences are used.
If were made extremely small (a sharply curved lens), how does the pattern change?
Since , small makes all rings tiny and tightly packed — the wedge steepens, thickness climbs fast with , so orders pile up quickly near the centre.
What does the pattern look like the instant you switch from reflected to transmitted viewing?
Every bright becomes dark and vice versa — the transmitted pattern is the exact photometric complement, with a bright centre, because the odd flip present in reflection is absent in transmission.
If white light is used, what appears at the centre and just around it?
The centre is dark for all colours (path difference scales with each , staying destructive), but the first few rings show coloured fringes since each colour meets its bright condition at a slightly different .
Recall Quick self-test
One flip or two at the centre, and what does it force? ::: Exactly one flip → net at → dark centre. Reflected vs transmitted centre? ::: Reflected dark, transmitted bright (complementary patterns). Which measurement quantity cancels the centre error? ::: Differences of squared diameters, .