8B · Interference — Newton Rings

Characteristics of Newton rings, formation with Stokes half-wave step, Dn squared derivation, and the 2019 wavelength numeric with full arithmetic.

A curved lens rests on a flat glass plate. The thin air film between them gives round dark and bright rings. Daily example: the colour rings you see when a lens presses on glass. Centre is dark.

Formula 1. Air-film thickness and path difference — $t = \dfrac{r^2}{2R},\qquad \Delta = 2t + \dfrac{\lambda}{2}$ (reflected, normal incidence)

Formula 2. Ring diameters — dark: $D_n^2 = 4n\lambda R$ ($t^2$ dropped); bright: $D_n^2 = 4\!\left(n-\tfrac{1}{2}\right)\!\lambda R$

Here $R$ = radius of curvature of the lens surface. $t$ = air-film thickness at ring radius $r$. $D_n = 2r_n$ = diameter of the $n$-th ring. $\lambda$ = wavelength. $\Delta$ = effective path difference (includes one Stokes $\lambda/2$).

Plano-convex lens of radius R on glass plate with air film thickness t and circular fringes

Fig: Newton rings — film thickness $t = r^2/2R$; centre dark because of one $\lambda/2$ reversal.

Thin air film showing Stokes half-wave reversal at the denser reflection

Fig: Why one $\lambda/2$ appears — reflection at the denser (air→glass) face flips phase by $\pi$ (= $\lambda/2$ path); the glass→air face has no flip.

Formation and derivation (no jumps)

Step 1 — Setup. A plano-convex lens of large radius $R$ sits on a flat glass plate. An air film of thickness $t$ (zero at contact, growing outward) lies between them. Monochromatic light falls from above at near-normal incidence.

Step 2 — Two reflections. Ray reflects (a) at the top of the film — glass→air: denser→rarer — NO phase flip, and (b) at the bottom of the film — air→glass: rarer→denser — phase flip of $\pi$, i.e. extra path $\lambda/2$ by Stokes’ rule.

Step 3 — Effective path difference. Geometric round trip is $2t$ (down and up, normal incidence so $\cos r = 1$). Add the one flip: $\Delta = 2t + \lambda/2$.

Step 4 — Bright/dark conditions. Bright: $\Delta = n\lambda \Rightarrow 2t = (n-\tfrac12)\lambda$. Dark: $\Delta = (2n+1)\lambda/2 \Rightarrow 2t = n\lambda$. At the centre $t = 0$, $\Delta = \lambda/2$ — destructive, hence centre is dark.

Step 5 — Film geometry $t = r^2/2R$. From the lens circle: $R^2 = r^2 + (R-t)^2 = r^2 + R^2 - 2Rt + t^2$. Cancel $R^2$: $r^2 = 2Rt - t^2 \approx 2Rt$ (since $t \ll R$, drop $t^2$). So $t = r^2/2R$.

Step 6 — Ring diameters. Dark: $2(r_n^2/2R) = n\lambda \Rightarrow r_n^2 = n\lambda R \Rightarrow D_n^2 = 4n\lambda R$ ($t^2$ dropped). Bright: $r_n^2 = (n-\tfrac12)\lambda R \Rightarrow D_n^2 = 4(n-\tfrac12)\lambda R$.

Unit check: $[4n\lambda R] = L\cdot L = L^2$, an area — correct for $D_n^2$. Limit: $R \to \infty$ (flat plate) gives no rings ($t$ constant) — sensible.

Trap: transmitted rings are opposite (centre bright — no flip there). If the question says “reflected”, centre is dark.

Solved PYQs

Write down the characteristics of Newton's rings.

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Recall

Rings come from $t = r^2/2R$ growing outward, with $\Delta = 2t + \lambda/2$ in reflected light.

Steps

Rings are circular and concentric, centred on the contact point — because $t$ depends only on radius $r$.
Centre is dark in reflected light — at $t = 0$, $\Delta = \lambda/2$ which is destructive.
Rings crowd together outward — width falls as $1/\sqrt{n}$ because $r_n \propto \sqrt{n}$.
With white light, coloured rings appear (few) — violet inside, red outside — because $r_n \propto \sqrt{\lambda}$.
$D_n^2$ grows linearly with $n$ — a straight-line graph of $D_n^2$ vs $n$ with slope $4\lambda R$.

Answer

$\boxed{\text{Circular + dark centre + crowd outward + white-light colours + }D_n^2\propto n.}$

Exam tip

Give 4–5 points as short lines — each point is roughly one mark.

Explain the formation of Newton's rings. Deduce the expression for the diameter (radius) of the $n$-th bright and dark ring by reflection in an air film.

201920222023×3

Recall

Formation: a plano-convex lens on a flat plate traps an air film; one reflection is at a denser medium (Stokes $\lambda/2$). Derivation uses $t = r^2/2R$.

Steps

Setup + two reflections: $\Delta = 2t + \lambda/2$ because the air→glass face adds $\lambda/2$ (Stokes).
Bright $2t = (n-\tfrac12)\lambda$, dark $2t = n\lambda$, centre dark (substitute $n = 0$ in dark condition).
Geometry: $R^2 = r^2 + (R-t)^2$, drop $t^2$ because $t \ll R$, giving $t = r^2/2R$.
Substitute: $D_n^2 = 4n\lambda R$ for dark rings ($t^2$ dropped); bright rings use $n \to n-\tfrac12$.
State validity: normal incidence, monochromatic light, air film ($\mu = 1$), $R \gg t$.

Answer

$\boxed{D_n^2 = 4n\lambda R\ \text{(dark, }t^2\text{ dropped); bright: }D_n^2 = 4(n-\tfrac12)\lambda R.}$

Exam tip

This is a repeat (×2) — practise the 6-step chain until it takes under 6 minutes. Never skip the reason for dropping $t^2$.

In Newton's rings $R = 100$ cm; the diameter of the 3rd ring is $0.181$ cm and the diameter of the 13th ring is $0.501$ cm. Find the wavelength of light used.

2019

Recall

Subtracting two rings cancels the half-order: $\lambda = (D_m^2 - D_n^2)/[4(m-n)R]$.

Steps

Given: $R = 100$ cm, $D_3 = 0.181$ cm, $D_{13} = 0.501$ cm, $m-n = 13-3 = 10$.
Difference form: $D_{13}^2 - D_3^2 = 4(13-3)\lambda R = 40\lambda R$.
Rearrange: $\lambda = (D_{13}^2 - D_3^2)/40R$.
Arithmetic: $D_{13}^2 = 0.501^2 = 0.251001$; $D_3^2 = 0.181^2 = 0.032761$; difference $= 0.218240$ cm$^2$; denominator $40 \times 100 = 4000$ cm; $\lambda = 0.218240/4000 = 5.456 \times 10^{-5}$ cm.
Sense: $5.456 \times 10^{-5}$ cm $= 545.6$ nm $\approx 546$ nm — green-yellow light, visible (400–700 nm).

Answer

$\boxed{\lambda \approx 5.46\times 10^{-5}\ \text{cm} \approx 546\ \text{nm}}$

Exam tip

Square BEFORE subtracting. The common wrong answer uses $(D_{13}-D_3)^2$ — that is wrong; it must be $D_{13}^2 - D_3^2$.

Describe Newton's ring method to determine the wavelength of light.

2019

Recall

Measure two ring diameters $D_n$, $D_m$; then $\lambda = (D_m^2-D_n^2)/[4(m-n)R]$ with known $R$.

Steps

Set up Newton rings with monochromatic light and measure the diameters $D_n$ and $D_m$ of two well-separated dark rings (say $n = 5$, $m = 15$) with a travelling microscope.
Use the difference formula $D_m^2-D_n^2 = 4(m-n)\lambda R$ — subtracting kills the uncertain ring-numbering at the centre.
Find $R$ separately (spherometer) or use a lens of known $R$.
Compute $\lambda = (D_m^2-D_n^2)/[4(m-n)R]$; check it falls in 400–700 nm for visible light.
Why the difference form: the exact centre ring number is hard to fix (contact spot), but $m-n$ is exact.

Answer

$\boxed{\lambda = \dfrac{D_m^2-D_n^2}{4(m-n)R}}$ — measure two diameters, subtract squares, divide by $4(m-n)R$.

Exam tip

Write the difference formula first — it is the whole method. The worked numeric is Q3.

In Newton's rings experiment, the diameters of 5th and 15th rings are $0.336$ cm and $0.590$ cm respectively. Find the radius of curvature of the plano-convex lens if the wavelength of light is $5890$ Å.

2019

Recall

Rearrange the difference formula: $R = (D_m^2-D_n^2)/[4(m-n)\lambda]$.

Steps

Given: $D_5 = 0.336$ cm, $D_{15} = 0.590$ cm, $m-n = 10$, $\lambda = 5890$ Å $= 5.89\times 10^{-5}$ cm.
Difference: $D_{15}^2-D_5^2 = 0.590^2-0.336^2 = 0.3481-0.112896 = 0.235204$ cm$^2$.
Denominator: $4(m-n)\lambda = 4\times 10\times 5.89\times 10^{-5} = 2.356\times 10^{-3}$ cm.
Divide: $R = 0.235204/2.356\times 10^{-3} = 99.83$ cm $\approx 100$ cm $= 1.0$ m.
Sense: $R \approx 1$ m is a standard long-focus plano-convex lens — matches the Q3 setup in reverse.

Answer

$\boxed{R \approx 100\ \text{cm} = 1.0\ \text{m}}$

Exam tip

This is Q3 inverted (there $\lambda$ was asked, here $R$). Same formula, rearranged — learn one, get both.

In Newton's rings experiment, the diameter of the 4th dark ring is $0.4$ cm and that of the 12th dark ring is $0.7$ cm. Find the wavelength of incident light, if the radius of curvature of the plano-convex lens is $0.5$ m.

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Recall

$\lambda = (D_m^2-D_n^2)/[4(m-n)R]$ — same as Q3/Q4.

Steps

Given: $D_4 = 0.4$ cm, $D_{12} = 0.7$ cm, $m-n = 8$, $R = 0.5$ m $= 50$ cm.
Difference: $D_{12}^2-D_4^2 = 0.49-0.16 = 0.33$ cm$^2$.
Denominator: $4(m-n)R = 4\times 8\times 50 = 1600$ cm.
Divide: $\lambda = 0.33/1600 = 2.0625\times 10^{-4}$ cm $= 2062.5$ nm.
Note: this lies in the infrared — the paper's numbers give a non-visible wavelength; the method is what is being tested, so present the arithmetic as-is.

Answer

$\boxed{\lambda = 2.06\times 10^{-4}\ \text{cm} \approx 2060\ \text{nm}}$

Exam tip

Repeat (×2). Keep $R$ in cm throughout ($0.5$ m $= 50$ cm) — mixing m and cm is the standard trap.

Measuring refractive index with Newton rings

Fill the air film with the liquid whose index $\mu$ is wanted. The wavelength inside the film becomes $\lambda/\mu$, so every ring shrinks: $D_n^2(\text{liquid}) = 4n(\lambda/\mu)R$. Taking the ratio with the air-film rings for the same $n$,

Formula 3. Liquid index from ring diameters — $\mu = \dfrac{D_n^2(\text{air})}{D_n^2(\text{liquid})}$.

Measure the same ring (say the 10th dark) with air and then with the liquid; the ratio of the squared diameters is $\mu$ directly — no $R$ or $\lambda$ needed. $R$ cancels because the same lens is used.

Reference. Brij Lal, Subrahmanyam & Avadhanulu — Optics Ch-15 (thin films, Newton rings, Stokes rule); Ajoy Ghatak — Optics Ch-14.