Exercises — Newton's rings — derivation

The raw geometric path is $2\mu t$, but reflection at the air→glass boundary adds a Stokes flip worth $\lambda /2$. So the total path difference is $\Delta =2\mu t+\lambda /2$. Destructive interference needs $\Delta =(m+\frac{1}{2})\lambda$. Substitute: $2\mu t+\frac{\lambda }{2}=(m+\frac{1}{2})\lambda \Rightarrow 2\mu t=m\lambda .$ The extra $\lambda /2$ swapped the usual roles. So $2\mu t=m\lambda$ is dark. ✓

$r_1=\sqrt{m\lambda R}=\sqrt{1\times 600\times 10^{-9}\times 1.00}=\sqrt{6.0\times 10^{-7}}$. $r_1=7.75\times 10^{-4}\text{m}=0.775\text{mm}$.

$r_{10}=\sqrt{10\times 589\times 10^{-9}\times 0.90}=\sqrt{5.301\times 10^{-6}}$. $r_{10}=2.302\times 10^{-3}\text{m}\approx 2.30\text{mm}$.

Use $r_m=\sqrt{\frac{(2m-1)\lambda R}{2\mu }}$ with $m=5$, $\mu =1$. $2m-1=9$, so numerator $=9\times 550\times 10^{-9}\times 1.5=7.425\times 10^{-6}$. Divide by $2$: $3.7125\times 10^{-6}$. Root: $r_5=1.927\times 10^{-3}\text{m}\approx 1.93\text{mm}$.

Bright: $2t=(m-\frac{1}{2})\lambda \Rightarrow t=\frac{(2m-1)\lambda }{4}$. $t=\frac{9\times 550\times 10^{-9}}{4}=1.2375\times 10^{-6}\text{m}\approx 1.24\mu \text{m}$. Sanity check with $t=r^2/2R=(1.927\times 10^{-3}{)}^2/(2\times 1.5)=1.238\times 10^{-6}$ m ✓.

$\lambda =\frac{D_n^2-D_m^2}{4(n-m)R}$. $D_{20}^2=(6.00\times 10^{-3}{)}^2=3.600\times 10^{-5}{\text{m}}^2$. $D_{10}^2=(4.24\times 10^{-3}{)}^2=1.7978\times 10^{-5}{\text{m}}^2$. Numerator $=1.8022\times 10^{-5}$; denominator $=4(10)(1.00)=40$. $\lambda =4.505\times 10^{-7}\text{m}\approx 451\text{nm}$.

$r_m=\sqrt{m\lambda R}$, so $\Delta r=r_{m+1}-r_m=\sqrt{\lambda R}(\sqrt{m+1}-\sqrt{m})$. The $\sqrt{\lambda R}$ prefactor cancels in a ratio. $1\to 2$: $\sqrt{2}-\sqrt{1}=0.41421$. $20\to 21$: $\sqrt{21}-\sqrt{20}=4.58258-4.47214=0.11044$. Ratio $=0.41421/0.11044=3.751$. The inner gap is about 3.75× wider — rings crowd outward. ✓

For fixed $m$: $D_m^2=\frac{4m\lambda R}{\mu }$. So $D^2\propto 1/\mu$, giving $\mu =\frac{D_{\text{air}}^2}{D_{\text{liq}}^2}=\frac{(5.10{)}^2}{(4.42{)}^2}=\frac{26.01}{19.5364}=\mathrm{1.331.}$ $\mu \approx 1.33$ (water). Ring shrinkage is the measurement.

With a medium in the gap: $D_n^2-D_m^2=\frac{4(n-m)\lambda R}{\mu }$, so $\mu =\frac{4(n-m)\lambda R}{D_n^2-D_m^2}.$ $D_{15}^2=(3.60\times 10^{-3}{)}^2=1.2960\times 10^{-5}$; $D_5^2=(2.05\times 10^{-3}{)}^2=4.2025\times 10^{-6}$. Difference $=8.7575\times 10^{-6}{\text{m}}^2$. Numerator $=4(10)(589\times 10^{-9})(1.20)=2.8272\times 10^{-5}$. $\mu =2.8272\times 10^{-5}/8.7575\times 10^{-6}=3.228$?? That is unphysical — a check that our fabricated diameters are inconsistent. Lesson: always sanity-test $\mu \gtrsim 1$; if $\mu >2$ for a common liquid, a diameter was misread. (We keep this to show the method; the arithmetic itself is correct given the inputs.)

From $D_m^2=4m\lambda R$: $R=\frac{D_m^2}{4m\lambda }$. Formally yes — but this uses the raw $D^2$, which secretly assumes the ring is centred perfectly on the true contact point. If the centre is offset, $D^2$ carries an error you cannot see. The trustworthy route is $R=\frac{D_n^2-D_m^2}{4(n-m)\lambda }$ from two rings.

Coincidence of radii means equal $r^2$: $m_1{\lambda }_{1}R=m_2{\lambda }_{2}R\Rightarrow m_1{\lambda }_1=m_2{\lambda }_2$. $\frac{m_1}{m_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{600}{480}=\frac{5}{4}$. Smallest integers: $m_1=5$, $m_2=4$. So the 5th dark ring of 480 nm sits on the 4th dark ring of 600 nm. Common radius: $r=\sqrt{5\times 480\times 10^{-9}\times 1.0}=\sqrt{2.40\times 10^{-6}}=1.549\times 10^{-3}\text{m}\approx 1.55\text{mm}$.

Transmitted: both relevant reflections happen at the same type of boundary (or none nets a $\lambda /2$), so $\Delta =2\mu t$ with no extra half-wave. Dark now needs $\Delta =(m+\frac{1}{2})\lambda \Rightarrow 2\mu t=(m+\frac{1}{2})\lambda$. At $t=0$: $\Delta =0=0\cdot \lambda$ → constructive → bright centre. Transmitted rings are the exact photographic negative of the reflected ones. ✓

Now the gap is $t=t_0+\frac{r^2}{2R}$. Dark: $2(t_0+\frac{r^2}{2R})=m\lambda$. $\frac{r^2}{R}=m\lambda -2t_0\Rightarrow r^2=R(m\lambda -2t_0)$. $m\lambda =600\times 10^{-9}=6.0\times 10^{-7}$; $2t_0=6.0\times 10^{-7}$. $m\lambda -2t_0=0$ → $r=0$: the raised gap already contributes the full order, so the "first dark ring" collapses to the centre. Physically the lift has pushed the whole pattern inward by one order. Taking $m=2$: $r^2=1.0\times (1.2\times 10^{-6}-6.0\times 10^{-7})=6.0\times 10^{-7}$, $r=7.75\times 10^{-4}$ m $=0.775$ mm.