Aim#

To show Newton’s rings, and that its color sequence is not a rainbow.

Subjects#

• 6D30 (Thin Films)

Diagram#

Equipment#

• Newton’s rings apparatus (convex lens pressed against flat glass plate; pressure can be adjusted by screws in the ring-mount).

• Hg-lamp, with power-unit.

• Objective lens.

• Two lenses $f=150\mathrm{mm}/diam.=70\mathrm{mm}$.

• Flat surface mirror.

• Black screen.

Safety#

• The Hg-lamp needs some time to come to its full light intensity. It also becomes very hot! Do not touch it.

Presentation#

Set up the equipment as shown in Diagram. Images are projected on the wall (see Figure 634A)

After the lamp is heated up, situation of Diagram A is presented to the students, to indicate that there will be a reflected and a transmitted beam of light. Then the transmitted beam is blocked (black screen) and using the mirror and a $+150\mathrm{mm}$-lens the reflection image is projected (Diagram B). Clearly Newton’s rings are observed. Observe the central dark spot (see also: Remarks) observe the colored rings, the color-sequence and observe the diminishing distance between the rings when moving away from the centre. Changing the pressure on the Newton’s rings apparatus will change/move the reflected image. Then the black screen is removed and using the second $+150\mathrm{mm}$-lens the transmitted image is projected next to the reflected image (see Diagram C and Figure 634). It is clearly visible that both images are complementary.

At first glance, the observed colors look rainbowlike, but careful observation shows that it differs from a rainbow (see Figure 635; reality is much better than this photograph).

Observing the reflected image shows, when moving away from the central dark spot, at first a rainbow, but already in the next ring the color purple appears; in the next rings white and orange are dominating; around ring 10 there is a repeating sequence of blue and orange and around ring 16 repeating bands of dark violet and yellowish rings are visible giving form a distance the impression of a continuity of black and white fringes.

Explanation#

See Figure 634 B. Looking at the two red rays drawn in this figure, we see that it is the height $d$ that introduces the phasedifference. $d=R-{(R^2-x^2)}^{1/2}$.

The two rays, one reflecting from the hemisphere and the other reflecting from the plane, will have a phasedifference of $\mathrm{\Delta }\varphi =k(2d)-\pi$ ( $\pi$ at reflection off the plane).

Maximum, constructive interference will occur at $\mathrm{\Delta }\phi =\frac{4\pi d}{\lambda }-\pi =m2\pi$, so when $d=1/2\lambda (m+1/2)$.

This result translated to the distance $x$ (because $x$ lies in the plane we are watching/projecting) yields $1/2\lambda (m+1/2)=R-{(R^2-x^2)}^{1/2}$, giving $x=\{\lambda R(m+1/2)$ ${1/4{\lambda }^2(m+1/2{)}^2\}}^{1/2}$. And $R$ being much larger than $\lambda$ will give $x=\{\lambda R(m+1/2){\}}^{1/2}$. First conclusion is that $x$ is proportional to the squareroot of wavelength. So a higher wavelength yields a higher $x$. blue is on the inside, red on the outside. Second, the proportionality in $(m+1/2{)}^{1/2}$ shows that the sequence of the bright fringes follows a square root: moving away from the centre the fringes come closer and closer together.

Finally, we calculated for a number of $m$-values $x$. Figure 636 shows the calculated results ( ${10}^{-5};\mathrm{R}=1\mathrm{m}$ ) for the red, green and blue line of ${\mathrm{H}}_0$-light. In this way it is clear that the colours observed are the result of different combinations. Only near the centre a rainbow pattern appears.

It is not difficult now to show that for destructive interference we get $x=(\lambda R\mathrm{m}{)}^{1/2}$. This yields that the centre of the reflected Newton’s rings must be a dark spot. Figure 636 shows the minima as dashed lines for red, green and blue.

Remarks#

• Using filters, it is possible to show a monochromatic interference pattern. Especially in the yellow line of Hg the pattern is bright.

• In the projected reflection image the central area should be dark. But usually there is a coloured spot instead. This is probably due to trapped dirt in the contact area between the two surfaces.

Sources#

• Giancoli, D.G., Physics for scientists and engineers with modern physics, pag. 878-879

• Hecht, Eugene, Optics, pag. 398-399

• Young, H.D. and Freeman, R.A., University Physics, pag. 1152-1153