Length — Question 6

Question 6

Inferring a small width from a large pattern

Light passing through a narrow slit produces a pattern on a distant screen. For this exercise use the small-angle model $$a=\frac{\lambda D}{y},$$ where $a$ is slit width, $\lambda$ is the light wavelength, $D$ is slit-to-screen distance, and $y$ is the distance from the pattern centre to either first dark minimum. The two first minima are symmetric about that centre. No derivation from wave theory is required.

Take $\lambda=650\ \mathrm{nm}$ and $D=2.00\ \mathrm{m}$. Treat the reported data as exact within the supplied model; the model itself is an approximation.

1. Find the centre coordinate and $y$ from the two minimum coordinates. Determine $a$ in metres and micrometres.

2. One student uses the full separation of the minima as $y$. Another uses the right-hand coordinate $17\ \mathrm{mm}$ as $y$. Find both inferred widths and identify the distinct error in each method.

3. If $D$ is doubled while the slit and wavelength are unchanged, predict $y$ and the separation of the minima. Does a wider pattern now imply a wider slit? Explain using the model.

Question 6 - Solution

Inferring a small width from a large pattern

(a) Locate the centre from symmetry.

The midpoint is $x_c=[(-9+17)/2]\ \mathrm{mm}=4.0\ \mathrm{mm}$. Thus $$y=\frac{17-(-9)}{2}\ \mathrm{mm}=13.0\ \mathrm{mm}=0.0130\ \mathrm{m}.$$ The slit width is $$a=\frac{(650\times10^{-9}\ \mathrm{m})(2.00\ \mathrm{m})}{0.0130\ \mathrm{m}} =\boxed{1.00\times10^{-4}\ \mathrm{m}=100\ \mu\mathrm{m}}.$$

(b) Distinguish a full width from an offset coordinate.

Using $y=26\ \mathrm{mm}$ gives $a=50\ \mu\mathrm{m}$, half the correct inference: the model needs centre-to-minimum distance, not the full central-pattern width.

Using $y=17\ \mathrm{mm}$ gives $a=(1300/17)\ \mu\mathrm{m}\approx76.5\ \mu\mathrm{m}$. This incorrectly assumes the ruler zero coincides with the pattern centre. Using both minima removes that origin offset.

(c) A larger optical scale need not mean a larger object.

$$y=\frac{\lambda D}{a}\ \Longrightarrow\ \frac{y_{\mathrm{new}}}y=\frac{D_{\mathrm{new}}}D=2\quad(a,\lambda\text{ fixed}).$$

Rearranging gives $y=\lambda D/a$. Doubling $D$ doubles $y$ to $26\ \mathrm{mm}$ and the full separation to $52\ \mathrm{mm}$. Substituting the new $D$ and $y$ gives the same inferred slit width, $100\ \mu\mathrm{m}$.

At fixed $D$ and wavelength, a wider pattern would instead indicate a narrower slit. Pattern size must therefore be interpreted with the imaging geometry; the pattern is not a direct, same-size picture of the slit.