Subfield: Optics

Figure 1.1

<img_4357>

A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by

$$
y=a \cos [2 \pi(f t-x / \lambda)]
$$

where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).
Context question:
(i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

$$
A=2 a \cos \theta
$$

where

$$
\beta=\frac{\pi}{\lambda} d \sin \theta
$$
Context answer:
\boxed{证明题}


Context question:
(ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by

$$
R=\frac{a}{2 \sin \beta}
$$

Deduce that the resultant amplitude is

$$
\frac{a \sin N \beta}{\sin \beta}
$$



and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.
Context answer:
\boxed{证明题}


Context question:
(iv) Determine the intensities of the principal intensity maxima.
Context answer:
\boxed{$N^{2} a^{2}$}


Context question:
(v) Show that the number of principal maxima cannot exceed

$$
\left(\frac{2 d}{\lambda}+1\right)
$$
Context answer:
\boxed{证明题}


Question: (vi) Show that two wavelengths $\lambda$ and $\lambda+\delta \lambda$, where $\delta \lambda<<\lambda$, produce principal maxima with an angular separation given by

$$
\Delta \theta=\frac{n \Delta \lambda}{d \cos \theta} \quad \text { where } n=0, \pm 1, \pm 2 \ldots . \text { etc }
$$

Calculate this angular separation for the sodium $\mathrm{D}$ lines for which

$$
\begin{aligned}
& \lambda=589.0 \mathrm{~nm}, \quad \lambda+\Delta \lambda=589.6 \mathrm{~nm}, \quad n=2, \text { and } d=1.2 \times 10^{-6} \mathrm{~m} . \\
& {\left[\text { reminder: } \quad \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]}
\end{aligned}
$$

Solution:
(vi)

$$
\begin{aligned}
& \beta=n \pi \text { for a principle maximum } \\
& \text { i.e. } \frac{\pi}{\lambda} d \sin \theta=n \pi \quad n=0, \pm 1, \pm 2 \ldots \ldots \ldots . .
\end{aligned}
$$

Differentiating w.r.t, $\lambda$

$d \cos \theta \Delta \theta=n \Delta \lambda$

$$
\Delta \theta=\frac{n \Delta \lambda}{d \cos \theta}
$$

Substituting $\lambda=589.0 \mathrm{~nm}, \lambda+\Delta \lambda=589.6 \mathrm{~nm} . \mathrm{n}=2$ and $d=1.2 \times 10^{-6} \mathrm{~m}$.

$\Delta \theta=\frac{n \Delta \lambda}{d \sqrt{1-\left(\frac{n \lambda}{d}\right)^{2}}}$ as $\sin \theta=\frac{n \lambda}{d}$ and $\cos \theta=\sqrt{1-\left(\frac{n \lambda}{d}\right)^{2}}$

$\Rightarrow \Delta \theta=5.2 \times 10^{-3} \mathrm{rads}$ or $0.30^{\circ}$

Final answer: $0.30$ ^{\circ}