Diffraction

(b) plane

In Fraunhofer diffraction the source and screen are effectively at infinity (a convex lens is used), so the wavefront falling on the obstacle or slit is plane.

(b) 4

Grating element $d = \dfrac{1}{400}$ mm $= 2.5\times10^{-6}$ m. Since $\sin\theta \le 1$, $n_{\max} \le \dfrac{d}{\lambda} = \dfrac{2.5\times10^{-6}}{6\times10^{-7}} = 4.17$. The order must be a whole number, so the highest order is 4. (Counting both sides and the central maximum, $2\times4 + 1 = 9$ bright lines are seen.)

i) Diffraction of light: The bending of light around the edges of an obstacle or a narrow aperture, so that light spreads into the region of the geometrical shadow, is called diffraction of light.

ii) Width of central maximum

Given: $D = 4$ m, slit width $a = 0.01$ m, $\lambda = 600$ nm $= 6\times10^{-7}$ m.

The first minima lie at $y = \pm\dfrac{\lambda D}{a}$, so the width of the central maximum is

$$W = \frac{2\lambda D}{a} = \frac{2 \times 6\times10^{-7} \times 4}{0.01} = 4.8\times10^{-4} \text{ m}$$

Width of central peak = 0.48 mm.

Given: $\lambda = 5890$ Å $= 5.89\times10^{-7}$ m, $N = 6000$ lines/cm $= 6\times10^5$ lines/m.

Grating element:

$$d = \frac{1}{N} = \frac{1}{6\times10^5} = 1.667\times10^{-6} \text{ m}$$

i) Angle of second order image (grating equation $d\sin\theta = n\lambda$, $n = 2$):

$$\sin\theta_2 = \frac{2\lambda}{d} = 2 \times 5.89\times10^{-7} \times 6\times10^5 = 0.7068$$ $$\theta_2 = \sin^{-1}(0.7068) \approx 45.0^\circ$$

ii) Third order: For $n = 3$,

$$\sin\theta_3 = \frac{3\lambda}{d} = 3 \times 5.89\times10^{-7} \times 6\times10^5 = 1.06$$

Since $\sin\theta$ cannot exceed 1, the third order image is not possible. (Maximum order $= d/\lambda = 2.83$, i.e. 2.)

Given: $N = 500$ lines/mm $= 5\times10^5$ lines/m, order $n = 2$, $\theta = 30^\circ$.

Grating element:

$$d = \frac{1}{N} = \frac{1}{5\times10^5} = 2\times10^{-6} \text{ m}$$

Grating equation:

$$d\sin\theta = n\lambda \;\Rightarrow\; \lambda = \frac{d\sin\theta}{n}$$

Substitution:

$$\lambda = \frac{2\times10^{-6} \times \sin 30^\circ}{2} = \frac{2\times10^{-6} \times 0.5}{2} = 5\times10^{-7} \text{ m}$$

Wavelength = $5\times10^{-7}$ m = 500 nm (5000 Å).

Given: $N = 400$ lines/mm $= 4\times10^5$ lines/m, $\lambda = 600$ nm $= 6\times10^{-7}$ m.

i) Grating spacing

$$d = \frac{1}{N} = \frac{1}{4\times10^5} = 2.5\times10^{-6} \text{ m}$$

ii) First order angle ($d\sin\theta = n\lambda$, $n = 1$):

$$\sin\theta_1 = \frac{\lambda}{d} = \frac{6\times10^{-7}}{2.5\times10^{-6}} = 0.24 \;\Rightarrow\; \theta_1 = 13.9^\circ$$

iii) Number of maxima: The largest possible $\sin\theta$ is 1:

$$n_{\max} = \frac{d}{\lambda} = \frac{2.5\times10^{-6}}{6\times10^{-7}} = 4.17$$

So the highest order seen is $n = 4$. Counting orders 1 to 4 on each side plus the central maximum:

$$\text{Total maxima} = 2 \times 4 + 1 = 9$$

Given: $N = 5000$ lines/cm, so the grating element is

$$d = \frac{1}{5000}\ \text{cm} = 2\times10^{-6}\ \text{m}$$

For the second order, $n = 2$ and $\theta = 30^\circ$. From $d\sin\theta = n\lambda$:

$$\lambda = \frac{d\sin\theta}{n} = \frac{2\times10^{-6}\times0.5}{2} = 5\times10^{-7}\ \text{m}$$

The wavelength is $5\times10^{-7}$ m $= 5000$ Å.

(ii) Angle of second-order maximum

Given: $\lambda = 546$ nm $= 5.46\times10^{-7}$ m, $N = 3000$ lines/cm $= 3\times10^5$ lines/m, $n = 2$.

$$d = \frac{1}{N} = \frac{1}{3\times10^5} = 3.33\times10^{-6} \text{ m}$$ $$\sin\theta_2 = \frac{2\lambda}{d} = 2 \times 5.46\times10^{-7} \times 3\times10^5 = 0.3276$$ $$\theta_2 = \sin^{-1}(0.3276) \approx 19.1^\circ$$

(iii) More accurate experiment: The diffraction grating gives a more accurate value of $\lambda$. In the double-slit experiment the fringes are only about 0.87 mm apart and are broad and dim, so measuring their spacing has a large percentage error. A grating has thousands of slits, so its maxima are very sharp and bright and are spread at large angles ($19.1^\circ$ here), which can be measured precisely with a spectrometer.

(d) All bright fringes, except the central one, become colourful.

At the centre the path difference is zero for every wavelength, so all colours overlap and the central maximum is white. For other maxima the position depends on $\lambda$, so the colours separate and the fringes are coloured (violet nearer the centre, red farther out).