3.3.2.2
The maximum visible order is limited by sin theta = 1
Interference & Diffraction — AQA A-Level Physics
- The maximum angle of diffraction is 90 degrees, where sin $\theta = 1$.
- The highest visible order is found by:
$$n_{\text{max}} = \frac{d}{\lambda}$$
- Since n must be a whole number, round down if the result is not an integer.
- Example: if d/$\lambda = 2.7$, then $n_{\max} = 2$.
Worked Example
Light of wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). 550 nm passes through a diffraction gratingAn optical component with many equally spaced parallel slits that produces sharp interference maxima at specific angles. with slit spacing 1.7 micrometres. Calculate the angle alpha between the two second-order lines.
Show Solution
1
Write the grating equation for n = 2
$$d \sin \theta = n\lambda$$
2
Solve for sin theta
$$\sin \theta = \frac{n\lambda}{d} = \frac{2 \times 550 \times 10^{-9}}{1.7 \times 10^{-6}} = 0.647$$
3
Find theta
$$\theta = \sin^{-1}(0.647) = 40.5^{\circ}$$
4
Calculate the angle between the two second-order lines
The two second-order maxima are on opposite sides of the central beam:
$$\alpha = 2\theta = 2 \times 40.5 = 81^{\circ} \text{ (2 s.f.)}$$Answer
$\alpha = 81^{\circ}$
Examiner Tips and Tricks
- Take care that theta is measured from the centre (the normal), not between two orders.
- The angle between two symmetrical orders is 2 theta.