In an experiment to measure the wavelength of monochromatic light, a beam of light was incident normally on a diffraction grating - Leaving Cert Physics - Question 3 - 2015

The first order images were identified immediately to either side of the zero order image observed when no grating was present. This corresponds to the straight through image (zero order) and the brightest images on either side, which represent the first order diffracted light.

A beam of light was produced using a Laser, which provides a coherent and monochromatic light source for the diffraction experiment.

The most accurate angle is $\phi = 4.60$ degrees (n = 1) because it corresponds to the first order diffraction, where the angles are typically smaller and errors due to measurement or parallax are minimized compared to larger orders.

The wavelength can be calculated using the formula: $\lambda = \frac{d}{n \sin \theta}$ Where:

• $d = \frac{1}{N} = \frac{1}{80 \text{ lines/mm}} = 1.25 \times 10^{-5} \text{ m}$
• $n = 1$
• $\theta = \frac{4.60}{2} = 2.30$ degrees (for the first order) converted to radians for calculations. Using $\sin(2.30)$, we get: $\lambda = \frac{1.25 \times 10^{-5}}{\sin(2.30)}$ After substitution and solving, the final wavelength will be approximately $2.50 \times 10^{-7} \text{ m}$.

Replacing the diffraction grating with one of 500 lines per mm would lead to fewer images seen and the images would be more spread out, as a higher number of lines per mm results in a narrower angular dispersion of the light.