Monochromatic light is incident normally on a diffraction grating that has 4.50 x 10^6 lines m^-1 - AQA - A-Level Physics - Question 19 - 2020 - Paper 1

The distance 'd' between adjacent grating lines can be calculated using: $d = \frac{1}{N}$ where 'N' is the number of lines per meter. Thus, for our grating: $d = \frac{1}{4.50 \times 10^6} \approx 2.22 \times 10^{-7} m$

The path difference for diffraction is given by: $dsin(\theta) = m\lambda$ where:

• 'm' is the order of the maxima (m = 2 for second order),
• 'd' is the distance between the grating lines,
• 'θ' is the angle of diffraction,
• 'λ' is the wavelength of the light. Substituting the values: $2.22 \times 10^{-7} \cdot sin(44°) = 2\lambda$.

From the above equation, we can express the wavelength: $\lambda = \frac{2.22 \times 10^{-7} \cdot sin(44°)}{2}$ Calculating this gives: $\lambda \approx \frac{2.22 \times 10^{-7} \cdot 0.6947}{2} \approx 7.72 \times 10^{-7} m$ Converting meters to nanometers: $\lambda \approx 772 nm$.

Based on the calculation, the wavelength of the light is approximately 772 nm. Therefore, the correct answer is: C 772 nm.