A diffraction grating is illuminated normally - AQA - A-Level Physics - Question 13 - 2021 - Paper 1

To solve the problem, we use the diffraction grating equation:

$d \sin \theta = m \lambda$

Here:

• $d$ is the grating spacing (distance between slits),
• $\theta$ is the angle of diffraction,
• $m$ is the order of the maximum (an integer),
• $\lambda$ is the wavelength of light.

Since the second-order maximum for 650 nm wavelength occurs at the same angle as the third-order maximum for wavelength $\lambda$, we can express this with two equations for the respective orders.

For the second-order maximum (where $m=2$) with $\lambda_1 = 650 \text{ nm}$:

$d \sin \theta = 2 \times 650 \text{ nm}$

For the third-order maximum (where $m=3$) with $\lambda_2 = \lambda$:

$d \sin \theta = 3 \lambda$

Setting the two equations equal (since they occur at the same angle):

$2 \times 650 = 3 \lambda$

Now, solving for $\lambda$ gives:

$\lambda = \frac{2 \times 650}{3} \text{ nm} = \frac{1300}{3} \text{ nm} \approx 433.33 \text{ nm}$