A student carries out two experiments to investigate the spectra produced from a ray of white light - Scottish Highers Physics - Question 9 - 2015

To find the speed of red light in the glass prism, we can use the formula:

$v = \frac{c}{n}$

Where:

• $c = 3.00 \times 10^8 \text{ m/s}$ (speed of light in a vacuum)
• $n = 1.54$ (refractive index for red light)

Substituting the values: $v = \frac{3.00 \times 10^8 \text{ m/s}}{1.54} \approx 1.95 \times 10^8 \text{ m/s}$

We can use the formula for diffraction:

$d \sin(\theta) = m\lambda$

Where:

• $d$ is the distance between slits,
• $\theta$ is the angle (19.0°),
• $m$ is the order of the maximum (for second order, $m = 2$),
• $\lambda$ is the wavelength.

First, we need to find the wavelength ($\lambda$) of red light using the frequency: $\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{4.57 \times 10^{14} \text{ Hz}} \approx 6.56 \times 10^{-7} \text{ m}$

Now substituting back into the diffraction formula:

$d \sin(19.0°) = 2 \times (6.56 \times 10^{-7} \text{ m})$

Solving for $d$: $d = \frac{2 \times 6.56 \times 10^{-7} \text{ m}}{\sin(19.0°)} \approx 4.03 \times 10^{-6} \text{ m}$

The angle to the second order maximum for blue light is different from that for red light due to the fact that different colors of light have different wavelengths and refractive indices. Blue light has a shorter wavelength than red light, which means it will be refracted more when passing through the grating. This results in blue light having a different path difference and consequently a different angle at which the second order maximum appears.