A ray of monochromatic light is incident on a grating as shown - Scottish Highers Physics - Question 13 - 2016

The grating equation is given by:

$d imes ext{sin}( heta) = n imes ext{λ}$

Where:

• $d$ is the slit separation,
• $heta$ is the angle for the first order maximum (in this case, 36°),
• $n$ is the order of maximum (first order, so $n=1$),
• $ext{λ}$ is the wavelength of the light (633 nm, which is $633 imes 10^{-9}$ m).

Rearranging the equation for $d$ gives:

$d = \frac{n imes ext{λ}}{ ext{sin}( heta)}$

Substituting the values:

$d = \frac{1 imes 633 imes 10^{-9}}{ ext{sin}(36°)}$

Calculating:

First, calculate $\text{sin}(36°)$, which is approximately 0.5878.

Then,

$d = \frac{633 imes 10^{-9}}{0.5878} \approx 1.075 imes 10^{-6} ext{ m}$

Converting this value to scientific notation, we find:

$d \approx 1.08 imes 10^{-6} ext{ m}$

Therefore, the separation of the slits on the grating is approximately $1.08 imes 10^{-6} ext{ m}$, which corresponds to option B.