Certainly! Let's solve this step by step.
Given:
- The dark fringe for the zeroth-order ([tex]$m=0$[/tex]) is located at an angle [tex]$\theta = 15^\circ$[/tex].
To Find:
- The angle that locates the dark fringe for the first order ([tex]$m=1$[/tex]).
### Solution:
1. Convert the given angle to radians:
The angle at [tex]$m=0$[/tex] is [tex]$\theta_0 = 15^\circ$[/tex]. To proceed, we need to convert this angle from degrees to radians since trigonometric functions in physics are often used in radian measure.
[tex]\[
\theta_0 = 15^\circ
\][/tex]
Using the conversion factor [tex]\( \pi \text{ radians} = 180^\circ \)[/tex]:
[tex]\[
\theta_0 \text{ (in radians)} = 15^\circ \times \frac{\pi \text{ radians}}{180^\circ} = \frac{15\pi}{180} = \frac{\pi}{12}
\][/tex]
Therefore:
[tex]\[
\theta_0 \approx 0.2618 \text{ radians}
\][/tex]
2. Find the angle for the first dark fringe ([tex]$m=1$[/tex]):
In Young's double-slit experiment, the angle for the [tex]$m$[/tex]-th dark fringe can be derived from the principle where the path difference is directly proportional to [tex]\(m\)[/tex]. This means that if the angle for [tex]$m=0$[/tex] is [tex]$\theta_0$[/tex], the angle for [tex]$m=1$[/tex] would be effectively doubled (based on linearity for small angles).
Thus, the angle in radians for the first dark fringe, [tex]$\theta_1$[/tex], is:
[tex]\[
\theta_1 = 2 \times \theta_0 = 2 \times 0.2618
\][/tex]
Therefore:
[tex]\[
\theta_1 \approx 0.5236 \text{ radians}
\][/tex]
3. Convert the angle for [tex]$m=1$[/tex] back to degrees:
Now, convert [tex]$\theta_1$[/tex] from radians to degrees:
[tex]\[
\theta_1 \text{ (in degrees)} = 0.5236 \text{ radians} \times \frac{180^\circ}{\pi \text{ radians}} = 0.5236 \times \frac{180}{\pi} \approx 30^\circ
\][/tex]
Thus, the angle that locates the dark fringe for [tex]\( m=1 \)[/tex] is approximately [tex]\( 30^\circ \)[/tex].
