Answer:
Approximately 0.00760 m (that's 7.60 mm.)
Explanation:
Refer to the first diagram attached. For a double-slit diffraction, the angle (angular separation) between the m-th maximum and the central maximum satisfies the following equation:
[tex]\displaystyle \sin{\theta} = \frac{m\cdot \lambda}{d}[/tex], where
- [tex]\lambda[/tex] is the wavelength of the light, and
- [tex]d[/tex] is the separation between the two slits.
(Young’s Double Slit Experiment, OpenStaxCollege)
If [tex]d[/tex] is much larger than [tex]\lambda[/tex], the value of [tex]\theta[/tex] will be considerably small. The value of [tex]\theta[/tex] could thus be approximately as:
[tex]\displaystyle \theta \approx \frac{m\cdot \lambda}{d}[/tex].
For this problem,
- [tex]m = 1[/tex] for a first-order maximum.
- [tex]d= \rm 0.100\;mm = 0.100\times 10^{-3}\; m[/tex];
- [tex]\lambda = \rm 633\; nm = 633\times 10^{-9}\; m[/tex].
Approximate the value of [tex]\theta[/tex]:
[tex]\displaystyle \theta \approx \frac{m\cdot \lambda}{d} \approx \rm 0.00633\;radians[/tex].
Separation between the first and central maximum:
[tex]\displaystyle L \cdot \tan{\theta} \approx L\cdot \theta = 0.00760\; \rm m[/tex]
