Question

Light whose wavelength is 633nm falls on a double slit with spacing of 0.100mm. What is the separation between the 0th and 1st order peak on a screen, L=1.20m from the double slit?

Answer 1

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:

$\displaystyle \sin{\theta} = \frac{m\cdot \lambda}{d}$, where

• $\lambda$ is the wavelength of the light, and
• $d$ is the separation between the two slits.

(Young’s Double Slit Experiment, OpenStaxCollege)

If $d$ is much larger than $\lambda$, the value of $\theta$ will be considerably small. The value of $\theta$ could thus be approximately as:

$\displaystyle \theta \approx \frac{m\cdot \lambda}{d}$.

For this problem,

• $m = 1$ for a first-order maximum.
• $d= \rm 0.100\;mm = 0.100\times 10^{-3}\; m$;
• $\lambda = \rm 633\; nm = 633\times 10^{-9}\; m$.

Approximate the value of $\theta$:

$\displaystyle \theta \approx \frac{m\cdot \lambda}{d} \approx \rm 0.00633\;radians$.

Separation between the first and central maximum:

$\displaystyle L \cdot \tan{\theta} \approx L\cdot \theta = 0.00760\; \rm m$