Question

In Young's double-slit experiment, the angular width of a fringe formed on a distant screen is 1 degree. The wavelength of light used is 6000 Armstrong. What is the spacing between the slits?

Answer 1

Step 1: Given

Wavelength of light = $6000A^0$

Angular width = $1^{\circ }$

Step 2: Formula

$$\begin{gathered} n\lambda =d\sin \theta \\ wherenisorderoffringe,disspacingbetweenslits \\ if\sin \theta <<1 \\ then\sin \theta =\theta \\ thereforen\lambda =d\theta \end{gathered}$$

Step 3: Calculation:

to calculate the spacing between the consicuitve slits ( n=1)

$$\begin{gathered} n\lambda =d\theta \\ n=1 \\ =\lambda =d\theta \\ d=\frac{\lambda }{\theta } \end{gathered}$$

$$\begin{gathered} \lambda =6000\times {10}^{-10}m \\ =6\times {10}^{-7}m \\ \theta =1^{\circ } \\ =\frac{\mathrm{\pi }}{180} \\ now \\ putthevalueof\lambda ,\theta inbelowequation \\ d=\frac{\lambda }{\theta } \end{gathered}$$

$$\begin{gathered} =\frac{6\times {10}^{-7}}{{\displaystyle \frac{\pi }{180}}} \\ =3.4\times {10}^{-5}m \end{gathered}$$

Hence, the spacing between the slits is $3.4\times {10}^{-5}m$