Question

If the distance between the first and tenth minima of a double-slit pattern is 18.0 mm and the slits are separated by 0.150 mm with the screen 50.0 cm from the slits, what is the wavelength of the light used?

Short answer

Thus, the wavelength of light is $600nm$.

Step-by-step solution

The separation of slits.

The condition for a minimum in the two-slit interference pattern is$\mathbf{dsin\theta }\mathbf{=}\left(m+\frac{1}{2}\right)\mathbf{\lambda }$,where $\mathbf{d}$is the slit separation, $\mathbf{\lambda }$is the wavelength,$\mathbf{m}$is an integer, and $\mathbf{\theta }$is the angle made by the interfering rays with the forward direction. If $\mathbf{\theta }$is small, $\mathbf{sin\theta }$may be approximately by$\mathbf{\theta }$in radians.

Then, , $\theta =\left(m+\frac{1}{2}\right)\frac{\lambda }{d}$and the distance from the minimum to the central fringe is:

$$\begin{gathered} y=D\tan \theta \\ \approx D\sin \theta \\ \approx D\theta \\ =\left(m+\frac{1}{2}\right)\frac{D\lambda }{d} \end{gathered}$$

Where $D$is the distance from the slits to the screen. For the first minimum$m=0$ for the tenth one$m=9$ . The separation is:

$$\begin{gathered} \Delta y=\left(9+\frac{1}{2}\right)\frac{D\lambda }{d}-\frac{1}{2}\frac{D\lambda }{d} \\ =\frac{9D\lambda }{d} \end{gathered}$$

Calculate the wavelength of light. 

Solve for the wavelength as follows:

$$\begin{gathered} \lambda =\frac{d\Delta y}{9D} \\ =\frac{\left(0.15\times {10}^{-1}\text{m}\right)\left(18\times {10}^{-1}\text{m}\right)}{9\left(50\times {10}^{-2}\text{m}\right)} \\ =6.0\times {10}^{-7}\text{m} \\ =600\text{nm} \end{gathered}$$

Hence, the wavelength of light is $600nm$.