(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

The smallest wavelength of the visible light spectrum is

$\mathrm{\lambda }=380\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=3.80\times {10}^{-7}\mathrm{m}$

The largest wavelength of the visible light spectrum is

$\mathrm{\lambda }=760\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=7.60\times {10}^{-7}\mathrm{m}$

The order of the maximum is 2.

(a)

A formula for the angle of the second-order maximum can be expressed as,

$\mathrm{dsin\theta }=2\mathrm{\lambda }.........................................\left(1\right)$

The smallest value of d corresponds to ${}^{\mathrm{\theta }={90}^0}$.

Substituting the given values in the equation (1) will give,

$$\begin{gathered} \mathrm{d}=\frac{2\mathrm{\lambda }}{\mathrm{sin\theta }} \\ =\frac{2\times 3.80\times {10}^{-7}\mathrm{m}}{\sin \left({90}^0\right)} \\ =7.60\times {10}^{-7}\mathrm{m} \end{gathered}$$

Therefore, the smallest separation is ${}^{7.60\times {10}^{-7}\mathrm{m}}$.

(b)

Now, to evaluate the smallest slit separation for all visible light. We will determine the value of ${}^{\mathrm{d}}$the largest possible wavelength of the visible light which is ${}^{760\mathrm{nm}}$.

Substituting the given values in the equation (1) will give,

$$\begin{gathered} \mathrm{d}=\frac{2\mathrm{\lambda }}{\mathrm{sin\theta }} \\ =\frac{2\times 7.60\times {10}^{-7}\mathrm{m}}{\sin \left({90}^0\right)} \\ =1.52\times {10}^{-6}\mathrm{m} \end{gathered}$$

Therefore, the smallest separation for all visible light is${}^{1.52\times {10}^{-6}\mathrm{m}}$.