91Ó°ÊÓ

The smallest wavelength of the visible light spectrum is

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

The largest wavelength of the visible light spectrum is

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

The order of the maximum is 2.

(a)

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

$\mathrm{»å²õ¾\pm ²Ôθ}=2\mathrm{λ}.........................................\left(1\right)$

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

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

$$\begin{gathered} \mathrm{d}=\frac{2\mathrm{λ}}{\mathrm{²õ¾\pm ²Ôθ}} \\ =\frac{2×3.80×{10}^{-7}\mathrm{m}}{\sin \left({90}^0\right)} \\ =7.60×{10}^{-7}\mathrm{m} \end{gathered}$$

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

(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,

$$\begin{gathered} \mathrm{d}=\frac{2\mathrm{λ}}{\mathrm{²õ¾\pm ²Ôθ}} \\ =\frac{2×7.60×{10}^{-7}\mathrm{m}}{\sin \left({90}^0\right)} \\ =1.52×{10}^{-6}\mathrm{m} \end{gathered}$$

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