Answer
a. $D=\lambda$
b. $D=\lambda_{min}=400nm$
Work Step by Step
Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum.
$$sin\theta_1=\frac{\lambda}{D}$$
$$\theta_1=sin^{-1}(\frac{\lambda}{D})$$
a. For a given wavelength, there will be no solution to that equation, and therefore no diffraction minima, if the argument for the inverse sine function is greater than 1. We see that the critical value is when $D=\lambda$.
b. In order for no wavelength of visible light to exhibit a diffraction minimum, let the slit width D equal the minimum wavelength that is in the visible region.
$$D=\lambda_{min}=400nm$$
For all visible wavelengths, where $\lambda\gt \lambda_{min}=D$, there will be no diffraction minimum.
