In a diffraction experiment with waves of wavelength $\mathit{\lambda }$,there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width for which this occurs? Explain your answer.

When monochromatic light sent through a narrow slit, produces a diffraction pattern on a distant screen which is single slit diffraction.

Condition of dark fringes (destructive interference) in single slit diffraction is as follow,

$\mathbf{asin\theta }\mathbf{=}\mathbf{n\lambda }$

Where is slit width,$\mathbf{\theta }$ is the diffraction angle of minima; is the order of minima andis wavelength of the light.

If the diffraction angle for 1^st minima is $90°$then there will be no minima in field of view as for 1^st minima diffracted light become parallel to slit and screen.

No minima occur when$\theta =90°$ . From this one can say,

$$\begin{gathered} \mathrm{asin}90°=1\times \mathrm{\lambda } \\ \mathrm{a}=\mathrm{\lambda } \end{gathered}$$

The above equation is the criterion for “no-minima”.

Therefore, Maximum slit width for which there will be no minima is equal to the wavelength of light.