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Homework Statement
What is the maximum slit width so that no visible light exhibits a diffraction minimum?(Visible light has wavelengths from 400 nm to 750 nm.)
Homework Equations
D*sin(theta)=m*lambda ; where D = slit width,theta in degrees, m =1,2,3,..., and lambda is wavelength. The maximum slit width occurs at sin(90) =1 and if we take m=1 the equation reduces to D=lambda. When the slit width is equal to the wavelength of light passing through it then no diffraction minima are observed.
The Attempt at a Solution
For visible light, I just assumed that the maximum slit width had to be equal to the largest wavelength contained in visible light (750nm), but this does not make sense because then the wavelength at 450nm would still be defracted. I know for a fact that a slit width of 750nm is wrong, so I am guessing that the answer might be 450nm or maybe the average wavelength of visible light but am not sure
Any help is greatly appreciated.