Recent content by Decelerate

Let me enter it in and see if it is right. EDIT: yes 400nm is right Thanks!

so i guess d has to be 400nm for the range of 400-750 to not have a minimum

When lambda is larger than d...I still don't really understand how that will help me determine the width for visible light from 400-750

I guess my question is what wavelength do you use for visible light? It is a mixture of different wavelengths so How do I determine D when there are multiple wavelengths of light is it just the smallest wavelength (400) or the average...?

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...