Please show how you got those answers. It is hard to guess what you did wrong otherwise.Plasmosis1 said:I tried two answers, 4.4mm and 2.2mm and they are both wrong. Any suggestions?
Plasmosis1 said:Homework Statement
A 0.42 -mm-diameter hole is illuminated by light of wavelength 510nm. What is the width (in mm) of the central maximum on a screen 1.8m behind the slit?
Homework Equations
f=λx/s
The Attempt at a Solution
I tried two answers, 4.4mm and 2.2mm and they are both wrong. Any suggestions?
A single slit problem is a type of diffraction problem in which light from a single source passes through a small opening or slit and is then observed on a screen. This phenomenon can be used to calculate the width of the central maximum, which is the brightest part of the diffraction pattern.
The width of the central maximum can be calculated using the formula w = λL/d, where w is the width of the central maximum, λ is the wavelength of the light, L is the distance from the slit to the screen, and d is the width of the slit. This formula is known as the single slit diffraction equation.
The width of the central maximum can be affected by several factors, including the wavelength of the light, the distance between the slit and the screen, and the width of the slit. Additionally, the intensity of the light source and any interference patterns created by multiple slits can also impact the width of the central maximum.
Yes, it is possible for the width of the central maximum to be larger than the width of the slit. This can occur when the distance between the slit and the screen is large enough, causing the diffraction pattern to spread out and the central maximum to become wider.
The single slit problem has many practical applications, such as in the design of optical instruments like cameras and telescopes. It is also used in the study of wave behavior and can be seen in everyday phenomena like the diffraction of light through a window or the blurring of shadows due to light diffracting around objects.