Quick single slit diffraction question

[StudentofPhysics]

In a single-slit diffraction pattern, the central fringe is 430 times as wide as than the slit. The screen is 17500 times farther from the slit than the slit is wide. What is the ratio /W, where is the wavelength of the light shining through the slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that sin tan .
wrong check mark



2. sin theta = m x lamba/w
tan theta = sin theta



3.since the central fringe is 430w and the length is 17500w, i figured the ratio would be 430w/17500w = 430/17500 = lamba/w. This is not correct.

Any thoughts?

 

[lippyka]

The correct answer is lamba/w = 430/17500W. To get this, use the equation sin theta = m x lamba/w, where m is the order of the fringe and theta is the angle between the central maximum and the dark fringe. Since the central fringe is 430 times wider than the slit, m = 430. Since the screen is 17500 times farther from the slit than the slit is wide, tan theta = 17500. Therefore, sin theta = 430 x lamba/w and by rearranging the equation, lamba/w = 430/17500W.

 

[m2003]

I would approach this question by using the equation for single-slit diffraction pattern, which is given by: sin theta = m * lambda / W. Here, theta represents the angle that locates a dark fringe on the screen, m is the order of the fringe, lambda is the wavelength of the light, and W is the width of the slit.

Using the given information, we can rearrange the equation to solve for the ratio lambda/W. We know that the central fringe is 430 times wider than the slit, so m = 0 (since the central fringe is the first bright fringe), and theta = 0 (since the angle is small). Plugging these values in, we get:

sin 0 = 0 * lambda / W
0 = 0

This means that the ratio lambda/W is equal to 0. Therefore, the width of the slit is not a factor in determining the wavelength of the light. This may seem counterintuitive, but it is because the angle of diffraction is small and the central fringe is very wide compared to the slit.

In conclusion, the ratio lambda/W is equal to 0, regardless of the values for the central fringe width and the distance from the slit to the screen. This is a unique result of the small angle approximation for single-slit diffraction.