Presentation is loading. Please wait.

Presentation is loading. Please wait.

Diffraction Applications Physics 202 Professor Lee Carkner Lecture 28.

Similar presentations


Presentation on theme: "Diffraction Applications Physics 202 Professor Lee Carkner Lecture 28."— Presentation transcript:

1 Diffraction Applications Physics 202 Professor Lee Carkner Lecture 28

2 PAL #27  What is in the first side pattern in a double slit set-up with a = 0.08 mm and d = 0.25 mm and l = 650 nm?  a sin   = and a sin   = 2 sin   = /a and sin   = 2 /a sin  2 = (2)(650 X 10 -9 )/0.08 X 10 -3 =1.625 X 10 -2

3 PAL #27  What interference maxima are between the two angles? d sin  1 =m 1 and d sin  2 = m 2 m 1 = (0.25 X 10 -3 )(8.125 X 10 -3 )/650 X 10 -9 = 3.13 m 2 = (0.25 X 10 -3 )(1.625 X 10 -2 )/650 X 10 -9 = 6.25 

4 PAL #27  Middle interference fringe is m = 5    = (  a/  sin  = [(  )(0.08 X 10 -3 ) /(650 X 10 -9 )] (0.013) = 5.026 rad   = (  d/ ) sin  = [(  )(0.25 X 10 -3 ) /(650 X 10 -9 )] (0.013) = 15.71 rad  I = I m cos 2  (sin  /  ) 2 = I m (1)(0.0358) = 0.036 I m

5 Diffraction Gratings  For double slit interference the maxima are fairly broad   If we increase the number of slits (N) to very large numbers (1000’s) the individual maxima (called lines) become narrow   A system with large N is called a diffraction grating and is useful for spectroscopy

6 Maxima From Grating

7 Diffraction Grating

8 Grating Path Length

9 Location of Lines  d sin  = m   The m=0 maxima is in the center, and is flanked by a broad minima and then the m=1 maxima etc.  For polychromatic light each maxima is composed of many narrow lines (one for each wavelength the incident light is composed of)

10 Grating Orders

11 Line Width   The half-width (angular distance from the peak to zero intensity) of a line is given by:   where N is the number of slits and d is the distance between 2 slits

12 Line Profile

13 Spectroscopy  A hot material (e.g. a gas) composed of atoms will emit light due to electrons changing energy levels   Gratings are used to separate light into its constituent wavelengths in order to identify this light as spectral lines   This is how the chemical compositions of stars are determined

14 Spectroscope   This will produce a series of orders, each order containing lines (maxima) over a range of wavelengths   The wavelength of a line corresponds to its position angle   We measure  with a optical scope mounted on a vernier position scale 

15 Spectroscope

16 Spectral Type   This is how the temperature of stars is determined  Examples:   Stars like the Sun (T~5500 K) can be identified by the Ca h and k doublet which is only produced at moderate temperatures

17

18

19 Using Spectroscopy  What properties do we want our spectroscope grating to have?    How can we achieve this?

20 Dispersion  The size of the spectrum (from short to long ) produced by a grating is a function of the dispersion, D:  The dispersion can also be written:  For larger m and smaller d the resulting spectra takes up more space

21 Dispersed Spectra

22 Resolving Power  The most important property of a grating is the resolving power, a measure of how well closely separated lines (in ) can be distinguished R = av /    For example, a grating with R = 10000 could resolve 2 blue lines (  = 450 nm) that were separated by 0.045 nm

23 Dispersion and Resolving Power

24 Resolving Power of a Grating  R = Nm  Gratings with large N are the best for spectral work 


Download ppt "Diffraction Applications Physics 202 Professor Lee Carkner Lecture 28."

Similar presentations


Ads by Google