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**Lecture 16 Diffraction Chp. 37**

Topics Young’s double slit interference experiment Diffraction and the wave theory Single slit diffraction Intensity of single slit diffraction Circular aperture and double slit diffraction Diffraction grating Dispersion and resolving power Warm-up problem Demos Diffraction grating and slits Inverted mirage Measuring diameter of a strand Debra’s hair

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**Young’s Double Slit Interference Experiment**

q m=0 m=1 m=2 D

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**Maxima Minima m ym +/- 1 2 3 Dl/d 2Dl/d 3Dl/d m ym +/- 1 2 3 Dl/2d**

1 2 3 Dl/d 2Dl/d 3Dl/d m ym +/- 1 2 3 Dl/2d 3Dl/2d 5Dl/2d 7Dl/2d

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**What about the intensity of light along the screen?**

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**From the table on the previous slide we see that the separation **

13E Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes? From the table on the previous slide we see that the separation between bright fringes is

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**Diffraction of a single slit**

Find minimum ym Find maximum First maximum lies on the axis at q = 0 or ym=0. Other maxima lie in between the minima. To find them we need to find the intensity along the screen.

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**Intensity of single slit diffraction**

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**Maxima/Minima conditions**

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**Exact solution for maxima**

To find maxima of a function, take derivative and set equal to 0 Transcendental equation. Solve graphically

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**8. A 0. 10-mm-wide slit is illuminated by light of wavelength 589nm**

8. A 0.10-mm-wide slit is illuminated by light of wavelength 589nm. Consider a point P on a viewing screen on which the diffraction patters of the slit is viewed; the point is at 30o from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point P from the top and midpoint of the slit? (Hint: see Eq ) We note that nm = 10-9 m = 10-6 mm. From Eq. 37-4, This is equivalent to = 2.8 rad = 160o

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6. Sound waves with frequency 3000 Hz and speed 343 m/s diffract through a rectangular opening of a speaker cabinet and into a large auditorium. The opening, which has a horizontal width of 30.0 cm, faces a wall 100 m away. Where along that wall will a listener be at the first diffraction minimum and thus have difficult hearing the sound? (Neglect reflections). Let the first minimum be a distance y from the central axis which is perpendicular to the speaker. Then q y Therefore,

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**Diffraction and Interference by a double slit**

I = I (double slit interference) x I(diffraction)

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Sample problem 37-4 How many bright interference fringes fall within the central peak of the diffraction envelope? The idea here is to find the angle where the first minimum occurs of the diffraction envelope. Given We have m=0 and m=1,2,3 and 4 on both sides of central peak. Ans is 9

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**Diffraction by a circular aperature**

The first minimum for the diffraction of light from a circular aperature is given by: where d is the diameter of the aperature. Our ability to resolve two distant point like objects is determined when the first minimum of one objects diffraction pattern overlaps the central maximum of another. This is called Raleigh’s criterion. Example

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**Example 15E. The two headlights of an approaching automobile are 1**

Example 15E. The two headlights of an approaching automobile are 1.4 m apart. Assume the pupil diameter is 5.0 mm and the wavelength of light is 550 nm. (a) At what angular distance will the eye resolve them and (b) at what distance? (a) (b) s D

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**Diffraction Grating Double slit -- N slits or rulings. w d = w/N**

where w is the entire width of the grating

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**Can be used to measure wavelength of light**

Measure angles of diffracted lines with a spectroscope using formula below. Then relate to wavelength Resolving power of grating. Measure of the narrowness of lines Highest R Highest D Dispersion of a grating. Measure of how well lines are separated

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**Show diffraction gratings with increasing N and single slit diffraction with varying slit width a.**

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**Babinets Complementarity Principle**

In the diffraction region the intensity is the same whether you have an aperature or opaque disk. You can also replace a slit with a wire or hair strand. A compact disk is an example of a diffraction grating in reflection instead of transmission. Experiment: Measure diameter of a strand of hair from Debra.

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Mirage eye sky 1,09 1.09 1.08 1.08 1.07 1.07 1.06 Hot road causes gradient in the index of refraction that increases as you increase the distance from the road In the demo before you the gradient is in the opposite direction

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Warm-up HRW6 37.CQ.02. [73994] You are conducting a single-slit diffraction experiment with light of wavelength l. (a) What appears, on a distant viewing screen, at a point at which the top and bottom rays through the slit have a path length difference equal to 5l.? the m = 5 maximum the m = 5 minimum the m = 4 minimum the m = 4 maximum (b) What appears, on a distant viewing screen, at a point at which the top and bottom rays through the slit have a path length difference equal to 4.5l.? the minimum between the m = 4 and m = 5 minima the minimum between the m = 4 and m = 5 maxima the maximum between the m = 4 and m = 5 maxima the maximum between the m = 4 and m = 5 minima

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Lecture 16 Diffraction Ch. 36 Topics –Newtons Rings –Diffraction and the wave theory –Single slit diffraction –Intensity of single slit diffraction –Double.

Lecture 16 Diffraction Ch. 36 Topics –Newtons Rings –Diffraction and the wave theory –Single slit diffraction –Intensity of single slit diffraction –Double.

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