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Resolution. Resolution. Sketch the variation with angle of diffraction of the relative intensity of light emitted by two point sources that has been diffracted at a single slit. State the Rayleigh criterion for images of two sources to be just resolved including a circular aperture)

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resolution1
Resolution
  • Sketch the variation with angle ofdiffraction of the relative intensity oflight emitted by two point sourcesthat has been diffracted at a singleslit.
  • State the Rayleigh criterion for imagesof two sources to be just resolved including a circular aperture)
  • Describe the significance of resolutionin the development of devicessuch as CDs and DVDs, the electronmicroscope and radio telescopes.
diffraction from a single slit
Diffraction from a single slit

n = 2

n = 1

θ

b

bsinθ = nλ

diffraction from a single slit1
Diffraction from a single slit

bsinθ = nλ

(θb = λ) (small angle approximation)

diffraction from a circular aperture
Diffraction from a circular aperture

θ = 1.22λ (angle of first minimum radians)

b

rayleigh criterion
Rayleigh criterion
  • Two sources are just resolved if the central maximum of the diffraction pattern of one source falls on the first minimum of the other. i.e. θ = 1.22λ/b (radians)
two distant stars
Two distant stars
  • Images merge
observing two stars
Observing two stars
  • Two stars separated by distance s and lie a distance d from earth

s

θ

d

observing two stars1
Observing two stars
  • Angular separation = s/d (radians)

s

θ

d

observing two stars2
Observing two stars
  • The stars can be resolved by a telescope provided Angular separation = s/d (radians) = 1.22λ/b (small angle approximation in radians)

s

θ

d

example
Example
  • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm)
example1
Example
  • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm)

Angular separation = s/d = 1.22λ/b

example2
Example
  • The camera of a spy satellite orbiting 200km above the earth’s surface has a diameter of 35cm. What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm)

Angular separation = s/d = 1.22λ/b

example3
Example
  • The camera of a spy satellite orbiting 200km (2 x 105 m) above the earth’s surface has a diameter of 35cm (0.35m) . What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm (5 x 10-7 m))

Angular separation = s/d = 1.22λ/b

example4
Example
  • The camera of a spy satellite orbiting 200km (2 x 105 m) above the earth’s surface has a diameter of 35cm (0.35m) . What is the smallest distance the camera can resolve on the surface of the earth (assume λ = 500 nm (5 x 10-7 m))

s = 1.22dλ/b =1.22(2 x 105 x 5 x 10-7)/0.35

= 0.34m

another example
Another example

The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct?

another example1
Another example

The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct?

Angular separation = s/d = 1.22λ/b

another example2
Another example

The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm. Supposing wavelength of light of 500nm what is the maximum distance at which two headlights are seen as distinct?

Angular separation = s/d = 1.22λ/b

another example3
Another example

The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm (2 x 10-3 m). Supposing wavelength of light of 500nm (5 x 10-7 m) what is the maximum distance at which two headlights are seen as distinct?

Angular separation = s/d = 1.22λ/b

another example4
Another example

The headlights of a car are 2m apart. The pupil of the human eye has a diameter of 2mm (2 x 10-3 m). Supposing wavelength of light of 500nm (5 x 10-7 m) what is the maximum distance at which two headlights are seen as distinct?

d = sb/1.22λ = (2x2 x 10-3)/1.22x5 x 10-7

d= 7000m