Lecture 18- Dispersion and chromatic aberrations

Question 1

What is dispersion ?

Answer 1

the variation in refractive index with the wavelength of light

Question 2

What are examples of most transparent materials?

Answer 2

(e.g., glasses, liquids, plastics, crystals)

Question 3

What are transparent material described as?

Answer 3

Dielectric materials

Question 4

What is the common property that dielectric materials share?

Answer 4

they all show an increase in refractive index as the wavelength of light decreases

Question 5

What happens with the refractive indices of these dielectric materials?

Answer 5

The refractive indices of most transparent materials that are used in optical instruments vary throughout the visible spectrum in approximately the same way

Question 6

How do we quantify dispersion?

Answer 6

need to introduce spectral lines and materials

Question 7

What are spectral lines?

Answer 7

discrete wavelengths which are used for the measurement of refractive index and also too quantify the dispersive properties of glass materials

Question 8

What are spectral lines?

Answer 8

discrete wavelengths which are used for the measurement of refractive index and also too quantify the dispersive properties of glass materials

Question 9

What does the F line of hydrogen show?

Answer 9

blue light at 486.1nm

Question 10

What does the C line of hydrogen show?

Answer 10

red

-656.3nm

Question 11

What does the d line show in helium?

Answer 11

yellow

-587.6nm

Question 12

What shows in mercury line?

Answer 12

546nm

-green

Question 13

What shows in the sodium line?

Answer 13

Yellow

589 and 589.6nm

Question 14

What parameter is important?

Answer 14

refractive index - nd

- which is the refractive index of the corresponding material for the helium d line

Question 15

What is another important parameter?

Answer 15

Difference in refractive index between the nF- nC

- difference between the F line and C line nm of light of hydrogen material

Question 16

What is the V value parameter?

Answer 16

-quantifies the change in power in a thin lens as a result of dispersion and -varies about 25 to 65

Question 17

What are these parameters of?

Answer 17

dispersion parameters for principal material - check powerpoint

Question 18

Why are these parameters of great practical importance in making optical instrumentation in the description of dispersion and the effects on the optical proeprties of these materials?

Answer 18

• the reason why these particular wavelengths (red blue yellow light) are chosen is because they are easily produced from low pressure discharged lamps and
• they cover the EM spectrum to which the eye is sensitive to

Question 19

Where is the maximum sensitivity of the eye?

Answer 19

555nm

Question 20

What happens as a result of changes in Refractive index ?

Answer 20

-going from red to blue light- the properties of optical systems change

Question 21

How do you calculate the power of a thin lens?

Answer 21

F= (n-1) x (c1-c2)
c1 - curvature of front surface
c2- curvature of back surface
the curvature is just the reciprocal of the radius- so when the radius is very large the curvature is very small.

Question 22

What is the power of the second surface ?

Answer 22

F2= (1-n)c2

Question 23

What is the combined power ?

Answer 23

F = F1+F2-(d/n)F1F2

Question 24

What is the power on assumption that d - 0?

Answer 24

F=(n-1)(c1-c2)

Question 25

What is the power on assumption that d - 0?

Answer 25

F=(n-1)(c1-c2)

Question 26

What is the formula used for these changes in refractive index in the wavlelgnths?

Answer 26

F(wavelength)=(n(wavelength)-1)(c1-c2)

Question 27

Question : Find the focal lengths of a single thin, equiconvex lens for red (nC) and blue (nF) light given that the power of the lens for the helium d-line is +5D. ?

Hence we can calculate a value for the longitudinal chromatic aberration of this lens for a distant object.

Answer 27

• given refractive indices for c,d,f lines of crown glass and dense flint glass
• given power of helium d line
• have dispersion- which causes change in power of lens for helium d line and c,F lines.
  5= (1.5169-1)(2c1) - as equiconvex
  2c1= 9.67
  c1= 4.78 - now use this for F line equation and C line equation
• Ff = (nF-1)(2c1)
  Answers:
  r1 = -r2 = 20.676 cm
  f’C = 20.101 cm; f’F = 19.774 cm;
  LCA = 3.27 mm
  LCA = f’c - f’F
  (repeat for dense flint glass)

Question 28

What do you use if it is a equiconvex lens ?

Answer 28

F=(n-1)(2c1)

so when finding out curvature you then divide by 2.

Question 29

will an identical lens made from dense flint glass will have greater LCA?

Answer 29

yes