Calculating Chromatic Aberration in Dense Flint Lenses: A Homework Solution

  • Thread starter Thread starter 05holtel
  • Start date Start date
  • Tags Tags
    Material
AI Thread Summary
The discussion focuses on calculating chromatic aberration in dense flint lenses, emphasizing that different wavelengths of light are refracted at varying degrees, leading to image distortion. The refractive index for violet-blue light is 1.80, while for red light it is 1.70, affecting the focal lengths for each color. Using the lens maker's formula, the focal lengths for white, blue, and red light are determined to be approximately 6.67 cm, 6.25 cm, and 7.14 cm, respectively. The standard lens equation is then applied to find the image distances for red and blue light, allowing for the calculation of the distance between the two images. This process illustrates the impact of chromatic aberration on lens performance.
05holtel
Messages
52
Reaction score
0

Homework Statement



The refractive index of a material is different for different wavelengths and colours of light. For most materials in the visible range of the electromagnetic spectrum, shorter wavelengths have larger refractive index compared to longer wavelengths.

The effect of this on lenses is that different colours from one object will be focused at different distances and thus it is impossible to have the whole object completely focused. This is known as chromatic aberration.

Dense flint is a refractive material for which the shortest wavelength of the visible spectrum at violet-blue (400 nm) has a refractive index of 1.80, while for the longest wavelength of the visible spectrum at red (800 nm) has a refractive index of 1.70

Consider a converging lens made out of dense flint with R1=10 cm and R2=-10 cm.

We place a white object at a distance of 119 cm from the lens. Since white light is composed of all visible colours, when it passes through the lens, the different colours will form images at different distances.

What is distance between the red image of the object and the blue-violet image?

Homework Equations



lens maker equation

The Attempt at a Solution




The focal length at any particular wavelength can be calculated using the "lens maker's formula"

For white light with n = 1.75 (a red-blue average),
1/f = (0.75)(1/10 + 1/10) = 0.15
f(white)= 6.67 cm
f(blue) = 5/0.8 = 6.25 cm
f(red) = 5/0.7 = 7.14 cm

Use the standard lens equation
1/f = 1/do + 1/di
to compute the difference in the image distances, di. Use do = 119 cm

What is the f in the equation
 
Physics news on Phys.org
f is the focal length, that is the length at which a lense focuses a given object.

using the infomation given you are able to input firstly into the lensmakers equation to find a focal length and then in the standard lens equation to find the image distance for both red and blue light, which enables you to find the difference in distances

try having a look at http://hyperphysics.phy-astr.gsu.edu/Hbase/geoopt/lenseq.html#c1
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top