Posted by Sandra on Friday, March 26, 2010 at 11:19pm.
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 108 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?
- Physics - drwls, Saturday, March 27, 2010 at 12:30pm
The lens focal length is inversely proportional to n-1. Thus
f(blue)/f(red) = 7/8 = 0.875
The focal length at any particular wavelength can be calculated using the "lens maker's formula" whch you can find at
In your case, 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 = 108 cm
- Physics - mm, Saturday, March 27, 2010 at 3:16pm
how did u get that do=108cm?
- Physics - mm, Saturday, March 27, 2010 at 3:17pm
oh nevermind i got it
- Physics - mm, Saturday, March 27, 2010 at 3:19pm
sorry, another question. what would f be in the len makers equation?
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