(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

lens maker equation

3. 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

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# Homework Help: Refractive material

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