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The index of refraction of a glass varies with the wavelength of the light passing through it. This has important consequences for the design of optical instruments. A type of glass known as "extra-dense flint" has an n = 1.7378 for blue light, and an n of 1.7130 for red light. Calculate the focal length of the lens above, for both red and blue light.
As just been pointed out to me the Lens must be from the question before it which is:
The so-called "lensmakers equation" is shown below. The focal
length of the lens (in cm) is f. The constant n is the index of
refraction of the glass, relative to the medium around it (which is
just the listed value of n unless the lens is submerged in water,
or some other liquid). R1 and R2 are the radii of curvature of the
two sides of the lens. The value of R for a side is considered to
be positive if that side is convex; otherwise, it is negative. If
the lens is flat on one side, the value of R for that side is
infinite, and the ratio (1/R) is equal to zero.
1/f=(n-1)(1/R_1 + 1/R_2 )
Assume the lens is biconvex; that is, that both sides are convex. The radii of curvature are both 30 cm. The coefficient of refraction is 1.60. What is the focal length of the lens, in cm?
The answer for this was given as:
1/f=(n-1)(1/R1 + 1/R2 )
=> 1/f=(1.6-1)(1/30 + 1/30 )
=>1/f= 0.6(1/15 )
=> f= 15/.6 = 25 cm