# Geometrical optics

1. Apr 14, 2005

### Guji_Gyal

:surprised Hey guys and gals...i am really stuff on dis question.. n i really need 2 knw hw 2 do dis.. so ne1 plz help..n post der views...cheers!

Design a converging achromatic lens of focal length 100cm. Refractive indices and dispersive powers for the two glasses available are 1.51 and 0.016 for crown glass and 1.61 and 0.026 for flint glass. The converging element of the achromat is to be bi-convex and the diverging element is to be plano-concave. Which glass is to be used for each of the two elements of the lens, and determine the radii of curvature of the lens surfaces?

2. Apr 15, 2005

### Andrew Mason

The diverging element would be the higher density glass. The dispersion in the concave lens has to correct for the dispersion in the double convex lens. With the higher index, it can do this with a lower power lens. Since the powers add and the power of the concave is negative, the result is a + power (converging) lens.

The rest of it is a little complicated.

The power / focal length of the biconvex lens, with surfaces having radii R_a and R_b, would be determined by the lensmaker's formula:

$$P_1 = \frac{1}{f_1} = (n_1-1)(\frac{1}{R_a}+ \frac{1}{R_b}) = (n_1 -1)k_{ab}$$

where $k_{ab} = (\frac{1}{R_a}+ \frac{1}{R_b})$

and the power/focal length of the concave surface (radius = -R_b) would be:

$$P_2 = \frac{1}{f_2} = (n_2-1)\frac{1}{-R_b} = (n_2-1)k_c$$

where $-R_b = 1/k_c$ is the radius of curvature of the concave surface

The focal length of the combined lens is

$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$

(where f_2 is negative).

The key to this problem is relating the dispersion to the radius of the lens.
For the convex lens the dispersion is:

$$d_{ab} = d_1k_{ab}$$ where $d_1$ is the dispersive power of the crown glass

For the concave lens the dispersion is:

$$d_c = d_2k_c$$ where $k_c$ is negative. $d_2$ is the dispersive power of the flint glass

The condition for 0 dispersion is $d_{ab} + d_c = 0$

So try to work out the values of $R_a and R_b$ from all that!

AM