Achromatic Objective $($cementea doublet$)$:

Type: Plano$-$interface effective focal length $=729mm$ diameter $=71\mathrm{.}5mm$

radius of curvature of the front convex surface $rl=187mm**$ thickness $=9mm$

glass$=$BK$7$

internal surface curvature is $0($flat$)$

second lens thickness $=6mm$

radius of curvature of the back concave surface $r3=470\mathrm{.}6mm$ glass$=$SF$5$

back focal distance $=710mm$

$*$Similar to our singlet design the front convex surface should be aspheric, to reduce the effect of spherical aberration, the conic constant $k=-0\mathrm{.}416.$

For this achromatic doublet evaluate the image sport size and compare it to the Airy disk size. Find the distance between the focal points in blue and red colour and compare it with the distance found for the previous design of the singlet$.$

The optical power of a single thin element at wavelength ${}_d$ with refractive index $n_d$ and curvatures $c_1=\frac{1}{r_1}$ and $c_2=\frac{1}{r_2}$ is given by ${}_d=(n_d-1)(c_1-c_2),$ note $c=0$ for plane surfaces. The power variation at the two wavelengths ${}_F$ and ${}_C$ becomes:

$={}_F-{}_C=(n_F-n_C)(c_1-c_2)=(n_F-n_C)\frac{{}_d}{n_d-1}=\frac{{}_d}{V_d},$ where $V_d=\frac{n_d-1}{n_F-n_C}$ is the Abbe number, which characterises the dispersion of glass.

For BK$7$ glass, $V_d=64\mathrm{.}2$ and $n_d=1\mathrm{.}5168$; for SF$5$ glass $V_d=32\mathrm{.}2,$ and $n_d=1\mathrm{.}6730.$

To make the doublet achromatic, the power variation of the two$-$lens system, must be zero:

$={}_1+{}_2=\frac{{}_1}{V_1}+\frac{{}_2}{V_2}=0$

Check that this condition is satisficed in the doublet objective above.