We are given:
$$\frac{n_1}{l_0}+\frac{n_2}{l_i}=\frac{1}{R}(\frac{n_2s_i}{l_i}-\frac{n_1s_0}{l_0})$$
$$l_0=\sqrt{R^2+(s_0+R)^2-2R(s_0+R)cos(\phi)}$$
$$l_i=\sqrt{R^2+(s_i-R)^2+2R(s_i-R)cos(\phi)}$$
$$ h= R\;sin(\phi) $$
The equation we want to derive is
$$\frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R}+h^2(\frac{n_1}{2s_0}(\frac{1}{s_0}+\frac{1}{R})^2+\frac{n_2}{2s_i}(\frac{1}{R}-\frac{1}{s_i})^2)$$
We are told to proceed by replacing $cos(\phi)$ with its third degree taylor polynomial, and to use the first two terms in the binomial series of $\frac{1}{l_i}$ and $\frac{1}{l_0}$ as well as approximating $\phi$ by $sin(\phi)$.
I can't seem to get this solution even using these hints. I am representing $cos(\phi)$ by $1-\frac{\phi^2}{2}$ and $\frac{1}{l_i}$ and $\frac{1}{l_0}$ as $2-l_i$ and $2-l_0$ respectively.
Does anyone have any tips?
As there is no optics ray diagram and space for comments is inadequate I am commenting as follows in the Answers space with some numerical values which may not be representative of the needed optical configuration:
R = 1.06384/.665179;
h=1.0;n1=1.1;n2=1.5;so =5.8;si=4.7;lo = 5.6;li= 3.2;
lhsEQ
lhs=n1/lo+n2/li
rhsEQ
rhs= (n2 si/li -n1 so/lo)/R
csph=1-(h/R)^2;
lo2=R^2+(so+R)^2-2 R (so+R)csph;li2=R^2+(so-R)^2+2 R (so-R)csph;
LHSEQ
LHS= n1/so + n2/si
aa= (1/so+1/R)^2 n1/(2 so);bb= (1/so-1/R)^2 n2/(2 si) ;
RHSEQ
RHS = (n2-n1)/R + h^2 ( aa +bb )
After execution, results of left and right hand sides tally in the Gaussian formula but not in its linearized version.