= Solution
Substituting the stated <refractive index> and thickness coefficient, $\alpha=2\,\mathrm{m^{-1}}$, gives
$$
\boxed{f=\frac1{2(1.5-1)(2\,\mathrm{m^{-1}})}=0.50\,\mathrm m.}
$$
The <thin lens> equation then gives
$$
\frac1v=\frac1f-\frac1u=2-\frac1{0.7}=\frac47\,\mathrm{m^{-1}},
\qquad
\boxed{v=1.75\,\mathrm m.}
$$
The positive image distance places a <real image> on the outgoing side of the <optical lens>, at $x=+1.75\,\mathrm m$. The object is beyond the <focal point>, so outgoing <light rays> actually converge there rather than only their backward extensions meeting.
Back to article page
