Question

Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?
(a) $\frac{1}{\nu }-\frac{1}{u}=\frac{t}{uf}$

(b) $\frac{t}{{\nu }^2}-\frac{1}{u}=\frac{1}{f}$

(c) $\frac{1}{\nu -t}-\frac{1}{u+t}=\frac{1}{f}$

(d) $\frac{1}{\nu }-\frac{1}{u}+\frac{t}{uv}=\frac{t}{f}$.

Answer 1

(c) $\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$

The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:
$$\begin{gathered} u_{\mathrm{new}}=u+t(1+\frac{1}{\mu })\mathrm{and}{v}_{\mathrm{new}}=v-t(1+\frac{1}{\mu }) \\ or{u}_{\mathrm{new}}=u+t+\frac{t}{\mu }\mathrm{and}{v}_{\mathrm{new}}=v-t+\frac{t}{\mu } \end{gathered}$$
As $\frac{t}{\mu }$ will be quite small, it can be ignored.
Thus, u_new = u + t and v_new = v − t
We get the new lens formula:
$\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$