Physics - Geometrical Optics Question with Solution | TestHub

Focal length of equiconcave lens

$\mathrm{f}=\frac{-\mathrm{R}}{2\left(\mathrm{\mu }-1\right)}$

Where$\mathrm{\mu }=\frac{{\mathrm{\mu }}_{\mathrm{lens}}}{{\mathrm{\mu }}_{\mathrm{medium}}}$

Therefore focal length in the medium

${\mathrm{f}}_{\mathrm{med}}=\frac{-\mathrm{R}}{2\left(\frac{1.5}{1.75}-1\right)}$

${\mathrm{f}}_{\mathrm{med}}=+\frac{\mathrm{R}}{2\left(\frac{0.25}{1.75}\right)}$

${\mathrm{f}}_{\mathrm{med}}=+3.5\mathrm{R}$

Positive sign implies convergent sign.