Physics - Geometrical Optics Question with Solution | TestHub

(i) P_eq= 2 P_ℓ1 + 2 P​_ℓ2+ P_m...(1)

For plano-convex lens :$\frac{1}{{\text{f}}_{\ell 1}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty }-\frac{1}{30}\right)$

${\text{f}}_{\ell 1}=-60\text{cm}\Rightarrow {\text{P}}_{\ell 1}=\frac{100}{{\text{f}}_{\ell 1}}\Rightarrow {\text{P}}_{\ell 1}=-\frac{5}{3}\text{D}$

For plano-convex lens : ​$\frac{1}{{\text{f}}_{\ell 2}}=\left(\frac{5}{4}-1\right)\left(\frac{1}{30}-\frac{1}{\mathrm{\infty }}\right)$$\mathrm{ \Rightarrow }$$\frac{1}{{\text{f}}_{\ell 2}}=\frac{1}{120}$

${\text{f}}_{\ell 2}=120\text{cm}\Rightarrow {\text{P}}_{\ell 2}=\frac{100}{{\text{f}}_{\ell 2}}\Rightarrow {\text{P}}_{\ell 2}=\frac{5}{6}\text{D}$

For plane mirror :${\text{P}}_{\text{m}}=0\text{since}{\text{f}}_{\text{m}}=\infty \Rightarrow {\text{P}}_{\text{eq}}=2{\text{P}}_{\ell 1}+2{\text{P}}_{\ell 2}+{\text{P}}_{\text{m}}$

${\text{P}}_{\text{eq}}=-\frac{10}{3}+\frac{10}{6}+0$

$\Rightarrow {\text{P}}_{\text{eq}}=-\frac{10}{6}\text{D}$$\because {\mathrm{P}}_{\mathrm{eq}}=-\frac{1}{{\mathrm{F}}_{\mathrm{eq}}}$

⇒$F_{eq}=60 \mathrm{cm}$