Chapter 7.4, Problem 35WE

a.

To determine

To prove:  $\frac{1}{OJ}+\frac{1}{OI}=\frac{1}{OF}$

Explanation of Solution

Given information: 

Figure of convex lens,

  

Proof: 

  $\begin{array}{l}\text{In }\Delta \text{TJO and }\Delta \text{OIM,}\\ \angle TJO=\angle OIM(\text{Since they both are of }90°.)\\ \angle TOJ=\angle IOM\text{ (Vertically opposite angles}\text{.)}\\ \text{By AA rule,}\\ \Delta TJO~\Delta \text{OIM,}\\ \frac{OJ}{OI}=\frac{JT}{MI}\mathrm{.....}\text{ (1)}\\ \text{In }\Delta POF\text{ and }\Delta \text{FIM,}\\ \angle POF=\angle FIM(\text{Since they both are of }90°.)\\ \angle PFO=\angle IFM\text{ (Vertically opposite angles}\text{.)}\\ \text{By AA rule,}\\ \Delta POF~\Delta \text{FIM,}\\ \frac{OF}{IF}=\frac{PO}{MI}\text{. }\\ \text{Since PO}=\text{JT,}\\ \frac{OF}{IF}=\frac{JT}{MI}\mathrm{.....}\text{ (2)}\\ \text{Thus, from (1) and (2),}\\ \frac{OF}{IF}=\frac{OJ}{OI},\text{ or }\\ \frac{IF}{OF}=\frac{OI}{OJ}.\\ \text{Adding 1 on both side yields,}\\ \frac{IF}{OF}+1=\frac{OI}{OJ}+1,\\ \Rightarrow \frac{IF+OF}{OF}=\frac{OI}{OJ}+1,\\ \Rightarrow \frac{OI}{OF}=\frac{OI}{OJ}+1.\\ \text{Dividing both sides by OI,}\\ \frac{1}{OF}=\frac{1}{OJ}+\frac{1}{OI}.\square \end{array}$

b.

To determine

To prove: $OF=\frac{OJ\cdot OI}{OJ+OI}$

Explanation of Solution

Given information: 

  $\frac{1}{OF}=\frac{1}{OJ}+\frac{1}{OI}$

Proof: 

  $\text{Since, }\frac{1}{OF}=\frac{1}{OJ}+\frac{1}{OI}$ , then −

  $\frac{1}{OF}=\frac{OI+OJ}{OJ\cdot OI}$ .

  $\text{Upon reciprocating both sides,}$ we have −

  $\text{OF}=\frac{OJ\cdot OI}{OI+OJ}.$ $\square$