An-object-is-placed-between-a-converging-lens-and-a-plane-mirror-Explain-how-two-real-images-of-the-object-may-be-produced-by-the-system-If-the-focal-length-of-the-lens-is-15cm-and-the-object-is-2

Question Number 13054 by chux last updated on 12/May/17

$$\mathrm{An}\:\mathrm{object}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{between}\:\mathrm{a}\: \\ $$$$\mathrm{converging}\:\mathrm{lens}\:\mathrm{and}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{mirror}. \\ $$$$\mathrm{Explain}\:\mathrm{how}\:\mathrm{two}\:\mathrm{real}\:\mathrm{images}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{may}\:\mathrm{be}\:\mathrm{produced}\:\mathrm{by}\:\mathrm{the}\: \\ $$$$\mathrm{system}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{focal}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lens}\:\mathrm{is}\:\mathrm{15cm} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{object}\:\mathrm{is}\:\mathrm{20cm}\:\mathrm{from}\:\mathrm{both}\: \\ $$$$\mathrm{the}\:\mathrm{lens}\:\mathrm{and}\:\mathrm{the}\:\mathrm{mirror}.\mathrm{Calculate} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{images}\: \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{lens}.. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{this}….\: \\ $$

Answered by ajfour last updated on 12/May/17

Commented by ajfour last updated on 12/May/17

$$\frac{\mathrm{1}}{{v}}−\frac{\mathrm{1}}{{u}}=\frac{\mathrm{1}}{{f}} \\ $$$$\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{\left(−\mathrm{20}\right)}+\frac{\mathrm{1}}{\mathrm{15}}\:\:=\:\frac{−\mathrm{15}+\mathrm{20}}{\mathrm{300}}\:=\frac{\mathrm{1}}{\mathrm{60}} \\ $$$${v}=\mathrm{60}{cm} \\ $$$$\frac{\mathrm{1}}{{v}'}−\frac{\mathrm{1}}{{u}'}=\frac{\mathrm{1}}{{f}} \\ $$$${image}\:{of}\:{object}\:{in}\:{plane}\:{mirror} \\ $$$${serves}\:{as}\:{the}\:{second}\:{object}\:{for} \\ $$$${the}\:{lens}. \\ $$$${its}\:{distance}\:\:{u}'=−\mathrm{60}{cm} \\ $$$$\frac{\mathrm{1}}{{v}'}=\frac{\mathrm{1}}{\left(−\mathrm{60}\right)}+\frac{\mathrm{1}}{\mathrm{15}}=\frac{−\mathrm{1}+\mathrm{4}}{\mathrm{60}}\:=\frac{\mathrm{1}}{\mathrm{20}} \\ $$$${v}'=\mathrm{20}{cm} \\ $$$${distance}\:{between}\:{these}\:{two}\:{real} \\ $$$${images}\:\:{is}\:{v}−{v}'=\mathrm{60}{cm}−\mathrm{20}{cm} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{40}{cm}\:. \\ $$

Commented by chux last updated on 12/May/17

$$\mathrm{wow}……\:\mathrm{this}\:\mathrm{is}\:\mathrm{wonderful}…\mathrm{I}\: \\ $$$$\mathrm{really}\:\mathrm{appreciate}. \\ $$

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