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

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 20cm from both the lens and the mirror.Calculate the distance of the two images from the lens.. pls help me with this....

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

Answered by ajfour last updated on 12/May/17

Commented by ajfour last updated on 12/May/17

(1/v)−(1/u)=(1/f) (1/v)=(1/((−20)))+(1/(15)) = ((−15+20)/(300)) =(1/(60)) v=60cm (1/(v′))−(1/(u′))=(1/f) image of object in plane mirror serves as the second object for the lens. its distance u′=−60cm (1/(v′))=(1/((−60)))+(1/(15))=((−1+4)/(60)) =(1/(20)) v′=20cm distance between these two real images is v−v′=60cm−20cm = 40cm .

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

Commented by chux last updated on 12/May/17

wow...... this is wonderful...I really appreciate.

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

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