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

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

Answered by ajfour last updated on 12/May/17

Commented by ajfour last updated on 12/May/17

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$$$\frac{1}{v}=\frac{1}{(-20)}+\frac{1}{15}=\frac{-15+20}{300}=\frac{1}{60}$$$$v=60cm$$$$\frac{1}{v^{\prime }}-\frac{1}{u^{\prime }}=\frac{1}{f}$$$$imageofobjectinplanemirror$$$$servesasthesecondobjectfor$$$$thelens.$$$$itsdistance{u}^{\prime }=-60cm$$$$\frac{1}{v^{\prime }}=\frac{1}{(-60)}+\frac{1}{15}=\frac{-1+4}{60}=\frac{1}{20}$$$$v^{\prime }=20cm$$$$distancebetweenthesetworeal$$$$imagesisv-v^{\prime }=60cm-20cm$$$$=40cm.$$

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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