A-convex-lens-of-focal-length-10cm-is-used-to-form-a-real-image-which-is-half-the-size-of-the-object-how-far-from-the-object-is-the-image-

Question Number 132991 by aurpeyz last updated on 17/Feb/21

$${A}\:{convex}\:{lens}\:{of}\:{focal}\:{length}\:\mathrm{10}{cm} \\ $$$${is}\:{used}\:{to}\:{form}\:{a}\:{real}\:{image}\:{which}\:{is} \\ $$$${half}\:{the}\:{size}\:{of}\:{the}\:{object}.\:{how}\:{far} \\ $$$${from}\:{the}\:{object}\:{is}\:{the}\:{image}??? \\ $$

Commented by aurpeyz last updated on 17/Feb/21

$${pls}\:{help}\:{me} \\ $$

Answered by mr W last updated on 17/Feb/21

Commented by mr W last updated on 17/Feb/21

u=−2v −(1/u)+(1/v)=(1/f) −(1/(−2v))+(1/v)=(1/f) (3/(2v))=(1/f) ⇒v=((3f)/2)=15cm ⇒u=−3f=−30 cm distance object from image =30+15=45 cm

$${u}=−\mathrm{2}{v} \\ $$$$−\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{{f}} \\ $$$$−\frac{\mathrm{1}}{−\mathrm{2}{v}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{{f}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}{v}}=\frac{\mathrm{1}}{{f}} \\ $$$$\Rightarrow{v}=\frac{\mathrm{3}{f}}{\mathrm{2}}=\mathrm{15}{cm} \\ $$$$\Rightarrow{u}=−\mathrm{3}{f}=−\mathrm{30}\:{cm} \\ $$$$ \\ $$$${distance}\:{object}\:{from}\:{image} \\ $$$$=\mathrm{30}+\mathrm{15}=\mathrm{45}\:{cm} \\ $$

Commented by aurpeyz last updated on 17/Feb/21

m=(1/2)=((−v)/u)≫u=−2v f=+10cm (1/(10))=(1/(−2v))+(1/v)⇒v=5cm..u=10cm v+u=15cm is it right?

$${m}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{−{v}}{{u}}\gg{u}=−\mathrm{2}{v} \\ $$$${f}=+\mathrm{10}{cm} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{1}}{−\mathrm{2}{v}}+\frac{\mathrm{1}}{{v}}\Rightarrow{v}=\mathrm{5}{cm}..{u}=\mathrm{10}{cm} \\ $$$${v}+{u}=\mathrm{15}{cm} \\ $$$${is}\:{it}\:{right}? \\ $$

Commented by aurpeyz last updated on 17/Feb/21

pls why does the formula(line 2) starts with −ve?

$${pls}\:{why}\:{does}\:{the}\:{formula}\left({line}\:\mathrm{2}\right) \\ $$$${starts}\:{with}\:−{ve}? \\ $$

Commented by mr W last updated on 17/Feb/21

$${i}\:{can}'{t}\:{and}\:{don}'{t}\:{want}\:{to}\:{teach}\:{you} \\ $$$${elementary}\:{things}.\:{if}\:{you}\:{have} \\ $$$${problems}\:{to}\:{remember}\:{the}\:{formula} \\ $$$${and}\:{the}\:{corresponding}\:{sign} \\ $$$${convention},\:{you}\:{should}\:{understand} \\ $$$${the}\:{physics}\:{itself}\:{as}\:{the}\:{picture} \\ $$$${shows}.\:{let}'{s}\:{forget}\:{the}\:{formula}\:{and} \\ $$$${sign}\:{convention},\:{we}\:{take}\:{f},\:{u},\:{v}\:{all} \\ $$$${as}\:{positive}.\:{what}\:{can}\:{we}\:{get}\:{from} \\ $$$${the}\:{picture}\:{above}?\:{we}\:{get}: \\ $$$$\frac{{A}'{B}'}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}},\:{since}\:{image}=\frac{\mathrm{1}}{\mathrm{2}}\:{object}\:{size} \\ $$$$\frac{{F}_{\mathrm{2}} {B}}{{OF}_{\mathrm{2}} }=\frac{{A}'{B}'}{{OC}}=\frac{{A}'{B}'}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{v}−{f}}{{f}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({i}\right) \\ $$$$\Rightarrow{v}=\frac{\mathrm{3}{f}}{\mathrm{2}}=\mathrm{15}\:{cm} \\ $$$$ \\ $$$$\frac{{OB}'}{{OB}}=\frac{{A}'{B}'}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{v}}{{u}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{u}=\mathrm{2}{v}=\mathrm{3}{f}=\mathrm{30}\:{cm} \\ $$

Commented by aurpeyz last updated on 18/Feb/21