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

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???

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

Commented by aurpeyz last updated on 17/Feb/21

pls help me

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

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: ((A′B′)/(AB))=(1/2), since image=(1/2) object size ((F_2 B)/(OF_2 ))=((A′B′)/(OC))=((A′B′)/(AB))=(1/2) ⇒((v−f)/f)=(1/2) ...(i) ⇒v=((3f)/2)=15 cm ((OB′)/(OB))=((A′B′)/(AB))=(1/2) ⇒(v/u)=(1/2) ...(ii) ⇒u=2v=3f=30 cm

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

Commented by aurpeyz last updated on 18/Feb/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mr W last updated on 18/Feb/21

Commented by aurpeyz last updated on 18/Feb/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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