Question Number 51637 by Necxx last updated on 29/Dec/18
$${An}\:{object}\:\mathrm{60}{cm}\:{tall}\:{is}\:{placed}\:{on} \\ $$$${the}\:{principal}\:{axis}\:{of}\:{a}\:{convex}\:{lens} \\ $$$${and}\:{its}\:\mathrm{30}{cm}\:{tall}\:{image}\:{is}\:{formed} \\ $$$${on}\:{the}\:{screen}\:{placed}\:{at}\:{a}\:{distance} \\ $$$${of}\:\mathrm{40}{cm}\:{from}\:{the}\:{object}.{What}\:{is} \\ $$$${the}\:{focal}\:{length}\:{of}\:{the}\:{lens}? \\ $$
Answered by aseerimad last updated on 29/Dec/18
Commented by aseerimad last updated on 29/Dec/18
pls verify...i am also a student of this level....
Commented by Kunal12588 last updated on 29/Dec/18
$${me}\:{too}\:\left({class}\:\mathrm{11}\right) \\ $$
Commented by aseerimad last updated on 29/Dec/18
is it ok?
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
$$\frac{\mathrm{1}}{{v}}+\frac{\mathrm{1}}{{u}}=\frac{\mathrm{1}}{{f}} \\ $$$$\mathrm{1}+\frac{{v}}{{u}}=\frac{{v}}{{f}} \\ $$$$\mathrm{1}+{m}=\frac{{v}}{{f}} \\ $$$${u}+{v}={d} \\ $$$${u}\left(\mathrm{1}+\frac{{v}}{{u}}\right)={d} \\ $$$$\mathrm{1}+{m}=\frac{{d}}{{u}} \\ $$$$\left(\mathrm{1}+{m}\right)^{\mathrm{2}} =\frac{{v}}{{f}}×\frac{{d}}{{u}}=\frac{{md}}{{f}} \\ $$$${f}=\frac{{md}}{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{40}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{20}×\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{80}}{\mathrm{9}} \\ $$