Question Number 134319 by aurpeyz last updated on 02/Mar/21
$${the}\:{radii}\:{of}\:{curvatures}\:{of}\:{front}\:{and} \\$$$${rear}\:{surfaces}\:{of}\:{a}\:{thin}\:{convex}\:{lens}\: \\$$$${are}\:\mathrm{30}{cm}\:{and}\:\mathrm{12}{cm}\:{respectively}.\: \\$$$${what}\:{is}\:{its}\:{focal}\:{length}\:{when}\:{placed} \\$$$${inside}\:{water}?\:\left({take}\:{the}\:{refractive}\right. \\$$$${indices}\:{of}\:{glass}\:{and}\:{water}\:{to}\:{be}\:\mathrm{1}.\mathrm{52} \\$$$${and}\:\mathrm{1}.\mathrm{33}\:{respectively} \\$$
Commented by aurpeyz last updated on 02/Mar/21
$${pls}\:{help} \\$$
Answered by ajfour last updated on 02/Mar/21
$$\frac{\mathrm{1}}{{f}}=\left(\frac{\mu_{{glass}} }{\mu_{{water}} }−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{R}_{\mathrm{1}} }−\frac{\mathrm{1}}{{R}_{\mathrm{2}} }\right) \\$$$$\:\:\:\:=\left(\frac{\mathrm{1}.\mathrm{52}×\mathrm{3}−\mathrm{4}}{\mathrm{4}}\right)\left(\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{12}}\right) \\$$$$\frac{\mathrm{1}}{{f}}=\frac{\mathrm{0}.\mathrm{14}×\mathrm{7}}{\mathrm{60}}=\frac{\mathrm{0}.\mathrm{98}}{\mathrm{60}} \\$$$${f}=\frac{\mathrm{60}}{\mathrm{0}.\mathrm{98}}\approx\mathrm{60}\left(\mathrm{1}.\mathrm{02}\right)=\mathrm{61}.\mathrm{2}{cm} \\$$
Commented by aurpeyz last updated on 02/Mar/21
$${okay}\:{sir}.\:{thanks}\:{alot} \\$$
Commented by aurpeyz last updated on 02/Mar/21
$$\left(\frac{\mu_{{glass}} }{\mu_{{water}} }−\mathrm{1}\right)=\frac{\mathrm{1}.\mathrm{52}−\mathrm{4}}{\mathrm{4}}?? \\$$$${you}\:{had}\:\frac{\mathrm{1}.\mathrm{52}×\mathrm{3}−\mathrm{4}}{\mathrm{4}} \\$$
Commented by aurpeyz last updated on 02/Mar/21
$${the}\:{options}\:{are} \\$$$$\left({a}\right)\mathrm{141}{cm}\:\left({b}\right)\:−\mathrm{0}.\mathrm{71}\:{cm}\:\left({c}\right)\:\mathrm{0}.\mathrm{71}{cm} \\$$$$\left({d}\right)−\mathrm{141}{cm}\:\left({e}\right)\mathrm{14}.\mathrm{1}{cm} \\$$
Commented by ajfour last updated on 02/Mar/21
$${cant}\:{say},\:{my}\:{answer}\:{seems} \\$$$${correct}\:{to}\:{me}! \\$$