Question Number 134323 by aurpeyz last updated on 02/Mar/21
$${A}\:{lens}\:{is}\:{required}\:{to}\:{have}\:{a}\:{power}\:{of} \\ $$$$−\mathrm{2}.\mathrm{5}\:{dioptres}\:{in}\:{air}.\:{the}\:{convex} \\ $$$${front}\:{surface}\:{has}\:{a}\:{radius}\:{of}\:{curvature} \\ $$$${of}\:\mathrm{30}{cm}.\:{calculate}\:{the}\:{radius}\:{of}\:{curvature} \\ $$$${of}\:{the}\:{rear}\:{surface} \\ $$
Answered by ajfour last updated on 02/Mar/21
$$\frac{\mathrm{1}}{{f}}=\left(\mu−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{R}_{\mathrm{1}} }−\frac{\mathrm{1}}{{R}_{\mathrm{2}} }\right) \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{100}}{\mathrm{30}}−\frac{\mathrm{100}}{{R}_{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{100}}{{R}_{\mathrm{2}} }=\mathrm{5}+\frac{\mathrm{100}}{\mathrm{30}}\:=\:\frac{\mathrm{25}}{\mathrm{3}} \\ $$$${R}_{\mathrm{2}} =\mathrm{12}{cm} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
$${appreciated}\:{thanks} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
$${what}\:{is}\:{the}\:{value}\:{of}\:\mu? \\ $$
Commented by ajfour last updated on 02/Mar/21
$${refractive}\:{index}\:{of}\:{lens}\left({glass}\right) \\ $$$${with}\:{respect}\:{to}\:{that}\:{of}\:{the} \\ $$$${medium}\:{surrounding}\:{the}\:{lens} \\ $$$$\left({here}\:{air}\right). \\ $$
Commented by ajfour last updated on 02/Mar/21
$${different}\:{glasses}\:{may}\:{have} \\ $$$${different}\:{values}\:{of}\:{refractive} \\ $$$${index},\:{if}\:{not}\:{given}\:{in}\:{quwstion} \\ $$$$\left({such}\:{as}\:{yours}\right)\:{we}\:{take}\:{it}\:{to}\:{be} \\ $$$$\mu_{{glass}} =\mathrm{1}.\mathrm{5}\:\:,\:\mu_{{air}} =\mathrm{1} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
$${you}\:{changed}\:\mathrm{1}\:{to}\:\mathrm{100}\:{in}\:{line}\:\mathrm{2} \\ $$
Commented by ajfour last updated on 02/Mar/21
$${Power}\:{in}\:{dioptres}\:{is}\:{obtained} \\ $$$${from}\:{P}=\frac{\mathrm{1}}{{f}}\:\:{only}\:{when} \\ $$$${f}\:{is}\:{in}\:{metres}.\:{So}\:{i}\:{converted} \\ $$$${took}\:\:{R}_{\mathrm{1}} =\frac{\mathrm{30}}{\mathrm{100}}{m}\:\: \\ $$
Commented by aurpeyz last updated on 02/Mar/21
$${thanks}\:{so}\:{much}.\:{i}\:{appreciate}\:{alot} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
$${thanks}\:{so}\:{much}.\:{i}\:{appreciate}\:{alot} \\ $$