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 30cm and 12cm respectively. what is its focal length when placed inside water? (take the refractive indices of glass and water to be 1.52 and 1.33 respectively](https://www.tinkutara.com/question/Q134319.png)
$${the}\:{radii}\:{of}\:{curvatures}\:{of}\:{front}\:{and} \\ $$$${rear}\:{surfaces}\:{of}\:{a}\:{thin}\:{convex}\:{lens}\: \\ $$$${are}\:\mathrm{30}{cm}\:{and}\:\mathrm{12}{cm}\:{respectively}.\: \\ $$$${what}\:{is}\:{its}\:{focal}\:{length}\:{when}\:{placed} \\ $$$${inside}\:{water}?\:\left({take}\:{the}\:{refractive}\right. \\ $$$${indices}\:{of}\:{glass}\:{and}\:{water}\:{to}\:{be}\:\mathrm{1}.\mathrm{52} \\ $$$${and}\:\mathrm{1}.\mathrm{33}\:{respectively} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
![pls help](https://www.tinkutara.com/question/Q134346.png)
$${pls}\:{help} \\ $$
Answered by ajfour last updated on 02/Mar/21
![(1/f)=((μ_(glass) /μ_(water) )−1)((1/R_1 )−(1/R_2 )) =(((1.52×3−4)/4))((1/(30))+(1/(12))) (1/f)=((0.14×7)/(60))=((0.98)/(60)) f=((60)/(0.98))≈60(1.02)=61.2cm](https://www.tinkutara.com/question/Q134348.png)
$$\frac{\mathrm{1}}{{f}}=\left(\frac{\mu_{{glass}} }{\mu_{{water}} }−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{R}_{\mathrm{1}} }−\frac{\mathrm{1}}{{R}_{\mathrm{2}} }\right) \\ $$$$\:\:\:\:=\left(\frac{\mathrm{1}.\mathrm{52}×\mathrm{3}−\mathrm{4}}{\mathrm{4}}\right)\left(\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}}{{f}}=\frac{\mathrm{0}.\mathrm{14}×\mathrm{7}}{\mathrm{60}}=\frac{\mathrm{0}.\mathrm{98}}{\mathrm{60}} \\ $$$${f}=\frac{\mathrm{60}}{\mathrm{0}.\mathrm{98}}\approx\mathrm{60}\left(\mathrm{1}.\mathrm{02}\right)=\mathrm{61}.\mathrm{2}{cm} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
![okay sir. thanks alot](https://www.tinkutara.com/question/Q134379.png)
$${okay}\:{sir}.\:{thanks}\:{alot} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
![((μ_(glass) /μ_(water) )−1)=((1.52−4)/4)?? you had ((1.52×3−4)/4)](https://www.tinkutara.com/question/Q134359.png)
$$\left(\frac{\mu_{{glass}} }{\mu_{{water}} }−\mathrm{1}\right)=\frac{\mathrm{1}.\mathrm{52}−\mathrm{4}}{\mathrm{4}}?? \\ $$$${you}\:{had}\:\frac{\mathrm{1}.\mathrm{52}×\mathrm{3}−\mathrm{4}}{\mathrm{4}} \\ $$
Commented by aurpeyz last updated on 02/Mar/21
![the options are (a)141cm (b) −0.71 cm (c) 0.71cm (d)−141cm (e)14.1cm](https://www.tinkutara.com/question/Q134362.png)
$${the}\:{options}\:{are} \\ $$$$\left({a}\right)\mathrm{141}{cm}\:\left({b}\right)\:−\mathrm{0}.\mathrm{71}\:{cm}\:\left({c}\right)\:\mathrm{0}.\mathrm{71}{cm} \\ $$$$\left({d}\right)−\mathrm{141}{cm}\:\left({e}\right)\mathrm{14}.\mathrm{1}{cm} \\ $$
Commented by ajfour last updated on 02/Mar/21
![cant say, my answer seems correct to me!](https://www.tinkutara.com/question/Q134364.png)
$${cant}\:{say},\:{my}\:{answer}\:{seems} \\ $$$${correct}\:{to}\:{me}! \\ $$