Physics - Geometrical Optics Question with Solution | TestHub

Given,$D=100 \mathrm{cm}, x=40 \mathrm{cm}$

$\therefore f=\frac{D^2-x^2}{4\text{D}}$

If$a$be his maximum upward acceleration, then$\left(70\mathrm{kg}\right)g=\left(60\mathrm{kg}\right)g+\left(60\mathrm{kg}\right)a$

$\therefore a=\frac{\left(70\mathrm{kg}\right)g-\left(60\mathrm{kg}\right)g}{60\mathrm{kg}}$

$=\frac{g}{6}=\frac{9.8 \mathrm{m} {\mathrm{s}}^{-2}}{6}=\mathrm{1.63 }\mathrm{m} {\mathrm{s}}^{-2}$

$f=\frac{{\left(100\right)}^2-{\left(40\right)}^2}{4\times 100}$

$=\frac{\left(100+40\right) \left(100-40\right)}{400}$

$=\frac{140\times 60}{400}=21 \mathrm{cm}$

We know that$P=\frac{100}{f\left(\mathrm{c}\mathrm{m}\right)}=\frac{100}{21}=5 \mathrm{D}$