Question

The distance between an object and the screen is $100cm.$ A lens produces an image on the screen when placed at either of two positions $40cm$ apart. The power of the lens is approximately

• A. $3D$
• B. $5D$
• C. $7D$
• D. $9D$

Answer 1

Answer: (B)

Given, $D=100cm,x=40cm$
$\therefore f=\frac{D^2-x^2}{4\text{D}}$
If $a$ be his maximum upward acceleration, then $\left(70kg\right)g=\left(60kg\right)g+\left(60kg\right)a$
$\therefore a=\frac{\left(70kg\right)g-\left(60kg\right)g}{60kg}$
$=\frac{g}{6}=\frac{9.8m{s}^{-2}}{6}=1.63m{s}^{-2}$
$f=\frac{{\left(100\right)}^2-{\left(40\right)}^2}{4\times 100}$
$=\frac{\left(100+40\right)\left(100-40\right)}{400}$
$=\frac{140\times 60}{400}=21cm$
We know that $P=\frac{100}{f\left(cm\right)}=\frac{100}{21}=5D$