Question

A point source of light is placed $60cm$ away from screen. Intensity detected at $pointP$ is $I.$ Now a diverging lens of focal length $20cm$ is placed $20cm$ away from $S$ between $S$ and $P.$ The lens transmits $75$% of light incident on it. Find the new value of intensity at $P.$

Answer 1

$u=-20,f=-20,givesv=-10$

let p= power of source,

$I=\frac{P}{4\pi \left({60}^2\right)}$ ....(1)

energy recieved by lens,

$E_2=\frac{P}{4\pi \left({20}^2\right)}{A}_1$

${\therefore I}_2=\frac{0.75E_2}{A_2}...\left(2\right)$

from similar triangles

$\frac{A_2}{A_1}=25$

$\therefore I_2=0.27I$

Hence the new value of intensity at P is $0.27I$.