A lens placed between a candle and a fixed screen forms a real triply magnified image of the candle on the screen. When the lens is moved away form the candle by `0.8m` without changing the position of the candle, a real image one-third the size of the candle is formed on the screen. Determine the focal length of the lens.

Answer

30 cm

Explanation

Given- Distance between two position of the lens for image is formed

`x = u_(2) - u_(1) = 0.8` m (Given)
`D = u_(1) + v_(1)`
`m_(1)= (v_(1))/(u_(1)) = 3 implies v_(1) = 3u_(1)`
`(v_(2))/(u_(2)) = (1)/(3) implies v_(2) = (u_(2))/(3)`
`x = v_(1) - u_(1) = 3u_(1) - u_(1) = 0.8`
`u_(1) = 0.4 therefore v_(1) = 3 xx 0.4 = 1.2`
`D = 1.2 + 0.4 - 1.6`
`f = (D^(2) - x^(2))/(4D) = ((1.6)^(2) - (0.8)^(2))/(4 xx 1.6) = 0.3 m = 30 cm`