Question

Find the distance at which an object should be placed in front of a convex lens of focal length 10cm to obtain an image of triple of its size? *

- 21cm or -13.33cm
- 13.33cm or - 6.95cm
- 6.66cm or -13.33cm
+13.33cm or +6.7cm

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ object \: distance \: (u) \\ \\ given : \\ focal \: length \: (f) = 10.cm \\ v = 3u \\ \\ we \: know \: that \\ according \: to \: lens \: formula \\ \\ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \frac{1}{3u} + \frac{1}{( - u)} = \frac{1}{10} \\ \\ \frac{1}{3u} - \frac{1}{u} = \frac{1}{10} \\ \\ \frac{ - 2u}{3u {}^{2} } = \frac{1}{10} \\ \\ 3u {}^{2} = - 20u \\ \\ 3u {}^{2} + 20u = 0\\ \\ u(3u + 20) = 0 \\ \\ u = 0 \: \: or \: \: u = \frac{ - 20}{3} \\ \\ ∵ \: distance \: is \: never \: zero\\ u = 0 \: \: is \: absurd \\ \\ u = \frac{ - 20}{3} \\ \\ = - 6.666 \\ \\ = - 6.66 \: cm \\ \\ here \: minus \: sign \: denotes \\ image \: is \: virtual \: and \: erect \: \: \\ and \: object \: is \: placed \: below \\ principal \: axis \: \:$