Notifications
Clear all
Light Reflection and Refraction
1
Posts
2
Users
0
Likes
393
Views
0
17/08/2021 3:31 pm
Topic starter
At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?
1 Answer
0
17/08/2021 3:35 pm
f = 12 cm
m = 1
m = \(\frac{v}{u}\) = 1
⇒ v = u
Lens formula: \(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
Putting the value of v, u and f
\(\frac{1}{u}\) - \(\frac{1}{-u}\) = \(\frac{1}{12}\) (image distance is negative)
\(\frac{2}{u}\) = \(\frac{1}{12}\)
u = 24 cm
The object should be placed at a distance of 24 cm to from the lens (on the left side).