Saved Bookmarks
| 1. |
A man with normal near point (25 cm) reads a book with small print using a magnifying glass, a thin con vex lens of focal length 5 cm. (a) What is the closest and farthest distance at which he can read the book when viewing through the magnifying glass? (b) What is the maximum and minimum magnifying power possible using the above simple microscope? |
|
Answer» Solution :(a) As for normal eye FAR and near points are `oo` and 25 cm respectively, so for magnifier `v_(max)=-ooandv_(min)=-25cm` However, for a lens as `1/v-1/u=1/f` i.e., `u=f/((f//v)-1)`, So u will be MINIMUM when v = min = -25 cm i.e., `(u)_(min)=5/(-(5//25)-1)=(-25)/6=-4.17cm` and u will be maximum when v = max = `oo` i.e., `(u)_(max)=5/((5//oo)-1)=-5cm` So the closest and FARTHEST distances of the book from the magnifier (or eye) for CLEAR viewing are 4.17 cm and 5cm respectively. (b) As in case of simple magnifier MP = (D/u). So MP will be minimum when u= max = 5 cm i.e., `(MP)_(min)=(-25)/(-5)=5" "[=D/f]` And MP will be maximum when u = min = (25/6)cm `(MP)_(min)=(-25)/(-(25//6))=6" "[=1+D/f]` |
|