Saved Bookmarks
| 1. |
A point object O is placed at a distance of 20 cm is front of a equiconvex lens (.^(a)mu_(g) = 1.5) of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in figure. Image will be formedat a distance h from lens the value of h is |
|
Answer» 5 cm `1/10 = (1.5-1)(1/R+1/R)=0.5xx2/R rArr R = 10 cm` `:. (mu_(2))/(V) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)` Refraction fromfirst surface, `(1.5)/(v_(1)) - (1)/(-20) = (1.5-1)/(+10) rArr v_(1) =oo` For the second surface, `2/v-(1.5)/(oo)=(2-1.5)/(-10) = v = - 40 cm :. h = 40 cm` |
|