InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A thin biconvex lens of refractive index 3//2and radius of curvature 50cm is placed on a reflecting convex surface of radius of curvature 100cm. A point object is placed on the principal axis of the system such that its final image coincides with itself. Now, few drops of a transparent liquid is placed between the mirror and lens such that final image of the object is at infinity. Find refractive index of the liquid used. Also, find the position of the object. |
|
Answer» POWER of LENS, `P_(L)=(1)/(f_(l))=((3)/(2)-1)((1)/(50)-(1)/(-50))=(1)/(50)` `:.` Power of equivalent mirror `P_(EM) =2P_(l)+P_(m)=(2)/(50)-(1)/(50)=(1)/(50)` i.e., Focal length of equivalent mirror `f_(CM)=-50` Since image and object coincides, hence position of the object is at the center of curvature of the equivalent mirror,i.e., at `2f_(cm)=100cm` from the lens. When the liquid of refractive index be isplaced between the lens and convex mirror, it forms a liquid lens of power, `p_(u)=((1)/(-50)-(1)/(100))=(3(l-mu))/(100)` Now, power of equivalent mirror `=2P_(l)+2P_(u)+P_(m)` `=(2)/(50)+(2xx3(1-mu))/(100)+(1)/(-50)=(3(1+mu)+1)/(50)` (i) Since final image is at infifnity, hence the object should be at its focal point i.e., `f_(EM)^(')=100` `rArr =-(1)/(f_(EM)^('))=(1)/(100)` (ii) From Eqs (i) and (ii), we get `(3(1-mu)+1)/(50)=(1)/(100)` `rArr mu=7//6` |
|