Saved Bookmarks
| 1. |
A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y. |
|
Answer» Solution :Let` mul `denotethe refractive index of the liquid when the IMAGE , of the needle COINCIDES with the LENS itself , its DISTANCE form the lens , equals the relevant focal length with liquids layer present ,the given set up , is equivalent to a combination of the given (convex ) lens and a concavo plane/ piano concave .liquid. lens and ` ( 1)/(f) =((1)/( f_1) +(1)/(f_2)) ` as per the given data , we then have ` (1)/( f_1) =(1)/( y) =( 1.5- 1 ) ((1)/( R) -( 1)/(-R)) ` = `(1)/(R) ` ` therefore (1)/(x) =( mu _1 -1 ) (-(1)/(R))+( 1)/( y) = ( -mu _t)/(y) + ( 2)/(y)` ` therefore ( mu _t)/( x) =(2)/( y) -(1)/( y ) =((2x-y)/( xy )) ` `mu l =((2x-y)/( x))` |
|