Test Index

CBSE Class 12 Physics 2018 Delhi Set 1 Paper

© examsnet.com
Question : 17 of 26
Marks: +1, -0
A symmetric biconvex lens of radius of curvature RR 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 xx. On removing the liquid layer and repeating the experiment, the distance is found to be yy. Obtain the expression for the refractive index of the liquid in terms of xx and yy.
Solution:  
  Ans. Lens maker’s formula\;\text{Ans. Lens maker's formula}
  Formula for ’combination of lenses’\;\text{Formula for 'combination of lenses'}
  Obtaining the expression for  μ\;\text{Obtaining the expression for}\;\mu
(a) Let μ1\mu_1 denote the refractive index of the liquid. When the image of the needle coincides with the lens itself ; its distance from the lens, equals the relevant focal length.
With liquid layer present, the given set up, is equivalent to a combination of the given (convex) lens and a concavo plane / plano concave 'liquid lens'.
We have   1f=(μ−1)(1R1−1R2)\;\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
and   1f=(1f1+1f2)\;\frac{1}{f}=\left(\frac{1}{f_1}+\frac{1}{f_2}\right)
as per the given data, we then have
  1f2    =  1y=(1⋅5−1)(1R−1(−R))\;\frac{1}{f_2}\;\;=\;\frac{1}{y}=(1\cdot5-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right)
  =  1R\;=\;\frac{1}{R}
  ∴    1x  =(μl−1)(−1R)+1y\;\therefore\;\;\frac{1}{x}\;= (\mu_l-1)\left(-\frac{1}{R}\right)+\frac{1}{y} =  −μly+2y=\;\frac{-\mu_l}{y}+\frac{2}{y}
  ∴    ∝ly  =  2y−1x=(2x−yxy)\;\therefore\;\;\frac{\propto_l}{y}\;=\;\frac{2}{y}-\frac{1}{x}= \left(\frac{2x-y}{xy}\right)
    or    μl  =(2x−yx)\;\;\text{or}\;\;\mu_l\;= \left(\frac{2x-y}{x}\right)
© examsnet.com
Go to Question:
Prev Question
Next Question