Figure  shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?.....

 

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eplain fully . also tell how to calculate the radius of curvature of the given system ( is there any radius of crvature of water ).....if so. how is it infinity .

The first measurement gives the focal length f of the combination of the convex lens and the plano-convex liquid lens.The second measurement gives the focal length f 1of the convex lens.The focal length f 2 of the plano-convex lens is then given by:

       1 / f = 1 / 45 - 1 / 30 = -1 / 90

 i.e. , f 2 = -90 cm

Using the Lensmaker's formula for the equiconvex lens,we get:

   1 / 30 = ( 1.5 - 1 ) (1 / R +1 / R) which gives R = 30 cm.

The same formula applied to the plano-convex lens lens gives 

      - 1 / 90 = ( n - 1) (1 / 30 + 0) from which the refractive index of liquid,n =1.33.