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Q.


A symmetric biconvex lens of radius of curvature Question Image 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 Question Image. On removing the liquid layer and repeating the experiment the distance is found to be Question Image. The expression for the refractive index of the liquid in terms of Question Image and Question Image:


Question Image

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a

2x-yx

b

4x-yx

c

2-yx

d

None of these 

answer is A.

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Detailed Solution

Concept- Given, refractive index of lens, Question ImageThe distance of the needle from the lens in the first case = The focal length of the combination of convex lens and plano- concave lens formed by the liquid, f =x
.And, the distance measured in the second case = Focal length of the convex lens, Question ImageIf the focal length of plano-concave lens formed by the liquid be Question Image, then
Question ImageQuestion Imageor Question ImageNow, refractive index of lens, Question Image, radius of curvature of one surface Question Image and radius of curvature of other surface Question ImageQuestion ImageQuestion ImageAnd, if refractive index of liquid is Question Image,
Radius of curvature on the side of plane mirror Question ImageRadius of curvature on the side of lens Question ImageQuestion ImageQuestion Imageor Question ImageQuestion ImageHence, the correct option is 1.
 
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