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Fig. 9.12 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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Answer» SOLUTION :From the first measurement we get the value of focal length/of the COMBINATION of equiconvex lens and a planoconcave LIQUID lens i.e f = 45.0 CM. From the second measurement we find the focal length fx of the equiconvex lens i.e `f_(1) = 30.0 cm` If `f/2`be the focal length of planoconcave liquid lens, then `1/f =1/f_(1) + 1/f_(2)` or `1/f_(2) =1/f -1/f_(1) =1/45 -1/30 =-1/90 rArr f_(2) =-90 cm` Using lens maker.s formula for equiconvex lens, we get `1/f_(2) =(n_(1)-1) [1/(-R) -1/infty]` `1/(-90) = (n_(1)-1) [1/(-30) -1/infty] =-(n_(1)-1)/30` or `(n_(1)-1) = 30/90 = 0.33 rArr n_(1) = 1.33` |
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