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 ?

Explanation

`rArr` Suppose, focal length of convex lens `f_1 = 30` cm and focal length of plano concave lens of liquid `f_2` and focal length of combination F = 45.0 cm
`rArr` Focal length of combination,
`1/F = 1/f_1 + 1/f_2`
`therefore 1/(f_2) = 1/F - 1/f_1`
`therefore 1/(f_2) = 1/45 -1 /30 = (2-3)/(90) = - (1)/(90)`
`therefore f_2 = -90 cm`
`rArr` For glass lens, suppose `R_1 = R, R_2 = - R`
`(1)/(f_1) = (mu -1) [ (1)/(R_1) - (1)/(R_2)]`
`(1)/(30) = [(3)/(2) -1][1/R - (1)/(-R)]`
`1/30 = 1/2 xx 2/R`
`therefore R = 30 cm `
`rArr` For lens of liquid,
`f_2 =- 90 cm , R_1 = -R = - 30 cm ` and `R_2 = oo`
`therefore` Lens maker.s formula,
`1/(f_2 ) = (mu_l-1)[(1)/(R_1) - (1)/(R_2)]`
`(1)/(-90) = (mu_l -1)[1/-R - 1/oo]`
`1/(-90) = (mu_1 -1) [ 1/-R][because 1/oo = 0]`
`therefore (1)/(-90) = (mu_1 -1)[1/-30]`
`therefore mu_l -1 = 1/3`
`therefore mu_l = 4/3 = 1.33`