When a biconvex lens of glass having refracti | vedclass

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Question

(D) The focal length of a lens in a medium is given by the Lens Maker's Formula:$\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2

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equal to that of glass

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(D) The focal length of a lens in a medium is given by the Lens Maker's Formula:$\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$where $\mu_l$ is the refractive index of the lens,$\mu_m$ is the refractive index of the surrounding medium,and $R_1, R_2$ are the radii of curvature.If the lens acts as a plane sheet of glass,its focal length $f$ becomes infinite $(f 	o \infty)$,which implies $\frac{1}{f} = 0$.Substituting this into the formula: $0 = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.Since the lens is biconvex,$(\frac{1}{R_1} - \frac{1}{R_2}) 
eq 0$.Therefore,$(\frac{\mu_l}{\mu_m} - 1) = 0$,which means $\frac{\mu_l}{\mu_m} = 1$,or $\mu_l = \mu_m$.Thus,the refractive index of the liquid must be equal to the refractive index of the glass $(1.47)$.

When a biconvex lens of glass having refractive index $1.47$ is dipped in a liquid,it acts as a plane sheet of glass. This implies that the liquid must have a refractive index:

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