A lens behaves as a converging lens in air bu | vedclass

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(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_{rel}

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$\mu < 4/3$

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(C) The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.$1$. In air,$\mu_{medium} = 1$. Since it acts as a converging lens,$f > 0$,which implies $\mu_{lens} > 1$.$2$. In water,$\mu_{medium} = 1.33$ (or $4/3$). Since it acts as a diverging lens,$f $3$. Therefore,$\frac{\mu_{lens}}{\mu_{water}} $4$. Given $\mu_{water} = 4/3$,the condition is $1 < \mu_{lens} < 4/3$.

$A$ lens behaves as a converging lens in air but a diverging lens in water,then the refractive index $(\mu)$ of its material is

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