The refractive index of a converging lens is | vedclass

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(B) According to the Lens Maker's Formula:$\frac{1}{f} = \left( \frac{\mu_{L}}{\mu_{S}} - 1 \right) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$W

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Infinite

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(B) According to the Lens Maker's Formula:$\frac{1}{f} = \left( \frac{\mu_{L}}{\mu_{S}} - 1 ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$Where $\mu_{L}$ is the refractive index of the lens and $\mu_{S}$ is the refractive index of the surrounding medium.Given that the lens is placed in a medium of the same refractive index,we have $\mu_{L} = \mu_{S}$.Substituting this into the formula:$\frac{1}{f} = \left( \frac{\mu_{L}}{\mu_{L}} - 1 ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$$\frac{1}{f} = (1 - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$$\frac{1}{f} = 0 	imes \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = 0$Since $\frac{1}{f} = 0$,the focal length $f$ becomes infinite $(f = \infty)$.Therefore,the lens behaves like a plane glass plate.

The refractive index of a converging lens is $1.4$. What will be the focal length of this lens if it is placed in a medium of the same refractive index? (Assume the radii of curvature of the faces of the lens are $R_{1}$ and $R_{2}$ respectively)

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