A bi-convex lens made of glass (refractive in | vedclass

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

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Increase and change sign

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(B) The focal length of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a bi-convex lens in air $(f_a)$,$\frac{1}{f_a} = (_a\mu_g - 1) \left( \frac{2}{R} ight)$,where $_a\mu_g = 1.5$.When placed in a liquid $(f_l)$,the refractive index becomes $_l\mu_g = \frac{\mu_g}{\mu_l} = \frac{1.5}{1.7} \approx 0.88$.Thus,$\frac{1}{f_l} = (_l\mu_g - 1) \left( \frac{2}{R} ight) = (0.88 - 1) \left( \frac{2}{R} ight) = -0.12 \left( \frac{2}{R} ight)$.Comparing the two: $\frac{f_l}{f_a} = \frac{_a\mu_g - 1}{_l\mu_g - 1} = \frac{1.5 - 1}{\frac{1.5}{1.7} - 1} = \frac{0.5}{\frac{1.5 - 1.7}{1.7}} = \frac{0.5 	imes 1.7}{-0.2} = -4.25$.Since the ratio is negative,the focal length changes sign (from convex to concave behavior) and its magnitude increases by a factor of $4.25$.

$A$ bi-convex lens made of glass (refractive index $1.5$) is put in a liquid of refractive index $1.7$. Its focal length will

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