A convex lens of focal length 0.15 m is made | vedclass

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(B) Given: Focal length in air $f_a = 0.15 \ m$,refractive index of glass $\mu_g = 1.5 = \frac{3}{2}$.When placed in liquid,the focal length increases

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$\frac{5}{4}$

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(B) Given: Focal length in air $f_a = 0.15 \ m$,refractive index of glass $\mu_g = 1.5 = \frac{3}{2}$.When placed in liquid,the focal length increases by $0.225 \ m$,so the new focal length $f_l = 0.15 + 0.225 = 0.375 \ m$.Using the lens maker's formula:$\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Dividing the two equations:$\frac{f_l}{f_a} = \frac{(\mu_g - 1)}{(\frac{\mu_g}{\mu_l} - 1)}$Substituting the values:$\frac{0.375}{0.15} = \frac{(1.5 - 1)}{(\frac{1.5}{\mu_l} - 1)}$$2.5 = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)}$$\frac{1.5}{\mu_l} - 1 = \frac{0.5}{2.5} = 0.2$$\frac{1.5}{\mu_l} = 1.2$$\mu_l = \frac{1.5}{1.2} = \frac{15}{12} = \frac{5}{4} = 1.25$.

$A$ convex lens of focal length $0.15 \ m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid,its focal length increases by $0.225 \ m$. The refractive index of the liquid is

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