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 $\mu_g = 1.5 = \frac{3}{2}$, and the increase in focal length is $0.225 \,m

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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 $\mu_g = 1.5 = \frac{3}{2}$, and the increase in focal length is $0.225 \,m$. Therefore, the new focal length in liquid is $f_l = f_a + 0.225 = 0.15 + 0.225 = 0.375 \,m$. Using the Lens Maker's formula: $\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$ For air $(\mu_a = 1)$: $\frac{1}{f_a} = (\frac{\mu_g}{1} - 1)(\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 - 1)K = 0.5K$, where $K = (\frac{1}{R_1} - \frac{1}{R_2})$. For liquid $(\mu_l)$: $\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)K$. Dividing the two equations: $\frac{f_l}{f_a} = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)}$. Substituting values: $\frac{0.375}{0.15} = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)} \Rightarrow 2.5 = \frac{0.5}{(\frac{1.5}{\mu_l} - 1)}$. $(\frac{1.5}{\mu_l} - 1) = \frac{0.5}{2.5} = 0.2$. $\frac{1.5}{\mu_l} = 1.2 \Rightarrow \mu_l = \frac{1.5}{1.2} = \frac{15}{12} = \frac{5}{4} = 1.25$.

$A$ 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 is increased by $0.225 \,m$. The refractive index of the liquid is

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