A bi-concave symmetric lens made of glass has | vedclass

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(D) Given:Refractive index of the glass lens,$n_2 = 1.5$Refractive index of the liquid medium,$n_1 = 1.75$Radius of curvature for a bi-concave lens: $

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(D) Given:Refractive index of the glass lens,$n_2 = 1.5$Refractive index of the liquid medium,$n_1 = 1.75$Radius of curvature for a bi-concave lens: $R_1 = -R$ and $R_2 = +R$.Using the Lens Maker's formula:$\frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Substituting the values:$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$$f = 3.5 \,R$Since the focal length $f$ is positive,the lens behaves as a converging lens with a focal length of $3.5 \,R$. Therefore,the correct option is $D$.

$A$ bi-concave symmetric lens made of glass has a refractive index of $1.5$. It has both surfaces with the same radius of curvature $R$. On immersion in a liquid of refractive index $1.75$,it will behave as a:

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