A bi-concave glass lens having refractive ind | vedclass

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(A) The focal length of a lens in a medium is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_

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convergent lens of focal length $3.5 R$

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(A) The focal length of a lens in a medium is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.For a bi-concave lens,$R_1 = -R$ and $R_2 = +R$.Given: $\mu_l = 1.5$,$\mu_m = 1.75$.Substituting the values: $\frac{1}{f} = (\frac{1.5}{1.75} - 1) (\frac{1}{-R} - \frac{1}{R})$.$\frac{1}{f} = (\frac{6}{7} - 1) (-\frac{2}{R}) = (-\frac{1}{7}) (-\frac{2}{R}) = \frac{2}{7R}$.Thus,$f = 3.5 R$.Since the focal length $f$ is positive,the lens behaves as a convergent lens.

$A$ bi-concave glass lens having refractive index $1.5$ has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

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