A concave lens of glass,refractive index 1.5, | vedclass

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

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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: $\frac{1}{f} = (\frac{n_l}{n_m} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$.For a concave lens,$R_1 = -R$ and $R_2 = +R$.Given: $n_l = 1.5$,$n_m = 1.75$.Substituting the values: $\frac{1}{f} = (\frac{1.5}{1.75} - 1) (\frac{1}{-R} - \frac{1}{R})$.$\frac{1}{f} = (\frac{6}{7} - 1) (-\frac{2}{R}) = (-\frac{1}{7}) (-\frac{2}{R}) = \frac{2}{7R}$.Therefore,$f = +3.5 R$.Since the focal length is positive,the lens behaves as a convergent (convex) lens.

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

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