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

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a concave lens in air,$R_1 =

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Convergent lens of focal length $3.5 R$

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a concave lens in air,$R_1 = -R$ and $R_2 = +R$,so $\frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = 0.5 \left( -\frac{2}{R} \right) = -\frac{1}{R}$,which means $f_a = -R$.When immersed in a medium of refractive index $\mu_m = 1.75$,the relative refractive index is $\mu_{rel} = \frac{\mu_g}{\mu_m} = \frac{1.5}{1.75} = \frac{6}{7}$.The new focal length $f_l$ is given by $\frac{1}{f_l} = \left( \frac{6}{7} - 1 \right) \left( -\frac{2}{R} \right) = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$.Thus,$f_l = +3.5 R$.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 same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

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