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(C) Consider the refraction at the first surface (from water to glass):$\frac{\mu_2 - \mu_1}{R_1} = \frac{\mu_1}{-u} + \frac{\mu_2}{v_1}$Given $\mu_1

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$3R/2$

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(C) Consider the refraction at the first surface (from water to glass):$\frac{\mu_2 - \mu_1}{R_1} = \frac{\mu_1}{-u} + \frac{\mu_2}{v_1}$Given $\mu_1 = 4/3$,$\mu_2 = 3/2$,$R_1 = R$,and $u = \infty$:$\frac{3/2 - 4/3}{R} = \frac{4/3}{\infty} + \frac{3/2}{v_1} \Rightarrow \frac{1/6}{R} = \frac{3/2}{v_1} \Rightarrow v_1 = 9R$Now consider the refraction at the second surface (from glass to air):$\frac{\mu_3 - \mu_2}{R_2} = \frac{\mu_2}{-v_1} + \frac{\mu_3}{v_2}$Given $\mu_2 = 3/2$,$\mu_3 = 1$,$R_2 = -R$,and $v_1 = 9R$:$\frac{1 - 3/2}{-R} = \frac{3/2}{-9R} + \frac{1}{v_2} \Rightarrow \frac{-1/2}{-R} = -\frac{1}{6R} + \frac{1}{v_2}$$\frac{1}{2R} + \frac{1}{6R} = \frac{1}{v_2} \Rightarrow \frac{3+1}{6R} = \frac{1}{v_2} \Rightarrow \frac{4}{6R} = \frac{1}{v_2} \Rightarrow v_2 = \frac{6R}{4} = \frac{3}{2}R$Since the incident rays are parallel,the final image distance $v_2$ is the focal length $f = 1.5R$.

There is an equiconvex glass lens with radius of each face as $R$,$_a{\mu _g} = 3/2$,and $_a{\mu _w} = 4/3$. If there is water in the object space and air in the image space,then the focal length is:

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