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(A) The lens is composed of two different materials with refractive indices $\mu_1$ and $\mu_2$. From the figure,the ray passing through the upper hal

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$\mu _1 > \mu _2$

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(A) The lens is composed of two different materials with refractive indices $\mu_1$ and $\mu_2$. From the figure,the ray passing through the upper half of the lens converges at point $I_1$,while the ray passing through the lower half converges at point $I_2$. The focal length $f$ of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Since the geometry of the lens is the same for both halves,the focal length is inversely proportional to the refractive index,i.e.,$f \propto \frac{1}{\mu - 1}$. From the figure,the image $I_1$ is formed closer to the lens than $I_2$,which means the focal length $f_1$ is less than $f_2$ $(f_1 Since $f \propto \frac{1}{\mu - 1}$,a smaller focal length corresponds to a larger refractive index. Therefore,$\mu_1 > \mu_2$.

$A$ lens made up of different materials forms two images of a point object $O$ as shown in the figure. Then:

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