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

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$A$ convergent and a divergent beam

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(D) The focal length $f$ of a lens in a medium is given by the lens maker's formula: $\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For the upper half of the lens,the surrounding medium has refractive index $\mu_2$. Since $\mu_2 > \mu_1$,the term $(\frac{\mu_1}{\mu_2} - 1)$ is negative. Thus,the upper half acts as a diverging lens.For the lower half of the lens,the surrounding medium has refractive index $\mu_3$. Since $\mu_1 > \mu_3$,the term $(\frac{\mu_1}{\mu_3} - 1)$ is positive. Thus,the lower half acts as a converging lens.Therefore,the lens produces a convergent beam from the lower half and a divergent beam from the upper half.

$A$ double convex lens,made of a material of refractive index $\mu_1$,is placed inside two liquids of refractive indices $\mu_2$ and $\mu_3$,as shown. Given $\mu_2 > \mu_1 > \mu_3$. $A$ wide,parallel beam of light is incident on the lens from the left. The lens will give rise to

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