Fig. Shows a concavo-convex lens. What is the condition on the refractive indices so that the lens is diverging ?
The refractive index of the lens is $μ_{2}$

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$μ_{3} > 2 (μ_{1} - μ_{2})$

$2 μ_{3} < μ_{1} + μ_{2}$

$2 μ_{3} > μ_{1} + μ_{2}$

None of these

answer is B.

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Detailed Solution

$\frac{μ_{2}}{V} - \frac{μ_{1}}{- \infty} = \frac{μ_{2} - μ_{1}}{- 2 R}$

$\frac{μ_{3}}{V_{f}} - \frac{μ_{2}}{V} = \frac{μ_{3} - μ_{2}}{- R}$

$\frac{μ_{3}}{V_{f}} + \frac{μ_{1}}{\infty} = \frac{μ_{1} - μ_{2}}{2 R} + \frac{μ_{2} - μ_{3}}{R}$

$\frac{{Rμ}_{3}}{V_{f}} = \frac{μ_{1} - μ_{2}}{2} + \frac{μ_{2} - μ_{3}}{1}$

$∵ V_{f} < 0$

$∴ \frac{μ_{1} - μ_{2}}{2} < μ_{3} - μ_{2}$

$\Rightarrow μ_{1} - μ_{2} < 2 μ_{3} - 2 μ_{2}$

$\Rightarrow μ_{1} + μ_{2} < 2 μ_{3}$

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