Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

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Answer 1

$Ths is a combination of convex, concave and convex lens. Number them 1, 2, 3 given f_{1} μ_{1} μ_{2} = f_{3} = 20 c m = μ_{3} = 1.5 = 4 / 3$
$μ_{2} = 4 / 3 Now densmaker formula \frac{1}{f} = (μ_{r} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) for (1) \frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{R _{1}} - (- \frac{1}{R}))$ $(equi convex lenS R_{1} = R_{2})$

$\frac{1}{2 0} = \frac{1}{2} \times \frac{2}{R} R = 20 c m for (2) \frac{1}{f _{2}} = (\frac{4}{3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) \frac{1}{f _{2}} = \frac{1}{3} (\frac{- 2}{R}) f_{2} = - 30 c m$

$Now we know that ieffective \frac{1}{f _{e q}} f_{eq} = 15 c m = P_{1} + P_{2} + P_{3} = \frac{1}{2 0} - \frac{1}{3 0} + \frac{1}{2 0} So it is a converging lens of focal length 15 c m .$

Answer 2

Ths is a combination of convex, concave and convex lens. Number them 1, 2, 3 given f_{1} μ_{1} μ_{2} = f_{3} = 20 c m = μ_{3} = 1.5 = 4 / 3

μ_{2} = 4 / 3 Now densmaker formula \frac{1}{f} = (μ_{r} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) for (1) \frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{R _{1}} - (- \frac{1}{R}))

(equi convex lenS R_{1} = R_{2})

\frac{1}{2 0} = \frac{1}{2} \times \frac{2}{R} R = 20 c m for (2) \frac{1}{f _{2}} = (\frac{4}{3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) \frac{1}{f _{2}} = \frac{1}{3} (\frac{- 2}{R}) f_{2} = - 30 c m

Now we know that ieffective \frac{1}{f _{e q}} f_{eq} = 15 c m = P_{1} + P_{2} + P_{3} = \frac{1}{2 0} - \frac{1}{3 0} + \frac{1}{2 0} So it is a converging lens of focal length 15 c m .