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Q.

A double convex lens has a radius of curvature 15 cm. The magnification of an object is 4/3 and the object is at a distance of 9 cm from the focus of the lens. The refractive index of the material of the lens is

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a

1.125

b

1.625

c

1.375

d

1.5

answer is B.

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Detailed Solution

$f = \sqrt{x y} magnification = \sqrt{\frac{y}{x}} = \frac{4}{3} . ∴ y = \frac{16}{9} x and x = 9 cm . ∴ y = 16 cm$

$∴ \sqrt{(16) (9)} = 12 cm But \frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} + \frac{1}{R_{2}}]$

$∴ \sqrt{(16) (9)} = 12 cm But \frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} + \frac{1}{R_{2}}]$

$∴ \frac{15}{24} = \frac{5}{8} = μ - 1$

$μ = 1.625$

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A double convex lens has a radius of curvature 15 cm. The magnification of an object is 4/3 and the object is at a distance of 9 cm from the focus of the lens. The refractive index of the material of the lens is