Question

The radius of curvature of two faces of a bouble convex lens are 10cm, 15cm and its focal length is 12cm, Then determine the refractive index of the lens. ​

Answer 1

Given:

Radius of curvature of two faces of a double convex lens are:

$\rm R_1 = 10 \: cm$

$\rm R_2 = -15 \: cm$

Focal length of the lens (f) = 12 cm

To Find:

Refractive index of the lens $\rm (\mu)$

Answer:

Lens Maker's Formula:

$\boxed{ \boxed{ \bf{ \dfrac{1}{f} = ( \mu - 1) \bigg( \dfrac{1}{R_1} - \dfrac{1}{R_2} \bigg) }}}$

By substituting values in the formula we get:

$\rm \implies \dfrac{1}{12} = ( \mu - 1) \bigg( \dfrac{1}{10} - \dfrac{1}{(-15)} \bigg) \\ \\ \rm \implies \dfrac{1}{12} = ( \mu - 1) \bigg( \dfrac{1}{10} + \dfrac{1}{15} \bigg) \\ \\ \rm \implies \dfrac{1}{12} = ( \mu - 1) \bigg( \dfrac{3 + 2}{30} \bigg) \\ \\ \rm \implies \dfrac{1}{12} = ( \mu - 1) \times \dfrac{5}{30} \\ \\ \rm \implies \dfrac{1}{12} = ( \mu - 1) \times \dfrac{1}{6} \\ \\ \rm \implies \mu - 1 = \dfrac{6}{12} \\ \\ \rm \implies \mu - 1 = \dfrac{1}{2} \\ \\\rm \implies \mu = \dfrac{1}{2} + 1 \\ \\ \rm \implies \mu = \dfrac{1 + 2}{2} \\ \\ \rm \implies \mu = \dfrac{3}{2} \\ \\ \rm \implies \mu = 1.5$

$\therefore$ $\boxed{\mathfrak{Refractive \ index \ of \ the \ lens \ (\mu) = 1.5}}$

Answer 2

1/f = (μ - 1)(1/R1 - 1/R2)

1/12 = (μ - 1)(1/10 - 1/-15)

μ = 3/2