Question

2. A biconvex lens made of glass (refractive index 1.5)
has two spherical surfaces having radii 20 cm and
30 cm. Calculate its focal length. (ISC 2018)​

Answer 1

Given:

Refractive index (µ) = 1.5

Radii of two spherical surfaces of biconvex lens:

$\sf R_1 = 20 \ cm$

$\sf R_2 = -30 \ cm$

To Find:

Focal length (f)

Answer:

Lens Maker's Formula

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

By substituting values we get:

$\rm \implies \dfrac{1}{f} = (1.5 - 1)\bigg(\dfrac{1}{20} - \dfrac{1}{( - 30)} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5\bigg(\dfrac{1}{20} + \dfrac{1}{ 30} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5 \bigg(\dfrac{3 + 2}{60} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5 \times \dfrac{5}{60} \\ \\ \rm \implies \dfrac{1}{f} = \dfrac{2.5}{60} \\ \\ \rm \implies \dfrac{1}{f} = \dfrac{1}{24} \\ \\ \rm \implies f = 24 \: cm$

$\therefore$ $\boxed{\mathfrak{Focal \ length \ (f) = 24 \ cm}}$

Answer 2

Given

• A biconvex lens made of glass (refractive index 1.5) has two spherical surfaces having radii 20 cm and 30 cm.

We Find

• It's Focal Length

We Used

• Lens Maker's Formula

According to the question

$\huge{By \: substituting \: values \: we \: get:} \\ \\ </p><p></p><p>\begin{gathered}\rm \implies \dfrac{1}{f} = (1.5 - 1)\bigg(\dfrac{1}{20} - \dfrac{1}{( - 30)} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5\bigg(\dfrac{1}{20} + \dfrac{1}{ 30} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5 \bigg(\dfrac{3 + 2}{60} \bigg) \\ \\ \rm \implies \dfrac{1}{f} = 0.5 \times \dfrac{5}{60} \\ \\ \rm \implies \dfrac{1}{f} = \dfrac{2.5}{60} \\ \\ \rm \implies \dfrac{1}{f} = \dfrac{1}{24} \\ \\ \rm \implies f = 24 \: cm\end{gathered}</p><p></p><p></p><p>$

Hence, Focal length is 24 cm