A biconvex thick lens is constructed with glass (μ = 1.50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens.

#### Solution

Given,

Biconvex lens with each surface has a radius (*R*_{1}_{ }= *R*_{2} *= R*) = 10 cm,

and thickness of the lens (*t*) = 5 cm

Refractive index of the lens (μ) = 1.50

Object is at infinity (∴ u = ∞ )

First refraction takes place at A.

We know that,

\[\frac{\mu_g}{v} - \frac{\mu_a}{u} = \frac{\mu_g - \mu_a}{R}\]

\[ \Rightarrow \frac{1 . 5}{v} - \left( - \frac{1}{\infty} \right) = \frac{1 . 5 - 1}{10}\]

\[ \Rightarrow \frac{1 . 5}{v} = \frac{0 . 5}{10}\]

\[ \Rightarrow v = 30 \text{ cm }\]

Now, the second refraction is at B.

For this, a virtual object is the image of the previous refraction.

Thus, *u* = (30 − 5) = 25 cm

\[\frac{\mu_a}{v} - \frac{\mu_g}{u} = \frac{\mu_a - \mu_g}{R}\]

\[ \Rightarrow \frac{1}{v} - \frac{1 . 5}{25} = \frac{1 - 1 . 5}{10}\]

\[ \Rightarrow \frac{1}{v} - \frac{1 . 5}{25} = \frac{- 0 . 5}{10}\]

\[ \Rightarrow v = 9 . 1 cm\]

Hence, the image is formed 9.1 cm far from the second refraction or second surface of the lens.