Question

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.

Answer 1

Given,
Biconvex lens with each surface has a radius (R₁ = R₂ = 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,

$$\begin{gathered} \frac{{\mathrm{\mu }}_g}{v}-\frac{{\mathrm{\mu }}_{\mathrm{a}}}{u}=\frac{{\mathrm{\mu }}_{\mathrm{g}}-{\mathrm{\mu }}_{\mathrm{a}}}{R} \\ \Rightarrow \frac{1.5}{v}-\left(-\frac{1}{\infty }\right)=\frac{1.5-1}{10} \\ \Rightarrow \frac{1.5}{v}=\frac{0.5}{10} \\ \Rightarrow v=30\mathrm{cm} \end{gathered}$$

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

$$\begin{gathered} \frac{{\mathrm{\mu }}_{\mathrm{a}}}{v}-\frac{{\mathrm{\mu }}_{\mathrm{g}}}{u}=\frac{{\mathrm{\mu }}_{\mathrm{a}}-{\mathrm{\mu }}_{\mathrm{g}}}{R} \\ \Rightarrow \frac{1}{v}-\frac{1.5}{25}=\frac{1-1.5}{10} \\ \Rightarrow \frac{1}{v}-\frac{1.5}{25}=\frac{-0.5}{10} \\ \Rightarrow v=9.1\mathrm{cm} \end{gathered}$$

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