Question

A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50cm . Using this lens, you observe that it forms an image of a very distant tree at a distance of 1.87cm from the lens. What is the index of refraction of the lens?

Short answer

1. The refractive index of the lens is 1.67

Step-by-step solution

(a) Concept of refractive index. 

The concept of refractive index is related to the refraction of light which states the phenomenon of bending of light when it travels from one medium to another medium. The refractive index is a dimensionless quantity which gives the measure by which a medium can bend light travelling through it compared to another standard medium. The measure of path bend is suggested from the value of refractive index of the medium in concern.

(b) Determination of the refractive index of the lens.

The lens formula is,

$\frac{1}{\mathrm{f}}=\left(\mathrm{n}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{\mathrm{R}2}\right)$

Here, f is the focal length, n is the refractive index and the R’s are the radii of curvature.

For a distantly placed object, the image is formed at the focal length. From question f = 1.87 cm. Now, for a bi-convex lens, ${\mathrm{R}}_1=+\mathrm{R}$and ${\mathrm{R}}_2=-\mathrm{R}$, where R is given to be So substitute all the values in above equation and solve for n,

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{R}}{2\mathrm{f}}+1 \\ =\frac{2.50\mathrm{cm}}{2\left(1.87\mathrm{cm}\right)}+1 \\ =1.67 \end{gathered}$$

Thus, the refractive index of the lens is 1.67 .