Concepts of Physics

IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions

Variation of refractive index with wavelength

Objective

To find the variation of refractive index with wavelength.

Introduction

Refractive index of a material varies with wavelength. The relation is approximately given as \(\mu=\mu_0+\frac{A}{\lambda^2}\). The goal of this experiment is to find \(\mu_0\) and \(A\) for water. To get the refractive index you will use a water prism and get angle of minimum deviation. The relation between refractive index and minimum deviation is \(\mu=\frac{\sin\frac{A+\delta}{2}}{\sin\frac{A}{2}}\).

Apparatus

A hollow acrylic prism, red and green Laser torches, a plastic box, clamp in which laser can be fixed horizontally, measuring steel tape, sine and tan tables

Procedure

The glass tumbler is to be used as a stand for placing the prism and the wall is used as a screen. Clamp the red laser at an appropriate height so that the beam can go through the prism placed on inverted glass tumbler.

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Make arrangement so that the laser, without the prism, falls perpendicularly on the wall. Prism may be placed about 1.5 metre away from the wall.

With the laser switch pressed on (by the clamp), and putting the hollow prism on the tumbler, mark the spot on the wall. Put water in the prism so that the laser beam goes through water. The spot will shift.

Rotate the tumbler about its axis to get the position of minimum deviation. Measure different distances and calculate the angle of minimum deviation. From this, calculate the refractive index of water.

Repeat the same with green laser. Using wavelengths \(\lambda_1=525\, \mathrm{nm}\) and \(\lambda_2=672\, \mathrm{nm}\) get \(\mu_0\) and \(A\).