Question

Determine the refractive index of a material for a prism using sodium source.

Answer 1

EXPERIMENT 5: Determination of the refractive index (µ) of the material of a prism using sprectometer ... a) Mercury lamp(as source of white light) b) Sprectometer.

Answer 2

Answer:

Refractive index: $n =\frac{sin((\delta _m + A)/2)}{sin(A/2)}$

Explanation:

We need a spectrometer to perform this experiment. Spectrometer need to be adjusted to minimum deviation codition and then required angles are measured.

From the figure:

Applying snell's law on each surface for light ray gives two equations,

$sin(i_1)=nsin(r_1)$ ----(1)

and, $n.sin(r_2)=sin(i_2)$----(2)

where $n$ is the refractive index.

Let $A$ be the angle of the prism.

Angle $r_1$ and $r_2$ are related by equation:

$A=r_1+r_2$----(3)

Applying exterior angle theorem, we find:

$\delta = (i_1-r_1)+(i_2-r_2)$ ----(4)

where $\delta$ is the angle of deviation.

from (3) and (4), we get,

$\delta = i_1+i_2-A$ ----(5)

Also (1)    $\implies i_1= sin^{-1} (n.sin(r_1))$ ----(6)

Also (2)   $\implies i_1= sin^{-1} (n.sin(A-r_1))$ ----(7)

Therefore, $\delta = sin^{-1}(n.sin(r_1))+sin^{-1}(n.sin(A-r_1))-A$

In the case of minimum deviation:  $\angle i_1=\angle i_2;$

So, $\angle r_1=\angle r_2=\angle r$ (say)

or, $\angle i = \frac{A}{2}$

$\implies i_2 = (\delta _m+A)/2$

substituting this in (2) gives,

$n =\frac{sin((\delta _m + A)/2)}{sin(A/2)}$

Refractive index is given by $n =\frac{sin((\delta _m + A)/2)}{sin(A/2)}$

#SPJ2