This power point explains about resolving power of prism

1. RESOLVING POWER OF A PRISM

2. RESOLVING POWER OF A PRISM
Let ABC be the section of a prism which is placed in the
minimum deviation position. A parallel beam of light consisting
of wavelengths (+ d) is incident on the prism. The telescope
objective of the spectrometer focuses the wave front CE at P1 and
CF at P2. P1and P2 are the positions of the central maxima of
diffracting pattern for  and + d respectively.

3.  The two spectrual line disappear just resolved if the central
maxima of one falls over the first minima of others. By the
Fermat principle the optical path between the incident and
emergent wavefronts of any wavelength must be the same.
 Hence the wavelenght  PA+AE = nBC
 For wavelenght ( +d) PA+AF=( n- dn )BC
 Here n and (n- dn) are the refractive indices of the material of
the prism for wavelenght  respectively.
 Sub eq 2 from eq 1
AE- AF =BC . dn =tdn

4.  Here BC = t = width of the base of the prism Let
and(+d)be the angle of deviation of avelengths
(+d) and  respectively.
 Then the angle between the two emergent wavefronts CE and CF
is d
AE- AF =EF =CE.d= D.d
 Here CE= D = the width of the emergent beam = the
diameter of the telescope objective. From eqs 3 and 4
D.d = t. dn
D=t dn / D

5.  But the limiting condition for resolution in the case of the telescope is d = /D
From eqs 5 and 6
/D = t dn /D
= t dn/ d
That is:-
RP  t
RP  ( dn /d).

6. THANK YOU