Question

The refractive indices of the crown glass for blue and red
light are 1.51 and 1.49 respectively and those of the flint
glass are 1.77 and 1.73 respectively. An isosceles prism of
angle 6${}^{\circ }$ is made of crown glass. A beam of white light is
incident at a small angle on this prism. The other flint glass
isosceles prism is combined with the crown glass prism such
that there is no deviation of the incident light.
(a) Determine the angle of the flint glass prism.
(b) Calculate the net dispersion of the combined system.

Answer 1

(a) When angle of prism is small and angle of incidence is
also small, the deviation is given by $\delta =(\mu -1)A.$.
Net deviation by the two prisms is zero. So
${\delta }_1+{\delta }_2=0$
or $({\mu }_1-1)A_1+({\mu }_2-1)A_2=0...(i)$
Here, ${\mu }_1$ and ${\mu }_2$ are the refractive indices for crown and
flint glasses respectively.
Hence,${\mu }_1=\frac{1.51+1.49}{2}=1.5$
and ${\mu }_2=\frac{1.77+1.73}{2}=1.75$
$A_1$=Angle of prism for crown glass = 6${}^{\circ }$
Substituting the values in Eq. (i), we get
$(1.5-1)(6^{\circ })+(1.75-1)A_2=0$
This gives $A_2$= - 4${}^{\circ }$
Hence, angle of flint glass prism is 4${}^{\circ }$ (Negative sign
shows that flint glass prism is inverted with respect to the
crown glass prism.)
(b) Net dispersion due to the two prisms is
$=({\mu }_{b_1}-{\mu }_{r_1})A_1+({\mu }_{b_2}-{\mu }_{r_2})A_2$
$=(1.51-1.49)(6^{\circ })+(1.77-1.73)(-4^{\circ })$
$=$ - 0.04${}^{\circ }$
$\therefore$ Net dispersion is - 0.04${}^{\circ }$