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let us now study about the angle of daeviation of light ray through a prizm. this is the
figure which is showing, the propagation of light which is being refracted at the 2 mat
surfaces of the prizm. if prizm angle is ay we already studied that ay is equal to r-1
plus r-2. and for the 2. surface where refraction is taking place by snell’s law we can write.
sine i is equal to mu sine r-1. and sine- e is equal to mu sine r-2. now in this situation
if prizm was not present the light ray would be going in this direction. finally this is
the emergent ray here we can write it as incident ray and this is the emergent ray which is
coming out in air. and it is going in this direction if we back extended we can see this
is the total angle by which the light ray is daeviated from the initial direction. due
to refraction through the prizm. so this angle we define as delta here. this angle delta
we can write as, angle of daeviation. of light which is refracted through the prizm. now
we can easily calculate the value of this daeviation angle as this exterior angle this
must be sum of these 2 angles which are the interior angles of this triangle. so here
if this angle is i it is r-1. this angle we can write it i minus r-1, and this angle we
can write as e minus r-2 because this total angle is e and this angle is r-2. so in this
situation we can directly write angle of daeviation. this delta we can write as sum of these 2
interior angles which are i minus r-1 plus. e minus r-2. and here r-1 plus r-2 we know
it is equal to prizm angle so daeviation angle we can write. i plus e minus ay. and in this
relation we can also substitute the value of e in terms of i by using these relations.
so here we can see the value of e is sine inverse of. mu sine. r-2, r-2 we can write
as, ay minus r-1. so this will be ay minus r-1. this is the way how we can write the
value of e and value of r-1 we can substitute from here. so here you can see the value of
delta in terms of i only we can say it is i minus, ay. Plus, e and, e we can write as,
sine inverse of. mu sine of. a-minus, r- 1 we can substitute as. sine inverse of sine
i by mu. and here you can see the value of daeviation angle is a function of angle of
incidence of light on the first mat surface. so here you can see. as the angle of incidence
of light on first mat surface changes the value of angle of daeviation also change.
now in this situation. we’ll continue on the next sheet and we’ll study how. the
value of angle of daeviation depends on. the angle of incidence i and how it varies. lets
continue on the next sheet. as recently we have studied that the, angle of daeviation
of light which passes through 2 mat surface or prizm depends on angle of incidence. the
function which we have deduce on the previous sheet if we plot. you can see the approximate
curve of. the variation of delta with i looks like this. it first decreases with i and,
at a particular value of i which we can term as i-not. the value of daeviation angle. is
minimum. and for 2 values of, incidence angle which are. i-1 and i-2, the value of daeviation
angle is, maximum. now in this situation the value of minimum daeviation we can easily
deduce, by maximum minima as here we can write. delta will be minimum. when. d delta over
d-i is equal to zero. i am not solving this expression and its derivative because it’ll
be too lengthy. for a discussion. so here directly i can write. on analysis. of this
expression. we find that. d delta over d-i is equal to zero when. light ray. passes through
the prizm. symmetrically. now in this situation symmetrically implies. when. the value of
i is equal to e and, the value of refracting angle r-1 is equal to r-2 and which can be
written as ay by 2 as sum of the 2. internal angles is equal to the prizm angle. and here
we can directly use by snell’s law. we have studied sine i is equal to mu sine r-1 if
we substitute the value of i as i-not. then this will be sine of i-not, is equal to mu
sine of, the value of r-1 we can write as ay by 2. and the value of minimum daeviation
we can calculate as. we know that daeviation angle is written as i plus e minus ay. and
as we have written that the daeviation, of light through the prizm is minimum when i
is equal to e here which is i-not. so here delta we can write as 2-i-not minus ay and
this will be delta minimum. so in this situation the value of incidence angle would be given
as delta minimum plus, a-by 2. if we substitute this, value of incidence angle here corresponding
to minimum daeviation. this will give us sine of delta minimum plus ay by 2. is equal to
mu sine of ay by 2. this is an expression by which we can calculate the minimum daeviation
produced. by the prizm if its prizm angle and refractive index is known so this quite
an important expression you just keep in mind. and if we just draw the path of light ray
through the prizm in the state of minimum daeviation. as we have written minimum daeviation
would be when the light ray is passing through the prizm symmetrically so if this angle is
i-not. the light ray which is. refracted into the prizm will be passing symmetrically or
it’ll be parallel to the base of prizm if this is an isoscelas prizm. and on the other
side light will also be coming out at the same angle i-not. now in this situation, the
daeviation angle produce by the light this angle delta will be minimum. so always keep
in mind whenever a light ray is passing through the prizm symmetrically for an isoscelas prizm
is refracted rays parallel to the base. then in this situation the daeviation angle of.
light between incident ray and emergent ray will be minimum, and it can be directly calculated
by this relation.