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dreive an expression for the refractive index of the material of a prism in terms of angle of minimum deviation and angle of prism
one year ago µ = sin i/sin r     (By Snell’s law)

The refracted ray LM is incident on the face AC at the point M where N2MO is the normal and ∠r2 is the angle of incidence. Since the refraction now takes place from denser to rarer medium, therefore, the emergent ray MN such that ∠i2 is the angle of emergence.

In the absence of the prism, the incident ray KL would have proceeded straight, but due to refraction through the prism, it changes its path along the direction PMN. Thus, ∠QPN gives the angle of deviation ‘δ’, i.e., the angle through which the incident ray gets deviated in passing through the prism.

Thus,   δ = i1 – r1 + i2 -r2     ….... (1)

δ = i1 + i2 – (r+ r)

Again, in quadrilateral ALOM,

∠ALO + ∠AMO = 2rt∠s                [Since, ∠ALO = ∠AMO = 90º]

So, ∠LAM +∠LOM = 2rt∠s           [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s]    ….... (2)

Also in ?LOM,

∠r1 +∠r2 + ∠LOM = 2rt∠s         …... (3)

Comparing (2) and (3), we get

∠LAM = ∠r1 +∠r2

A = ∠r1 +∠r2

Using this value of ∠A, equation (1) becomes,

δ = i1 + i2 - A

or i1 + i2 = A + δ                   …... (4) The angle of deviation of a ray of light in passing through a prism not only depends upon its material but also upon the angle of incidence. The above figure (2) shows the nature of variation of the angle of deviation with the angle of incidence. It is clear that an angle of deviation has the minimum value ‘δm’ for only one value of the angle of incidence. The minimum value of the angle of deviation when a ray of light passes through a prism is called the angle of minimum deviation.

The figure (3) shows the prism ABC, placed in the minimum deviation position. If a plane mirror M is placed normally in the path of the emergent ray MN the ray will retrace its original path in the opposite direction NMLK so as to suffer the same minimum deviation dm.

In the minimum deviation position, ∠i1 = ∠i2

and so ∠r1 = ∠r2 = ∠r    (say)

Obviously, ∠ALM = ∠LMA = 90º – ∠r

Thus, AL = LM

and so LM l l BC Hence, the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.

Since for a prism,

∠A = ∠r1 + ∠r2

So, A = 2r      (Since, for the prism in minimum deviation position, ∠r1 = ∠r2 = ∠r)

or r = A/2        …...(5)

Again, i1 + i2 = A + δ

or i1 + i1 = A + δm         (Since, for the prism in minimum deviation position, i1 = i2 and δ = δm)

2i1 = A + δm

or i1 = (A + δm) / 2           …... (6)

Now  µ = sin i1/sin r1 = sin i1/sin r

µ = sin [(A + δm) / 2] / sin (A/2)               …... (7) one year ago

When the prism is at minimum deviation position “Angle of incidence $\angle{i}$ = Angle of emergence $\angle{e}$ i.e ( $\angle{i}= \angle{e}$) and the angle of refractions at both the surfaces of the prism will be equal i.e $\angle{r_1}= \angle{r_2}$. From geometry we have
d=i+e-A    ; ${r_1}+{r_2}= A$
but when the prism is in minimum deviation d= $D_m$ , $D_m$=2i-A since $\angle{i}= \angle{e}$ $\implies$2i = A+ $D_m$ $\implies$ i = $\frac{A+D_m}{2}$ – – – – – – –  –  – – – – (a)
and 2r =A since $\angle{r_1}= \angle{r_2}= \angle{r}$ $\implies$ $r = \frac{A}{2}$ – – – – – – –  –  – – – – (b)
But according to snell’s law $\mu= \frac{Sini}{Sinr}$
substituting the values of  i and r in the above equation from equations (a) and (b) we get $\mu= \frac{sin\frac{(A+D_m)}{2}}{sin \frac{A}{2}}$
one year ago
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