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Derive the formula for angle of minimum deviation for the prism

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Was searching for this answer... thanks :)

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A prism is a wedge-shaped body made from a refracting medium bounded by two plane faces inclined to each other at some angle. The two plane faces are called are the refracting faces and the angle included between these two faces is called the angle of prism or the refracting angle.

In the below figure , ABC represents the principal section of a glass-prism having ∠A as its refracting angle.

A ray KL is incident on the face AB at the point F where $N_1$LO is the normal and $∠i_1$ is the angle of incidence. Since the refraction takes place from air to glass, therefore, the refracted ray LM bends toward the normal such that $∠r_1$ is the angle of refraction. If µ be the refractive index of glass with respect to air, then

Principal Section of a Glass Prism µ = $\frac {sin i}{sin r}$ ... (By Snell’s law)

The refracted ray LM is incident on the face AC at the point M where $N_2$MO is the normal and $∠r_2$ is the angle of incidence. Since the refraction now takes place from denser to rarer medium, therefore, the emergent ray MN such that $∠i_2$ 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, $δ = i_1 – r_1 + i_2 - r_2$ ….... (1)

$δ = i_1 + i_2 – (r_1 + r_2 )$

∠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,

$∠r_1 +∠r_2 + ∠LOM = 2rt∠s$ …... (3)

Comparing (2) and (3), we get

$∠LAM = ∠r_1 +∠r_2$

$A = ∠r_1 +∠r_2$

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

$δ = i_1 + i_2 - A$

or $i_1 + i_2 = A + δ$ …... (4)

Nature of Variation of the Angle of Deviation with the Angle of Incidence

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 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 $δ_m$ 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 below 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, $∠i_1 = ∠i_2$

and so $∠r_1 = ∠r_2 = ∠r$ (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,

$∠A = ∠r_1 + ∠r_2$

So, A = 2r (Since, for the prism in minimum deviation position, $∠r_1 = ∠r_2 = ∠r$)

or r = A/2 …...(5)

Again, $i_1 + i_2 = A + δ$

or $i_1 + i_1 = A + δ_m$ (Since, for the prism in minimum deviation position, $i_1 = i_2 and δ = δ_m$)

$2i_1 = A + δ_m$

or $i_1 = \frac {A + δ_m} 2$ …... (6)

Now $µ = \frac {sin i_1}{sin r_1} = \frac {sin i_1}{sin r}$

$\boxed {µ = \frac {sin [\frac {A + δm}2]} {sin (\frac {A}{2})}}$ …... (7)

This is the Relation Between Refractive Index and Angle of Minimum Deviation

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Thank you very much Just a quick question "❓ " you said that la=lm is it reasonable? I think not necessarily.