Optics Concept Page - 24

Formula

Variation in angle of deviation due to change in angle of incidence for a triangular prism

The angle of deviation dependent on the angle of incidence as:
δ=i+e−A$\delta =i+e-A$
where δ$\delta$ = angle of deviation
i$i$ = angle of incidence
e$e$ = angle of emergence
A$A$ = angle of the prism
A plot between the angle of deviation and angle of incidence is shown in the figure. In general, any given value of δ$\delta$, except for i=e$i=e$, corresponds to two values of i$i$ and hence of e$e$.

Definition

Deviation without Dispersion

Deviation without dispersion means an achromatic combination of two prisms in which net or resultant dispersion is zero and deviation is produced.

If A1$A_1$ and A2$A_2$ are the angles of the prism then the for mean ray (yellow ray) is

i)δ1+δ2=0${\delta }_1+{\delta }_2=0$; ω1.d1+ω2.d2=0${\omega }_1.d_1+{\omega }_2.d_2=0$  

ii) A1(μv1−μR1)+A2(μv2−μR2)$A_1({\mu }_{v1}-{\mu }_{R1})+A_2({\mu }_{v2}-{\mu }_{R2})$

iii) d1+d2≠0$d_1+d_2\ne 0$

A1A2=−(μv2−μR2)(μv1−μR1)$\frac{A_1}{A_2}=\frac{-({\mu }_{v_2}-{\mu }_{R_2})}{({\mu }_{v_1}-{\mu }_{R_1})}$

Example

Problem on Deviation without Dispersion

Problem:
A crown glass prism of refracting angle 6∘$6^{\circ }$ is to be used for deviation without dispersion with a flint glass of angle of prism a.Given: for crown glass μr=1.513${\mu }_r=1.513$ and μv=1.523${\mu }_v=1.523$, for flint glass μr=1.645${\mu }_r=1.645$ and μv=1.665${\mu }_v=1.665$. Find a.

Solution:

(μv1−μr1)a1=(μv2−μr2)a2$({\mu }_{v_1}-{\mu }_{r_1})a_1=({\mu }_{v_2}-{\mu }_{r_2})a_2$

ora2=(1.523−1.513)(1.665−1.645)6∘=3∘$or{a}_2={\displaystyle \frac{(1.523-1.513)}{(1.665-1.645)}}{6}^{\circ }=3^{\circ }$

Example

Calculate angle of minimum deviation of light ray passing through prism

The Angle of Deviation is the angle equal to the difference between the angle of incidence and the angle of refraction of a ray of light passing through the surface between one medium and another of different refractive index.

Example: A prism has a refractive index 32−−√${\displaystyle \sqrt{\frac{3}{2}}}$ and refracting angle 90o${90}^o$. Find the minimum deviation produced by prism.

Solution:
At minimum deviation, r1=r2=A/2=45∘$r_1=r_2=A/2={45}^{\circ }$
And, i=e=sin−1(32−−√)=60∘$i=e={\sin }^{-1}(\sqrt{{\displaystyle \frac{3}{2}}})={60}^{\circ }$
δmin=i+e−A=30∘${\delta }_{min}=i+e-A={30}^{\circ }$

Result

Factors affecting the angle of deviation

Angle of deviation depends on:

1. Angle of incidence i$i$.
2. Material of prism i.e. μ$\mu$.
3. Angle of prism A$A$.
4. Colour of wavelength used λ$\lambda$. 

Shortcut

Factors affecting on angle of deviation

Angle of deviation (δ$\delta$) is the angle between emergent ray and incident ray. 
 δ=(μ−1)A$\delta =(\mu -1)A$

1. Angle of incidence
2. Refractive angle of the prism
3. Refractive index of the prism material
   

Definition

Minimum angle of deviation for a prism

At the minimum deviation Dm$D_m$, the refracted ray inside the prism becomes parallel to its base.
i.e i=e⇒r1=r2=r$i=e\Rightarrow r_1=r_2=r$
Then r=A2$r={\displaystyle \frac{A}{2}}$
and Dm=2i−A$D_m=2i-A$
where i$i$ is the angle of incidence,
e$e$ is the angle of emergence.
r1,r2$r_1,r_2$ are the angles of refraction and A$A$ is the angle of the prism.

Example

Refractive Index of Prism using the angle of minimum deviation

Example: If the angle of minimum deviation produced by an equilateral prism is 30∘${}^{\circ }$ , then find the refractive index of material of prism.

Solution:
given,      A=60∘$A={60}^{\circ }$

δm=30∘$\delta m={30}^{\circ }$

μ= sinδm+A2sinA2$\mu ={\displaystyle \frac{sin{\displaystyle \frac{\delta m+A}{2}}}{sin{\displaystyle \frac{A}{2}}}}$

μ= sin60+302sin602$\mu ={\displaystyle \frac{sin{\displaystyle \frac{60+30}{2}}}{sin{\displaystyle \frac{60}{2}}}}$

=2√

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10  Page 11
Page 12  Page 13  Page 14  Page 15  Page 16  Page 17  Page 18  Page 19  Page 20  Page 21 
Page 22  Page 23  Page 24