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Wave Optics

Question
CBSEENPH12039122

(a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.

Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.

(b) Explain briefly how the phenomenon of total internal reflection is used in fiber optics.

Solution

a) The phenomenon of refraction for a ray of monochromatic light passing through a glass prism is shown as below: 

Let, PQR be the principal section of the prism.
The refracting angle of prism is A. Monochromatic light EF is incident on face PQ at angle of incidence i1. This ray enters from a rarer to denser medium and hence is refracted towards the normal FN. Refracted ray is FG. Angle of refraction for this face is r1. Refracted ray FG becomes incident on face PR and is refracted away from the normal GN2 and emerges in the direction GH. The angle of incidence on this face is r2 (into prism) and angle of refraction (into air) is i2. At point O, incident and the emergent ray meet. The angle between these two rays is called the angle of deviation ‘straight delta’. 
angle OFG space equals straight i subscript 1 space minus space straight r subscript 1 space space and space angle OGF space equals space straight i subscript 2 space minus space straight r subscript 2
increment F O G comma space delta space i s space t h e space e x t e r i o r space a n g l e.
delta space equals space angle O F G plus angle O G F space

italic space italic space equals space left parenthesis i subscript 1 space minus space r subscript 1 right parenthesis plus left parenthesis i subscript 2 space minus space r subscript 2 right parenthesis space equals space left parenthesis i subscript 1 plus i subscript 2 right parenthesis minus left parenthesis r subscript 1 plus r subscript 2 right parenthesis                    ... (1) 

The normal FN1 and GN2 on faces PQ and PR respectively, when produced meet at N.
Let, angle space F N G space equals space theta
r subscript 1 space plus space r subscript 2 space plus theta space equals space 180 to the power of o                              ... (2) 
I n space q u a d r i l a t e r a l space P F N G comma space angle P F N space equals space 90 to the power of o

therefore space A plus space 90 degree plus theta plus 90 degree space equals space 360 degree

space italic space italic space italic space A plus theta space equals space 180 degree space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis

Comparing space equations space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get space

straight r subscript 1 space end subscript plus space straight r subscript 2 space end subscript equals space straight A space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis space space space space space space space space space space space space space space
Putting space this space in space equation space left parenthesis 1 right parenthesis comma space we space get

delta space equals space i subscript 1 plus i subscript 2 space minus space A italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space space space... space left parenthesis 5 right parenthesis space space space

rightwards double arrow space i subscript 1 plus i subscript 2 space equals space A plus delta italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space... space left parenthesis 6 right parenthesis

Now comma space using space Snell ’ straight s space law comma space we space get

n space equals space fraction numerator sin i subscript 1 over denominator sin r subscript 1 end fraction equals fraction numerator sin i subscript 2 over denominator sin r subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 7 right parenthesis space
We can see that, for one angle of deviation we have two angles of incidence. But, angle of deviation is minimum for only one particular angle if incidence. Angle of minimum deviation is denoted by delta subscript m.

So let,                            i1 = i2 = i

And                               r1 = r2 = r

Therefore, we have (4) and (6), we have
space space space space space straight r plus straight r space equals space straight A space space

rightwards double arrow space r space equals space bevelled A over 2 a n d space i plus i equals space A plus delta subscript m

rightwards double arrow space i equals fraction numerator A italic plus delta subscript m over denominator italic 2 end fraction 
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b) An optical fiber is a device based on total internal reflection by which a light signal may be transmitted from one place to another with a negligible loss of energy. When a light ray is incident on one end at a small angle of incidence, it suffers refraction from air to quartz and strikes the quartz-coating interface at an angle more than the critical angle. Hence, total internal reflection happens and the ray of light strikes the opposite face again at an angle greater than critical angle. The phenomenon of total internal reflection takes place. Thus the ray within the fiber suffers multiple total internal reflections and finally strikes the other end at an angle less than critical angle for quartz-air interface and emerges in air.

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