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Physics Topics can help us understand the behavior of the natural world around us.
How do you Verify the Law of Refraction?
Dutch scientist Christian Huygens introduced a geometrical method suggesting the propagation of wavefront which is very useful in interpreting reflection, refraction, interference, diffraction etc., of waves. From the position and shape of a wavefront at a moment, we can easily determine its new position and shape at any subsequent time using Huygens’ principle.
Huygens’ principle: Each point on a wavefront acts as a new source called secondary source. From these secondary sources, secondary waves or wavelets generate and propagate in all directions at the same speed of the wave. The new wavefront at a later stage is simply the common envelope or tangential plane to these wavelets.
Let S be a point source of light from which waves spread in all directions. At some instant of time, AB is a spherical wavefront [Fig.(a)].
According to Huygens’ principle, points P1, P2, P3,….. etc. on wavefront AB act as new sources, and wavelets from these new sources spread out at the same speed. Small spheres each of radius ct (c = velocity of light) with centres P1, P2, P3 ….. respectively are imagined. Clearly, wavelets produced by the secondary sources P1, P2, P3 … reach surfaces of these imaginary spheres after a time interval t. Following Huygens’ principle, the common tangential plane A1B1, to the front surface of small spheres should be the new wavefront in the forward direction, after a time t [Fig.]. In this context, it is pointed out that A1B1 is also a spherical surface with centre at S.
Wavefronts, close to a source of light are spherical [Fig.(a)]. But when a wavefront is far enough from the source of light, the wavefront is taken to be a plane wavefront [Fig.(b)].
From Huygens’ method of tracing wavefronts, like wavefront A1B1, another wavefront A2B2 is also obtained behind AB, which can be called back wavefront. But in Huygens’ opinion, there is no existence of such back wavefront. Later this opinion was supported theoretically and also experimentally.
Verification of the Laws of Reflection
Using Huygens’ principle, laws of reflection of light can be proved. Reflection of a plane wavefront from a plane reflecting surface is shown in Fig.
Let AC, part of a plane wavefront, perpendicular to the plane of paper, be incident on the surface of a plane reflector XY [Fig.]. Note that the plane wavefront and the plane of paper intersect each other along line AC. The plane of paper and the plane of reflector are also perpendicular to each other. According to Huygen’s each point on wavefront AC acts as a source of secondary wavelet. Let at time t = 0, one end of the wavefront touches the reflecting surface at A. At the same moment, wavelets form from each point on the wavefront AC.
These wavelets gradually reach the reflector. After a time t say, wavelet from C reaches the point F of the reflector. In accordance with Huygens’ principle, all points from A to F on the reflecting plane in turn generate wavelets. Hence by the time the wavelet from C reaches F, wavelet generated at A reaches D in the same medium. Centering the point A, an arc of radius CF is drawn. Assuming c to be the speed of light in the medium under consideration, we have CF = ct. Tangent FD is drawn on the arc. FD is the reflected wavefront after time t.
The perpendiculars M1A and M2F drawn on the incident wavefront AC are the incident rays and the perpendiculars AM’1 and FM’2 drawn on the reflected wavefront DF are the corresponding reflected rays. NA and N’F are the normals drawn at the points of incidence on the reflector.
∴ ∠M1AN = angle of incidence (i)
and ∠M’2FN’ = angle of reflection (r)
According to Fig.,
∠CAF = 90° – ∠NAC = ∠M1AC – ∠NAC
= ∠M1AN = i
and ∠AFD = 90° – ∠N’FD = ∠M2‘FD – ∠N’FD
= ∠M’2FN’ = r
Now from ΔACF and ΔAFD, ∠ACF = ∠ADF = 90°, CF = AD = ct and AF is the common side.
Hence the triangles are congruent.
∴ ∠CAF = ∠AFD
Angle of incidence (i) = angle of reflection (r)
Thus one of the laws of reflection is proved.
AC, FD and AF are the lines of intersection of incident wavefront, reflected wavefront and plane of the reflector with the plane of paper respectively, which means AC, FD and AF lie on the plane of paper. Hence the perpendiculars to these lines, that is, incident ray (M1A or M2F), reflected ray (AM’1 or FM’2) and the perpendicular to the reflector (AN or FN’) at the point of incidence lie on the same plane. Thus, the other law of reflection is also proved.
Verification of the Laws of Refraction
Laws of refraction can also be proved using Huygens’ theory of wave propagation. Refraction of a plane wavefront in a plane refracting surface is shown in Fig.
Let XY be a plane refracting surface which is the surface of separation of medium 1 and medium 2 [Fig.], Speed of light in medium 1 is c1 and that in medium 2 is c2. Refractive indices of media 1 and 2 are µ1 and µ2 respectively.
Therefore, µ1 = \(\frac{c}{c_1}\) and µ2 = \(\frac{c}{c_2}\) where c is the speed of light in vacuum.
∴ Refractive index of the medium 2 with respect to that of medium 1,
1µ2 = \(\frac{\mu_2}{\mu_1}\) = \(\frac{c / c_2}{c / c_1}\) = \(\frac{c_1}{c_2}\) ….. (1)
Assume that, part of a plane wavefront normal to the plane of paper is incident on the surface of separation of two refractive media [Fig.]. The wavefront and the plane of paper intersect each other along the line AC. Plane of refraction XY is also at right angles to the plane of paper [Fig.].
According to Huygens’ principle, all the points on the wavefront AC act as sources of secondary wavelets. Suppose, at t = 0, one end of the wavefront touches the plane of the refracting medium at A. At the same moment, wavelets form from each point on the wavefront AC. These wavelets gradually reach the plane of the refracting surface. After a time t, say, the other end C of the wavefront touches F on the surface of separation.
Following Huygens’ principle, all points from A to F on the plane of separation will in turn generate wavelets. Hence, by the time the wavelet from C reaches F, wavelets generated at A advance toward medium 2. Centering the point A, an arc of radius c2t is drawn. Tangent FD is drawn to the arc from point F. FD is the refracted wavefront after time t.
Naturally, CF = c1t ….. (2)
and AD = c2t ….. (3)
The perpendiculars M1A and M2F drawn on the incident wavefront AC are the incident rays and the perpendiculars AM’1 and FM’2 drawn on the refracted wavefront DF are the corresponding refracted rays. NA and N’F are the normals drawn at the points of incidence on the surface of separation.
∴ ∠M1AN = angle of incidence (i)
and ∠M’2FN’ = angle of refraction (r)
According to Fig.,
∠CAF = 90° – ∠NAC = ∠M1AC – ∠NAC
= ∠M1AN = i
and ∠AFD = 90° – ∠N’FD = ∠M’2FD – ∠N’FD
= ∠M’2FN’ = r
∴ sini = sin∠CAF = \(\frac{C F}{A F}\)
and sinr = sin∠AFD = \(\frac{A D}{A F}\)
∴ \(\frac{\sin i}{\sin r}\) = \(\frac{\frac{C F}{A F}}{\frac{A D}{A F}}\) = \(\frac{C F}{A D}\) = \(\frac{c_1 t}{c_2 t}\) = \(\frac{c_1}{c_2}\) = 1µ2 = \(\frac{\mu_2}{\mu_1}\)
or, µ1 sin i = µ2 sin r
Hence Snell’s law of refraction is proved.
AC, FD and AF are the lines of intersection of incident wavefront, refracted wavefront, and the plane of refraction with the plane of paper respectively which means AC, FD, AF lie on the plane of the paper. Hence the perpendiculars on them, that is, incident ray (M1A or M2F), refracted ray (AM’1 or FM’2) and perpendicular to the refracting surface (AN or FN’) lie on the same plane. Thus the other law of refraction is also proved.
[The Fig.’s relating to the above discussion have been drawn for the refraction from a rarer to a denser medium. The entire proof can be done for the refraction from a denser to a rarer medium as well in the same manner.]
Velocity of light changes while passing from one medium to another keeping its frequency unchanged.
From equation (1) we get,
1µ2 = \(\frac{c_1}{c_2}\) = \(\frac{n \lambda_1}{n \lambda_2}\) = \(\frac{\lambda_1}{\lambda_2}\) …. (4)
[∵ c = nλ; n = frequency of light]
λ1 and λ2 are the wavelengths of light in media (1) and (2)
respectively. It means, wavelength of light changes due to refraction.
Numerical Examples
Example 1.
A plane wavefront, after being incident on a plane reflector at an angle of Incidence i, reflects from it. Show that the Incident wavefront and the reflected wavefront are Inclined at an angle (180° – 2i) with each other.
Solution:
AB and CD are the incident and reflected wavefronts respectively [Fig.]. On extending the wavefronts, they meet at E and the angle between them is θ. From figure, ∠BAC = i and ∠DCA = r.
From ΔACE, θ + i + r = 180°
or, θ = 180° – (i + r) or, θ = 180° – 2i [∵ i = r]
Example 2.
The wavelength of a light ray in vacuum is 5896A. What will be its velocity and wavelength when it passes through glass? Given, refractive Index of glass = 1.5 and velocity of light in vacuum = 3 × 108 m ᐧ s-1.
Solution:
Here, λ = 5896Å, c = 3 × 108 m ᐧ s-1, µg = 1.5
We know that, µg = \(\frac{c}{c_g}\)
or, cg = \(\frac{c}{\mu_g}\) = \(\frac{3 \times 10^8}{1.5}\) = 2 × 108 m ᐧ s-1
Again, nλ = c (in vacuum), nλg = cg (in glass)
∴ \(\frac{n \lambda_g}{n \lambda}\) = \(\frac{c_g}{c}\)
or λg = λ\(\frac{c_g}{c}\) = \(\frac{\lambda}{\mu_g}\) = \(\frac{5896}{1.5}\) = 3931Å
Example 3.
Refractive indices of glass with respect to water and air are 1.13 and 1.51 respectively. If velocity of light in air is 3 × 108 m ᐧ s-1, what will be its velocity in water?
Solution:
We know, aµg = \(\frac{c_a}{c_g}\)
∴ 1.51 = \(\frac{3 \times 10^8}{c_g}\) [cg = velocity of light in glass]
or cg = \(\frac{3 \times 10^8}{1.51}\) = 2 × 108m ᐧ s-1 (approx.)
Again, ωμg = \(\frac{c_w}{c_g}\) [cω = velocity of light in water]
or, 1.13 = \(\frac{c_w}{2 \times 10^8}\)
or, cω = 2.26 × 108 ᐧ s-1