Refraction Through Plane Surfaces

Refraction Through Plane Surfaces 

Glass Slab 

The lateral displacement (d) of the ray is the perpendicular distance between the incident and the emergent ray as shown in the fig.(15.27). It is given by 

d =   (15.10)

Since alpha; and beta; are small, therefore, tan α ≈ sin α and tan β ≈ sin β 

Thus,

Using Snell’s law 

μ sin β = sin α 

∴ sin α/sin β = μ

Now μ = h/h'

or h′ = h/μ

The apparent shift in the position of the source is 

s = h – h′ = h(1-1/μ)(15.11)

If there are n number of slabs with different refractive indices are placed between the observer and the object, then the total apparent shift is equal to the summation of all the individual shifts. 

s = s1 + s2 + ………..+sn

ors = h1 (15.12)

 

If the shift comes out to be positive, the image of the object shifts toward the observer, and vice-versa.

Example 15.7

Solution 

Example 15.8

Solution 

Using Snell’s law 

μsinθ = constant 

∴1 sin 60o =

Thus,sin r2 = or r2 = sin-1(1/√3) ≈ 35o

Important 

1.When the glass slab of thickness t and refractive index μ is placed in the path of a convergent beam as shown in the figure(15.31  a), then the point of convergence is shifted by 

s = t(1 - 1/μ)

2.When the same glass slab is placed in the path of a divergent beam as shown in fig.(15.31b), then the path of divergence also gets shifted by 

s = t(1 - 1/μ)

Example 15.9

If a point object is placed at distance 20 cm in front of a concave mirror of focal length 10 cm, then the final image coincides with the object itself. But, if we place a glass slab (μ = 1.5) of thickness 3 cm between the mirror and the object. Then find the new position of the object so that the final image again coincides with the object. Also, draw the ray diagram. 

Solution 

Total Internal Reflection 

Fig.(15.32) shows a boundary between two media with refractive indices μ1 and μ2 where μ2 > μ1. A ray approaching the boundary from the medium of higher refractive index is refracted away from the normal. For small angle of incidence, there is both a reflected and a refracted ray. But, at a particular angle of incidence (called critical angle θc) the refracted ray emerges parallel to the surface. For any angle of incidence greater than θc the light is totally reflected back into the medium of higher refractive index. The phenomenon is called total internal reflection and was first noted by Kepler in 1604. 

Example 15.10

In the fig.(15.33) shown, for an angle of incidence i at the top surface, what is the minimum refractive index needed for total internal reflection at the vertical face? 

Solution 

Applying Snell’s law at the top surface 

μ sin r = sin i (i)

For total internal reflection at the vertical face 

μ sin θc = 1 

Using geometry,  θc = 90o – r 

∴μ sin (90 – r) = 1 

orμ cos r = 1 (ii)

On squaring and adding equation (i) and (ii), we get 

μ2sin2r + μ2cos2r = 1 + sin2i 

or μ = √1 + sin2i

Example 15.11

Solution 

Rays coming out of the source and incidenting at an angle greater than θc will be reflected back into the liquid, therefore, the corresponding region on the surface will appear dark. 

As obvious from the figure, the radius of the bright spot is given by 

R = h tan θc =

orR =

Sincesinθc = 1/μ

∴R =

Example 15.12

Solution 

From the figure, 

α = 90o – A 

i = 90o - α = A (i)

Alsoβ = 90o – 2i = 90o – 2A 

andr = 90o - β = 2A 

Thus,γ = r = 2A

From geometry, 

A + γ + γ = 180o

A + 4A = 180o

orA = 180/5 = 36o 

PRISM 

Prism is a transparent medium bounded by any number of surfaces in such a way that the surface on which light is incident and the surface from which light emerges are plane and non-parallel. 

Important

1.Angle of prism or refracting angle (A) of prism means the angle between the faces on which light is incident and from which it emerges, as shown in the fig.(15.36). 

2.Angle of deviation (δ) means the angle between emergent and incident rays i.e., the angle through which incident ray turns in passing through a prism. It is represented by δ and is shown in fig.(15.36).

3.If the faces of a prism on which light is incident and from which it emerges becomes parallel (as in fig.(15.37)), angle of prism  will be zero and as incident ray will emerge parallel to itself, deviation will also be zero i.e., the prism will act as a slab. 

4.If μ of the material of the prism becomes equal to that of surroundings, no refraction at its faces will take place and light will pass through it undeviated.

i.e., δ = 0 

Condition of No Emergence 

The light will not emerge out of a prism for all values of angle of incidence, ifr2 > θc  for  i = 90o(1)

Applying Snell’s law,

1 (sin 90) = μ sin r1

orr1 = sin-1(1/μ)i.e.r1 = θc (2)

Thus, a ray of light will not emerge out of a prism (whatever be the angle of incidence) if A > 2theta;c,  that is, if mu; > cosec(A/2)

Condition of Grazing Emergence 

If a ray can emerge out of a prism, the value of angle of incidence i for which angle of emergence e = 90o is called the condition of grazing emergence. 

Sincesin θc = 1/μ

∴sini =

ori = (15.15)

Note that light will emerge out of a prism only if angle of incidence is greater than the angle of incidence i given by equation (15.15). 

Deviation Produced by a Prism 

As shown in the Fig.(15.40), the angle of deviation δ is given by

sin i = μ sin r1  ⇒i = μ r1

and sine = μ sin r2            ⇒e = μ r2

Thusδ = i + e – A = μ(r1 + r2) – A 

orδ = A(μ - 1)

Maximum deviation 

Deviation will be maximum when angle of incidence i is maximum (imax = 90o)

Thus,δmax = 90 + e – A (6)

However when i = 90o,  1 sin 90o = μ sin r1

orr1 = sin-1(1/μ)⇒r1 = θc

Thus, at the second surface,

μ sin r2 = 1 sine orsin e = μ sin(A - θc) 

ore = sin-1[μsin(A - θc)] (7)

So equation (6) gives maximum deviation while (7) angle of emergence in condition of maximum deviation i.e. i = 90o.

Minimum deviation 

Theory and experiment show that δ will be minimum when the angle of incidence is equal to the angle of  emergence. 

i = e 

Thus, 

δmin = (i + e) – A = 2i – A 

Applying Snell’s law at each interface, we get 

sin i  = μ sin r1

andμ sin r2 = sin e = sin i 

Thussin r1 = sin r2

orr1 = r2 = r (say)

Since r1 + r2 = A, therefore   r = A/2

The refractive index of the prism is given by 

μ = (15.16)

Note that if the prism is equilateral or isosceles then the ray inside the prism is parallel to its base. 

Dispersion of Light by a Prism

When a beam of composite light (consisting of several wavelengths) passes though a prism, it splits into its constituent colours. This phenomenon is called dispersion. The band of colours thus obtained on a screen is called the spectrum. If white light is used, seven colours are obtained as shown in the figure. The sequence of colours is VIBGYOR, from bottom to top.

 

The dispersion of light takes place because the refractive index μ of the medium depends on the wavelength of light. An approximate empirical relation is given by Cauchy’s formula,

where A and B are constants. The smaller the value of λ, the larger is the value of μ. Thus, μ is maximum for violet colour and minimum for red. The deviation of a ray depends on μ, it is larger for higher μ. Hence, violet suffers the maximum deviation and red the minimum.

Total angular dispersion is 

θ = δv – δr.

where δv and δr are the deviations for violet and red light respectively.

Dispersive Power of a Prism

The ratio of dispersion to mean deviation is called dispersive power,

where δy is the deviation of yellow light (whose wavelength is considered as mean of all the wavelengths present)

For a prism, δ = (μ – 1)A

∴(15.17)

The dispersive power has no units and no dimensions. Its value depends on the material of the prism. 

From a single prism, it is not possible to get deviation without dispersion, or to get dispersion without deviation.

Deviation Without Dispersion (Achromatic Prism)

where δ is the deviation for the mean ray.

Dispersion without Deviation

Two prisms can be combined in such a way that the deviation of the mean ray produced by one is equal and opposite to that produced by the other. Such a combination is called a direct vision prism. The required condition is  (μ – 1) A = (μ′ – 1) A′(15.19)