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REFRACTION

LAWS OF REFRACTION (AT ANY REFRACTING SURFACE) :

(i) The incident ray (AB), the normal (NN') to the refracting surface (II') at the point of incidence (B) and the refracted ray (BC) all lie in the same plane called the plane of incidence or plane of refraction . Refractive Index :

The refractive index (µ) of a medium is defined as the ratio of the speed of light in vacuum (c₀) to the speed of light in the medium (c).

µ =

c c₀

The relative refractive index of two media is equal to the ratio of their absolute refractive indices. µ₂₁ = 1 2   = 1 0 2 0 c / c c / c = 2 1 c c

The refractive indices of glass of water are

2 3 (= 1.5) and 3 4 (= 1.33), respectively.. (ii) Sin i Sin r = Constant :

for any two given media and for light of a given wave length. This is known as

SNELL'S Law . Sin i Sin r = 1n2 = n n 2 1 = v v 1 2 =   1 2

Note : Frequency of light does not change during refraction . Law of refraction in vector form :

If iˆ _{denotes a unit vector along incident light ray, rˆ a unit vector along refracted ray into a medium of}

refractive index m and nˆ a unit vector normal to the boundary of the medium directed towards incident medium, the laws of refraction can be expressed as

) nˆ rˆ ( nˆ iˆ _r i    

REFRACTION THROUGH A PLANE SURFACE Application of Snell's Law :

(i) When light propagates through a series of parallel layers of different medium as shown in figure, then according to Snell's law we may write µ₁ sin q₁ = µ₂ sin ₂ = µ₃ sin ₃ = ... = constant

In general,

µ sin  = constant

(ii) When light passes from rarer to denser medium it bends toward the normal as shown in figure. Using Snell's law

µ₁ sin ₁ = µ₂ sin  2 1 sin sin   = 1 2   > 1 Thus, if µ₂ > µ₁ then ₂ < ₁

When a light ray passes from denser to rarer medium it bends away from the normal as shown in figure. From Snell's law, we know that

2 1 sin sin   = 1 2   < 1

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Sir

Thus, if µ₂ < µ₁ then ₂ > ₁

Principle of Reversibility of Light Rays :

A refracted ray reversed to travel back along its path will get refracted along the path of the incident ray. Thus the incident and refracted ray are mutually reversible.

3. Relation between object and image distance(for normal incidence) :

An object O placed in medium 1 (refractive index µ₁) is viewed from the medium 2 (refractive index µ₂). It is important to note that the object and image both are formed on the same side of the boundary. The image distance y and the object distance x are related as

y = x 1 2        

• If µ₂ > µ₁, that is, when the object is observed from a denser medium, it appears to be farther away from the interface, i.e. y > x.

• If µ₂ < µ₁, that is, when the object is observed from a rarer medium, it appears to be closer to the interface, i.e. y < x.

Note : that the above formula is only applicable for normal view or paraxial ray assumptiion.

Ex. See the figure

(i) At what distance will the bird appear to the fish (ii) At what distance will the fish appear to the bird Sol.

(i) For fish : d_B = 36 + 48 = 84 cm d_B = 36 + 48 = 84 cm (ii) For bird : d_F = 27 + 36 = 63 cm.

Ex. A concave mirror is placed inside water with its shining surface upwards and principal axis of concave mirror. Find the position of final image.

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Sol. The incident rays will pass undeviated through the water surface and strike the mirror parallel to its principal axis. Therefore for the mirror, object is at . Its image A (in figure) will be formed at focus which is 20 cm from the mirror. Now for the interface

between water and air, d = 10 cm.

 d =       a w n n d =       1 3 / 4 10 = 7.5 cm.

Ex. A concave mirror is placed inside water with its shining surface upwards and principal axis vertical as shown. Rays are incident parallel to the principal axis of concave mirror. Find the position

of final image.

Sol. 7.5 cm above the water surface.

Refraction across multiple slabs :

In figure, an object is placed in front of two slabs in contact. The thickness and refractive indices of the slabs are t₁, µ_a and t₂, µ_b respectively. Where will the final image of the object appear to be ?

O

Slab 1

Slab 2

Eye

A object placed in front of two glass slabs in contact

A light ray emerging from O now refracts at three surfaces. The first is between air and µ_a, the second between µ_a and µ_b while the third is between µ_b and air. Let us solve the problem taking one step at a time. 1st interface O P1 Air µa X x Here , µ₁ = 1; µ₂ = µ_a d₁ = –x, d₂ = ? But 2 2 1 1 d d   

Therefore, the image distance d₂ = –µ_ax ... (8)

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2nd interface P2 µb X µa Y I1 Here, µ₁ = µ_a , µ₂ = µ_b d₁ = – (µ_ax + t₁) ; d₂ = ? Since 2 2 1 1 d d   

the image distance d₂ = –µ_{b }        a 1 µ t x _...(8) 3rd Interface µb X Y I2 Here, µ₁ = µ_b, µ₂ = 1 d₁ = –µ_b          a 1 t x _{+ t} 2, d2 = ?

and the final image distance from the 3rd interface is

d₂ = –           a 2 a 1 t t x

Therefore, the net shift in the position of the image is

s = – _           b 2 a 1 t t x _{–[–(x + t} 1 + t2) or s = t_{1 }         a 1 1 _{+ t} 2           b 1 1

Looking at the above result we realize that the net shift in the position of the image is simply the sum of the individual shifts at each of the slabs if they were independently placed in air.Apparent shift

Thus, the simple problem of refraction in a glass slab can be tackled in two ways :

1. By the method of interfaces : Here, the refraction formula, equation, is applied at each interface. 2. By the method of elements : Here, the slab itself is an element with a governing equation.

Most problems in ray optics can be solved by either of the two methods. In the following problems, we shall solve problems in both ways and highlight the method that gives a quicker solution.

Ex. A tank contains three layers of immiscible liquids (figure). The first layer is of water with refractive index 4/3 and thickness 8 cm. The second layer is an oil with refractive index 3/2 and thickness 9cm while the third layer is of glycerine with refractive index 2 and thickness 4 cm. Find the apparent depth of the bottom of the container.

µ = 2 µ = 3/2 µ = 4/3 4 cm 9 cm 8 cm O

Case-1 : Method of interfaces :

A ray of light from the object undergoes refraction at three interfaces. (1) Water-oil, (2) Oil-glycerine (3)

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Glycerine-air. The coordinate system for each of the interfaces is shown in figure. Interface 3 O Y Y Y X X X Interface 2 Interface 1 Water-Oil Interface : d₁ = –8cm, µ₁ = 4/3, µ₂ = 1.5 As 2 2 1 1 d d    or d₂ = 1 2   d₁we get d₂ = –9cm Oil-Glycerine Interface : d₁ = –(9 + 9) = –18cm, µ₁ = 1.5, µ₂ = 2 As 2 2 1 1 d d    or d₂ = 1 2   d₁ we get d₂ = –24 cm Glycerine-Air Interface : d₁ = –(4 + 24) = –28 cm, µ₁ = 2, µ₂ = 1 As 2 2 1 1 d d    or d₂ = 1 2   d₁ we get d₂ = –14 cm Thus the final image is 14cm below the glycerine - air interface. Case-2 : Method of elements :

The system now comprises three slabs - one of water, one of oil and one of glycerine. As discussed in this Section and given by equation, the net shift of the system in the sum of the individual shifts each of the slabs asuming they were in air.

Therefore, Net shift s = t₁         1 1 1 _{+ t} 2         2 1 1 _{+ t} 3         3 1 1 s = 8 _       4 3 1 _{+ 9 }       3 2 1 _{+ 4 }       2 1 1 = 7cm.

The direction of the shift is in the direction of the incident rays which is upwards. Therefore, final position of the object is = (4 + 9 + 8) – 7 = 14cm below the glycerine-air interface.

GLASS SLAB

Single glass slab in air

(i) W hen a glass slab of thickness and refractive index  is placed in the path of a convergent beam as shown in the figure, then the point of convergence is shifted by

s = t         1 1

(ii) When the same glass slab is placed in the path of a diverging beam, the point of divergence is shifted by

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s = t         1 1

Important It is important to note that the shift (s) is always on the direction of light.

If the slab is made of air and the surrounding medium is of refractive index , then the apparent shift would be

s = t ( – 1)

Concepts

The refracting surfaces of a glass slab are parallel to each other. When a light ray travels through a glass slab, it is refracted twice at the two parallel forces and finally emerges out parallel to its incident direction. The light ray undergoes zero deviation,  = O.

Angle of emergence = Angle of incidence

e = i

The lateral displacement of the ray is the perpendicular distance between the incident and the emergent ray and is given by

d =       r cos t sin (i – r)

Ex. A convergent beam is incident on two slabs placed in contact as shown in figure. Where will the rays finally converge ? µ = 3/2 µ = 2 A B O 6cm 4cm 14cm Air

Case-1 : Method of Interfaces

A ray of light from the object undergoes refraction at three interfaces. (1) Air-Medium A, (2) Medium A-Medium B (3) A-Medium B-Air. The coordinate system for each of the interfaces is shown in figure.

Air-Medium A Interface : A B O Y Y Y X X _X d₁ = +14cm, µ₁ = 1, µ₂ = 1.5 As 2 2 1 1 d d    or d₂ = 1 2   d₁ we get d₂ = +21 cm

Medium A-Medium B Interface :

d₁ = (21 – 6) = 15 cm, µ₁ = 1.5, µ₂ = 2 As 2 2 1 1 d d    or d₂ = 1 2   d₁ we get d₂ = +20 cm

Medium B-Air Interface :

d₁ = (20 – 4) = +16 cm, µ₁ = 2, µ₂ = 1

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As 2 2 1 1 d d    or d₂ = 1 2   d₁ we get d₂ = +8 cm Thus the final image is 8cm in front of the medium B-air interface. Case-2 : Method of Elements :

The system now comprises two slabs. The net shift is s = t₁         1 1 1 _{+ t} 2         2 1 1 = 6 _       3 2 1 _{+ 4 }       2 1 1 _{= 4 cm}

The direction of the shift is in the direction of the incident rays which is to the right. Therefore, the rays will finally converge to a point.

= (14 – 6 – 4) + 4 = 8 cm to the right of the Medium B-air interface.

Asking question

The figure shows three objects O₁, O₂ and O₃ located on a vertical line in a liquid of refractive index µ. These objects are observed from air along the vertical line. The distance between the images of O₁ and O₂ is

 2 h

and the distance between the

images of O₁ and O₃ is   ₃ 2 h h .

Note : that the distance between any two points in a denser medium is reduced by a factor µ.

eg. The image of an object kept at a distance 30cm in front of a concave mirror is found to coincide with itself. If a glass slab ( = 1.5) of thickness 3cm is

introduced between the mirror and the object, then

(i) identify, in which direction the mirror should be displaced so that the final image may again coincide with the object itself.

(ii) Find the magnitude of displacement.

Sol. (i) Since the apparent shift occurs in the direction of incident light, therefore, the mirror should be displaced away from the objects

(ii) The magnitude of displacement is equal to the apparent shift, i.e., s = t         1 1 _{= 3}        2 / 3 1 1 _{= 1 cm}

Relation between the velocities of object and image : (i) The figure shows an object O moving toward the plane boundary

of a denser medium. To an observer in the denser medium the

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object appears to be more distant but moving faster. If the speed of the object is v, then the speed of the image will be µv.

(ii) The figure shows an object O moving toward the plane boundary of a rarer medium. To an observer in the rarer medium the object appears to be closer but moving slowly. If the speed of the object is v, then the speed of the image will be

 v

.

ILLUSTRATIONS

Example:A bird in air is diving vertically over a tank with speed 6 cm/s. Base of the tank is silvered. A fish in the tank is rising upward along the same line with speed 4 cm/s. [Take: _water = 4/3]

(A) Speed of the image of fish as seen by the bird directly (B) Speed of image of bird relative to the fish looking upwards [Sol.

(A) Velocity of fish in air = 4 ×

4 3

= 3  Velocity of fish w.r.t bird = 3 + 6 = 9  (B) Velocity of bird in water = 6 ×

3 4

= 8  w.r.t fish = 8 + 4 = 12 

Total Internal Reflection :

The phenomenon of total internal reflection occurs when light travels from a medium of high refractive index to a medium of

lower refractive index.

At the critical angle (_c), the refracted ray just grazes the boundary between the two media. Using Snell's law, we get

µ₁ sin _c = µ₂ sin 90° or _c = sin–1         1 2

For an angle of incidence greater than _c the light is totally reflected back into the medium of higher refractive index. This phenomenon is called total internal reflection.

Medium c

Glass 42°

Glass 48.5°

7. The figure shows a ray of dichromatic light (consisting of two wavelength ₁ and ₂) incident from denser to rarer medium. The refractive indices of the medium corresponding to the two wavelengths are µ₁ and µ₂.

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If the critical angle of the medium is _c, then respective critical refractive index will be µ_c = c sin 1 

Suppose ₁ < _c < ₂, then comment on the transmission and refraction of the wavelengths ₁ and ₂. The wavelength ₂ will not be transmitted because ₂ > _c.

The wavelength ₁ will be transmitted because ₁ < _c. Conditions of T.I.R.

(a) Light is incident on the interface from denser medium.

(b) Angle of incidence should be greater than the critical angle (i > c). Figure shows a luminous object placed in denser medium at a distance h from an interface separating two media of refractive indices _r and _d. Subscript r & d stand for rarer and denser medium respectively.

In the figure ray 1 strikes the surface at an angle less than critical angle C and gets refracted in rarer medium. Ray 2 strikes the surface at critical angle and grazes the interface. Ray 3 strikes the surface making an angle more than critical angle and gets internally reflected. The locus of points where ray strikes at critical angle is a circle, called circle of illuminance. All light rays striking inside the circle of illuminance get refracted in rarer medium. If an observer is in rarer medium, he/she will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no light will get refracted in rarer medium and then the object can not be seen from the rarer medium. Radius of C.O.I. can be easily found.

Ex. What should be the value of refractive index n of a glass rod placed in air, so that the light entering through the flat surface of the rod does not cross the curved surface of the rod.

Sol. it is required that all possible r should be more than critical angle. This will be automatically fulfilled if minimum r is more than critical angle ... (A)

Angle r is minimum when r is maximum i.e. C (why ?). Therefore the minimum value of r is 90–C.

From condition (A) :

90° – C > C or C < 45° sin C < sin 45° ; n 1 < 2 1 or n > 2 Ex. A rectangular block of refractive index  is placed on a printed page lying on a horizontal surface as shown in figure. Find the minimum value of  so that the letter L on the page is not visible form any of the vertical

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sides.

Sol. The letter L will not be visible from the vertical sides if the light ray does not emerge through it. We can apply the condition of no emergence for a prism of angle A = 90°.

Thus it,  > 2 A sin 1 or  > 2 90 sin 1  or m > 2

Ex. What is the apex angle of the conical region through which a person within water views the outside object?

Sol. For water-air interface, the critical angle is given as i_c = sin–1 _{(3 / 4)}

= 48.6°.

 The required apex angle is 2 × 48.6°.

Optional eg. Fraction of light emerging from an isotropic point source through a conical region having semi

vertex angle  and with its apex at the source.

If 2 steradian be the solid angle for the cone then, the fraction of light passing through the cone will be f =   4 2 =   2 ... (i)

Let us consider a sphere of radius R with its centre at the source S.

Let AB be the section (circular) where the cone ASB intercepts the sphere. If S be the area of the spherical portion ACB (lying within the conical region) then, nthe solid angle

2 = ₂ R

S 

... (ii)

Let SC be the symmetry axis of the portion of the sphere ACB. If x be the distance of a thin circular strip then its area

dS = 2x Rd = 2 (R sin ) Rd  Total area S =

_



 From equation (ii),

2 = 2 (1 – cos )

 From equation. (i), the required fraction is

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f = (1 – cos ) / 2. Deviation

(i) The figure shows a light ray travelling from a denser to a rater medium at an angle  greater than the critical angle _c.

The deviation  of the light ray is given by  =  – 

Since  and  = sin , therefore  = sin–1 _{( sin ) – }

This is a non-linear increasing equation. The maximum value of  occurs when

 = _c , and is equal to _max =

2 

– _c

(ii) If the light is incident at an angle  > _c, then the angle of deviation is given by

 =  – 2

This is a linearly decreasing function. The maximum value of  occurs when  = _c and is equal to

_max = p – 2_c

The variation of  with the angle of incidence  is plotted in figure.

Continuous variation of refractive :

 sin  = constant dx dy = slope of tangent = cot i, if  = f(y) = tan t, if  = f(x)

In situation where there is continuous variation of refractive index, critical angle is approximately 90° and bending of ray takes place if angle of incidence approaches 90° while travelling successively from denser to rarer layers.

Bending of light ray

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Mirage

It is an optical illusion created due to the phenomenon of total internal reflection. This is seen in hot region. In hot areas like deserts surface of earth is very hot. So, air in the lower regions of atmosphere is hot as compared to that in higher regions. This result in variation of density with height and it increases as we go up. In this situation atmosphere can be assumed to be made of large number of thin layers of air. A beam of light starting from an object say a tree and travelling downward finds itself going from denser to rater medium. Therefore, its angle of incidence at consecutive layers goes on increasing gradually till it surpasses the critical value and is reflected back due to total-internal reflection. A virtual image of the object is seen by eye at E. Due to the disturbance of air, the mirage is wavy in nature, thus giving an illusion for the presence of water which is actually not there. This effect is also called inferior mirage.

Looming

This effect occurs when the density of air decreases much more rapidly with increasing height than it does under normal conditions. This situation sometimes happens in cold region particularly in the vicinity of the cold surface of sea or of a lake. Light rays starting from an object S (say a ship) are curved downward and on entering the eye the rays appear to come from S, thus giving an impression that the ship is floating in air. This effect is also called superior mirage.

PRISM

Prism is a transparent medium whose refracting surfaces are not parallel but are inclined to each other.

1. Basic Terms

(i) Angle of prism or reflecting angle (A) The angle between the faces on which light is incident and from which it emerges.

(ii) Angle of deviation ()

It is the angle between the emergent and the incident ray. It other words, it is the angle through which incident ray turns in passing through a prism.

 = (i – r₁) + (e – r₂) or  = i + e – (r₁ + r₂) or  = i + e – A 2. Condition of no emergence

A ray of light incident on a prism of angle A and refractive index  will

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not emerge out of a prism (whatever may be the angle of incidence) if A > 2_c, where _c is the critical angle.

i.e.  > _{sin (A/2)}

1

3. Condition of grazing emergence

By the condition of grazing emergence we mean the angle of incidence i at which the angle of emergence becomes e = 90°.

i = sin–1





Note : That the light will emerge out of a given prism only if the angle of incidence is greater than the condition of grazing emergence.

4. Condition of maximum deviation

Maximum deviation occurs when the angle of incidence is 90°. _max = 90° + e – A

where e = sin–1 _{[ sin (A – } c)]

5. Condition of minimum deviation

The minimum deviation occurs when the angle of incidence is equal to the angle of emergence, i.e i = e

_min = 2i – A Using Snell's law

 =              2 A sin 2 A sin min

Note : That in the condition of minimum deviation the light ray passes through the prism symmetrically, i.e. the light ray in the prism becomes parallel to its base.

Characteristic of a prism

(a) Variation of  versus i (shown in diagram).

For one  (except  min) there are two values of angle of incidence.

If i and e are interchanged then we get the same value of  because of reversibility principle of light.

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(b) There is one and only one angle of incidence for which the angle of deviation is minimum.

(c) When  = _min, the angle of minimum deviation, then i = e and r₁ = r₂, the ray passes symmetrically w.r.t. the refracting surfaces. We can show by simple calculation that _min = 2i_min – A where i_min = angle of incidence for minimum deviation, and r = A/2

 n_rel =              2 A sin 2 A sin m , where n_rel = gs surroundin prism n n

Also _min = (n – 1) A (for small values of  A)

eg. A prism with angle A = 60° produces a minimum deviation of 30°. Find the refractive index of the material.

Sol. We know that

 =              2 A sin 2 A sin min Here A = 60°, _min = 30°   =              2 60 sin 2 30 90 sin =   30 sin 45 sin = 2

Ex. Figure shows a triangular prism of refracting angle 90°. A ray of light incident at face AB at an angle  refracts at point Q with an angle of refraction 90°. (a) What is the refractive index of the prism in terms of  ? (b) What is the maximum value that the refractive index can have ? What happens to the light at Q if the incident angle at Q is (c) increased slightly and (d) decreased slightly ?

90° 4 Q 90° 1 3 2 A C B P 6. Thin prisms

In thin prisms the distance between the refracting surfaces is negligible and the angle of prism (A) is very small.

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Since A = r₁ = r₂, therefore, if A is small then both r₁ and r₂ are also small, and the same is true for i₁ and i₂.

According to Snell's law sin i₁ =  sin r₁ or i₁ = r₁ sin i₂ =  sin r₂ oqr i₂ = r₂ Therefore, deviation  = (i₁ – r₁) + (i₂ – r₂)

 = (r₁ + r₂) + (i₂ – r₂)  = A ( – 1)

Note : That the deviation for a small angled prism is independent of the angle of incidence. ILLUSTRATIONS

Ex. A thin prism of angle A = 6° produces a deviation  = 3°. Find the refractive index of the material of prism.

Sol. We know that  = A ( – 1) or  = 1 + A  Here A = 6°,  = 3°, therefore  = 1 + 6 3 = 1.5

Ex. Find the co-ordinates of image of the point object ‘O’ formed after reflection from concave mirror as shown in figure assumming prism to be thin and small in size of prism angle 2°. Refractive index of prism materal is 3/2. X O (0, 0) 5cm 20cm f = 30cm Y

Sol. Consider image formation through prism. All incident rays will be deviated by  = (µ – 1)A =       1 2 3 2° = 1° = 180  rad

Now as prism is thin so object and imae will be in same plane as shown in figure.

  O I d  5cm It is clear 5 d

= tan  _  ( is very small)

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or d =

36 

cm

Now this image will act as an object for concave mirror. u = –25 cm, f = –30 cm  v = f u uf  = 150 cm Also, m = u v  = +6

 Distance of image from principal axis =

36  × 6 = 6  cm

Hence, co-ordinates of image formed after reflection from concave mirror are 

      cm 6 , cm 175

R.G.

Sir