5- Refraction at curved surface Theory Module-5-1

1.1 PRISM : It is a transparent medium whose refracting surfaces are not parallel but are inclined to each other. 1.2 TERMS RELATED TO PRISM : (A) Angle of Prism or Refracting Angle (A) : It is the angle between the faces on which light is incident and from which it emerges. (B) Angle of Deviation (  ) : It is the angle between the emergent and the incident ray. In other words, it is the angle through which incident ray turns while passing through a prism.  = (i – r1) + (e – r2) = i + e – (r1 + r2) = i + e – A 1.3 CONDITION OF NO EMERGENCE : 1.4 CRITICAL ANGLE : It is the angle of prism above which incident light ray on first surface will not emerge out from the second surface for all possible values of angle of incidence A = 2 C 1.5 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 = 90° is called condition of grazing emergence. i.e., r2 = c  , r1 = A – r2 = A – c (As r1 + r2 = A) Now from Snell’s law at AB , sin i =  sin r1  sin i =  sin (A – c) =  [sin A cos  c – cos A sin c] A =  [sin A – cos A sin c] R i ( r r =  [sin A 1 –  cos A] 1  2 Q  =  2  1 sin A  cos A  B C The light will not emerge out of a prism for all values of angle of incidence if at face AB for i = max = 90º, at face AC  i = sin–1  Note : sin A  cos A  r2 > c From Snell’s law at AB, 1 × sin 90º =  sin r1  r1 = sin –1 (1/) , r1 = c  r1 + r2 > 2c but r1 + r2 = A The light will emerge out of a give n prism only if the angle of incidence is greater than the condition of grazing emergence. 1.6 CONDITION OF MAXIMUM DEVIATION : A > 2c Deviation will be maximum when i = 90° or, sin A/2 > sin c So,  max = i max + e – A max    1 sin A 2 or  > cosec (A/2) = 90° + e – A However, when i = 90°, 1 i.e., A ray of light will not emerge out of a prism (what ever be the angle of incidence) if A > 2c i.e., if  > cosec (A/2) For “AB ” 1× sin 90° =  sin r1  sin r1 =   r1 =  c Again, r1 + r2 = A  r2= A – c Now for ‘AC’ sin r2 = 1 sin e  sin e =  sin (A – c)  e = sin–1 [ sin (A – c)] Thus, max = 90° + sin –1[ sin (A □ c)]– A PROOF We know,  + A = i + e and r1 + r 2 = A Again sin i =  sin r1 and sin e =  sin r2 d For min , di = 0 d di de de  di = di + di = 1 + di = 0 i.e de = – di also dr1 = – dr2 cosi di =  cos r1 dr1 and cose de =  cos r2 dr2 cos i di  = cosr1 dr1 Note : This situation is reverse of grazing emergence and may also be viewed as deviation at grazing incidence . 1.7 CONDITION FOR MINIMUM DEVIATION cos e de cos2 i  cos2 e = cosr2 cos2 r1 cos2 r2 1 sin2 r1 dr2 1 1 sin2 i  2 = = 1 sin2 r 1  sin2 i 2  sin2 i 1 1 sin2 e  2 or, 1 sin2 e = 2  sin2 e The minimum deviation occurs when the angle of incidence is equal to the angle of emergence i.e., i = e  min = 2i – A [As  = i + e – A] Using Snell ‘s law 1 x sin i =  sin r1 and  sin r2 = 1 x sin i  sin r1 = sin r2  r1 = r2 = r (say)  min = (2i – A) where r = A/2  min  A   sin2 i = sin2 e  i = e 1.8 GRAPHICAL REPRESENTATION OF ANGLE OF DEVIATION : The deviation produced by a prism depends on  = sin i sin  = 2  sin r sinA / 2 (A) Angle of incidence , i (B) Angle of prism , A Note: 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. (C) Refractive index of material ,  and (D) Wavelength of light ,  (i)  is f irst decreasing and then increasing for i < e and i > e respectively. (ii)   A (iii)   (  – 1 ) 1 According to given problem A = 30°, i1 = 60° and  = 30° and as in a prism  = (i1 + i2) – A, 30° = (60 + i2) – 30 i.e., (iv)    1.9 THIN PRISMS : In thin prism the distance between the refracting surfaces is negligible and the angle of prism (A) is very small. Since A = r1 + r2 , therefore, r1 and r2 both are small and the same is true for i1 and i2 According to Snell’s law : sin i =  sin r1  i = r1 sin e =  sin r2  e =  r2  deviation,  = (i1 – r1) + (e – r2) = (r1 + r2) ( – 1) = A ( – 1) Note : The deviation for a small angled prism is independent of the angle of incidence. Refraction by Prism Ex.1 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) i2 = 0 So the emergent ray is perpendicular to the face from which it emerges. Now asi2 = 0, r2 = 0 But as r1 + r2 = A , r1 = A = 30° So at first face 1× sin 60° =  sin 30° i.e.,  = Ex.3 The refractive index of the material of a prism is 2 and its prism angle is 30º. One of its refracting faces is polished. The incident beam of light will return back for the angle of incidence (A) 60º (B) 45º (C) 30º (D) 0º Sol. The ray of light returns back from the polished face AC  ADE = 90º. From the figure it is clear that the angle of refraction at face AB is 30º. Hence from Snell’s law sini A  = sin r or  = 1 +  A   = or r = 30º sin i D Here A = 6 °,  = 3° , therefore  = sin 30º  = 1 + 3 6 = 1.5 or sin i = 2 B C 2 Ex.2 A ray of light is incident at an angle of 60° on one face of a prism which has an angle 1 = 2 = sin 45º Sol. of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism .  i = 45º Hence the correct answer will be (B) Ex.4 A ray of light incident on a prism surface at an angle of 40º in the minimum deviation position. If the angle prism is 60º then the values of m and  will be respectively – (A) 40º and 1.532 (B) 60º and 1.532 (C) 90º and 1.532 (D) 0º and 1.532 Sol. A = r1 + r2 = 60 (1) In minimum deviation position r1 = r2 (2) From eqs. (1) and (2) A = 2r1 = 60º  r1 = 30º  n = sini sinr sin50º = sin30º = 1.532 Note : 1. As  decreases with increase in , for visible light  is maximum for violet and sin A  m n = 2 sin A 2 or 1.532 = sin 60  m 2 sin 300 minimum for red. 2. Because of the different refractive indices, light of different colours bend through different angles on refraction . sin 60  m 2  m = 40º 1.532 = 2 = 0.766 2.3 ANGULAR DISPERSION Angular dispersion produced by a prism for white light is the difference in the angles of deviation of two extreme colours i.e., violet and red colours Hence the correct answer will be (A). Ex.5 A glass prism of refractive index 1.5 and angle of prism 6º is put in contact with another prism of refractive index 1.6 when a ray of light is made incident on this combination normally then it emerges out undeviated. The angle of second prism will be – (A) 6º (B) 5º (C) 4º (D) 3º Sol. 1 = 2 (1 – 1) A1 = (2 – 1) A2 (1.5 – 1) 6 = (1.6 – 1) A2 A2 = 5º Hence the correct answer will be (B) 2.1 PHENOMENON : It is the phenomenon of splitting of a beam of white light into its constituent colours on passing through a prism. (v – r) = (v – r) A DERIVATION If v and r are deviation of violet and red colours respectively then, angular dispersion = v – r Now, deviation,  through a thin prism of refracting angle A is  = ( -1) A  v = (v – 1) A and r = (r – 1) A Thus, v – r = (v -1) A – (r – 1) A = (v – r) A 2.4 DISPERSIVE POWER : The dispersive power of a prism is defined as the ratio of angular dispersion to the mean deviation produced by the prism. It is represented by  .  i.e.,  =   1 2.2 CAUSE : The refractive index of a material depends on the wavelength of light, given by DERIVATION The mean deviation produced by the prism is  = ( – 1) A Angular dispersion , v – r = (v – r) A As dispersive power = angular dispersion mean deviation CAUCHY’S FORMULA   = v  r bv  r gA =    =  = A + B + 2 C + ......... where A, B, C are 4  b 1gA      1 constants Here,  = v r  2 Note: Dispersive power of a prism depends only on nature of material of the prism. However, angular dispersion and mean deviation, both depend also on the angle of prism. 2.5 COMBINATION OF PRISMS : Two prisms of refracting angles A1 and A2 and dispersive power 1 and 2 are placed symmetrically (as shown in the figure) for a particular ray refractive indices of the two prisms are 1 and 2 respectively. Note: By choosing 1 and 2 different and the refracting angles we can get dispersion without average deviation 2.5 (B) AVERAGE DEVIATION WITHOUT DISPERSION : If , v – r = 0 (1y – 1) 1 A1 = (2y – 1) 2 A2 or, 1v –  1r) A1 = (2v –  2r) A2 So , net average deviation  = (1y – 1) A1 – (2y – 1) A2 = ( – 1) A M1 2y  1 A2 1y 1 MN 1y  1 A1 L   = ( – 1) A M 1 =  1  1 1y 1 2 1  1  2  Thus deviation produced by two prisms are, 1 = (1 – 1) A1 and 2 = (2 – 1) A2 As the two deviations are opposite to each other, Net deviation,  = 1 – 2 = (1 – 1) A1 – (2 –1) A2 If white light passes through the combination then, v = (1v – 1) A1– (2v –11) A2 and r = (1r – 1) A1 – (2r – 1) A2 Thus , the angular dispersion produced by the combination is v – r = ( 1v – 1r) A1 – (2v – 2 r) A2 The dispersive powers are Dispersion Ex.6 A crown glass prism of angle 5° is to be combined with a flint glass prism in such a way that the mean ray passes undeviated. Find (a) the angle of the flint glass prism needed and (b) the angular dispersion produced by the combination when white light goes through it. Refractive indices for red, yellow and violet light are 1.514, 1.517 and 1.523 respectively for crown glass and 1.613, 1.620 and 1.632 for flint glass. Sol. The deviation produced by the crown prism is  = ( – 1) A and by the flint prism is ’ = (’ – 1) A’ 1 = 1v  1 r 1y  1 and 2 = 2 v  2 r 2 y  1 The prisms are placed with their angles inverted with respect to each other . The deviations are also in opposite directions. So, net angular dispersion v – r = (1y – 1) 1 A1 – (2v – 1) 2 A2 Thus the average deviation (or net deviation for yellow ray) y = (1y – 1) A1 – (2y – 1) A2 2.5 (A) DISPERSION WITHOUT AVERAGE DEVIATION : Thus, the net deviation is D =  –  ’ = ( – 1) A – (’ – 1) A’ ...........(1) (a) If the net deviation for the mean ray is zero, ( – 1) A = (’ – 1) A’ b  1 If y = 0 or, A’ = b'1g A (1y – 1) A1 = (2y – 1) A2  v – r = (1y-1) A1 ( 1–2) = 1(1 – 2) = 1.517  1 x 5 = 4.2°. 1.620  1 (b) The angular dispersion produced by the crown prism is v – r = (v – r) A and that by the flint prism is ’v – ’r = (’v – ’r) A’ The net angular dispersion is ,  = (v – r) A – (’v – ’r) A’ = (1.523 –1.514) x 5° – (1.632–1.613) x 4.2° = – 0.0348°. The angular dispersion has magnitude Ex.8 Find the angle of a prism of dispersive power 0.021 and refractive index 1.53 to form an achromatic combination with prism of angle 4.2° and dispersive power 0.045 having refractive index 1.65. Find the resultant deviation. Sol Here,  = 0.021;  = 1.53; ’ = 0.045; ’ = 1.65; A’ = 4.2° For no dispersion,   + ’ ’ = 0 or  A ( – 1) + ’ A ’ ( – 1) = 0 0.0348°. Ex.7 Calculate the dispersive power for crown or A = – ' A'b'1  b  1g=– 0.045 x 4.2 x b1.65  1 0.021 x b1.53  1g glass from the given data v= 1.5230, r =1.5145 Sol. v = 1.5230, r = 1.5145,  = ? Mean refractive index, = – 11.04° Net deviation,  + ’ = A (–1) + A’ (’ – 1) = – 11.04 (1.53 –1) + 4.2 (1.65 –1)  = v  r 2  = 1.5187  = v  r   1 = 1.5230  1.5145 2 = – 11.04 x 0.53 + 4.2 x 0.65 = – 5.85 + 2.73 = 3.12° 1.5230  1.5145 = 1.5187  1 0.0085 = 0.5187 = 0.0163 SOLVED EXAMPLES Ex.1 Light is incident on the side of a 30°– 60° – 90° prism , as shown in the figure. A thin layer of a liquid is spread over the hypotenuse of the prism. If the index of refraction of the prism is 1.5, find the maximum refractive index of the liquid in order that a ray passing normally through the 60°– base may be totally reflected. Sol. The angle of incidence of the hypotenuse is 60°. Sol. At first face of the prism as i1 = 0, sin 0 = 1.5 sin r1 i.e., , r1 = 0 And as for a prism r1 + r2 = A so r2 = A (1) But at second face, as the ray just fails to emerge i.e., r2  C (2) So from Eqn,.(1) and (2) A = r2 =  L L But as C = sin –1 MN1 2 = sin –1 N3 = 42° The ray is totally reflected if 60°  C where C is the critical angle for the glass- liquid interface. or sin 60°  sin C It is known that A 1.5 sin C =  sin 90° Where  is the refractive index of the liquid  So A = 42° Ex.3 The refractive indices of material of a prism for blue and red colours are 1.532 and 1.514 respectively. Calculate angular dispersion produced by the prism if angle of prism is 8°. Sol. Here, b= 1.532 and r = 1.514 A = 8° . Angular dispersion = (b – r) A = (1.532 – 1.514) x 8 = 0.018 x 8 = 0.144° . Ex.4 Determine the angle of flint glass prism, which should be combined with a crown glass prism of 5° so as to give dispersion but no deviation. Given, for crown glass : v = 1.523, r = 1.515 ; for flint glass : v´ = 1.688, r´ = 1.650 . Sol. Crown glass v = 1.523 ; r = 1.515 ; A = 5° Flint glass v´ = 1.688 ; r´ = 1 1.650 ; A´ = ? So that the combination of the two prisms or sin C = 1.5 does not produce any deviation, Ab  1  the ray is totally reflected if 3 2   1.5 A´ = – '1    1.523  1.515 or   1.299 Now,  = v r 2 2 = 1.519 Ex.2 A ray of light fall normally on a refracting face of a prism of refractive index 1.5. Find and ´ = v 'r ' 2 1.688  1.650 = 2 = 1.669 the angle of the prism if the ray just fails to emerge from the prism.  A´ = – 5 b1.519  1 1.669  1 5 x 0.519 = – 0.669 = – 3.88° Ex.5 The dispersive powers of crown and flint glasses are 0.03 and 0.05 respectively. The refractive indices for yellow light for these glasses are 1.517 and 1.621 respectively . It is desired to form an achromatic combination of prisms of crown and flint glasses which can produce a deviation of 1° in the yellow Similarly , the angular dispersion produced by the flint prism is (´ – 1) ´ A´ For achromatic combination , the net dispersion should be zero. Thus , ( – 1)  A = (´ – 1) ´ A´ ray. Find the refracting angles of the two prisms needed. Sol. Suppose, the angle of the crown prism needed is A and that of the flint prism is A’. or , A' A b  1g = ('1) ' = 0.517 x 0.03 0.621 x 0.05 = 0.50 ......(1) We have  = v  r   1 or, v – r = ( – 1)  The angular dispersion produced by the crown prism is (v – r) A = ( –1)  A The deviation in the yellow ray produced by the crown prism is  = ( -1) A and by the flint prism is ´ = (´ – 1) A´. The net deviation produced by the combination is  – ´ = ( – 1) A – (´ – 1) A´ or 1° = 0.517 A – 0.621 A´ (2) Solving (1) and (2 ), A = 4.8° and A´ = 2.4°. Thus, the crown prism should have its refracting angle 4.8° and that of the flint prism should be 2.4°.