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Page 1 27 Refraction of Light Introductory Exercise 27.1 1. Let real depth of dust par ti cle is x and thick ness of slab is t From Ist surface m = Real depth App.depth 1.5 cm = x 6 Þ x = 9 cm …(i) From other face m =  t x 4 Þ t x  = ´ 4 1.5 Þ t x = + = + = 6 9 6 16 cm 2. 1 2 4 3 m = Þ m m 2 1 4 3 = …(i) 2 3 3 2 m = Þ m m 3 2 3 2 = …(ii) From Eqs. (i) and (ii), we get m m 3 1 4 3 3 2 2 = ´ = 3. Frequency re main same. Let v 1 is velocity in medium (1) and v 2 in Medium (2) We have m 1 1 = c v and m 2 2 = c v Þ m m 1 2 2 1 = v v Þ v v 2 1 2 1 = m m Similarly, wavelength l m m l 2 1 2 1 = 4. From v n a a = l Þ l a a a v n = = ´ ´ = ´ +  3 10 6 10 5 10 8 14 7 m = 50 nm m l l = = = = a m 500 300 5 3 1.67 Introductory Exercise 27.2 1. Since light rays are com ing from glass to air ap pl y ing m m m m 2 1 2 1 v u R  =  Þ 1 10 1 v  =   1.5 1.5 1.5 Þ 1 1 30 10 1 4 30 v =  + =  + 1.5 .5 Þ v = = 30 3.5 8.57 cm 2. m m m m 2 1 2 1 v u R  =  (a) 1.5 0.5 m   = 1 20 6 ( ) On solving v = 45 cm (b) 1.5 0.5 v   = 1 10 6 ( ) On solving we get v =  90 cm t x 2nd face 1st face t – x Page 2 27 Refraction of Light Introductory Exercise 27.1 1. Let real depth of dust par ti cle is x and thick ness of slab is t From Ist surface m = Real depth App.depth 1.5 cm = x 6 Þ x = 9 cm …(i) From other face m =  t x 4 Þ t x  = ´ 4 1.5 Þ t x = + = + = 6 9 6 16 cm 2. 1 2 4 3 m = Þ m m 2 1 4 3 = …(i) 2 3 3 2 m = Þ m m 3 2 3 2 = …(ii) From Eqs. (i) and (ii), we get m m 3 1 4 3 3 2 2 = ´ = 3. Frequency re main same. Let v 1 is velocity in medium (1) and v 2 in Medium (2) We have m 1 1 = c v and m 2 2 = c v Þ m m 1 2 2 1 = v v Þ v v 2 1 2 1 = m m Similarly, wavelength l m m l 2 1 2 1 = 4. From v n a a = l Þ l a a a v n = = ´ ´ = ´ +  3 10 6 10 5 10 8 14 7 m = 50 nm m l l = = = = a m 500 300 5 3 1.67 Introductory Exercise 27.2 1. Since light rays are com ing from glass to air ap pl y ing m m m m 2 1 2 1 v u R  =  Þ 1 10 1 v  =   1.5 1.5 1.5 Þ 1 1 30 10 1 4 30 v =  + =  + 1.5 .5 Þ v = = 30 3.5 8.57 cm 2. m m m m 2 1 2 1 v u R  =  (a) 1.5 0.5 m   = 1 20 6 ( ) On solving v = 45 cm (b) 1.5 0.5 v   = 1 10 6 ( ) On solving we get v =  90 cm t x 2nd face 1st face t – x (c) 1.5 0.5 v   = 1 3 6 ( ) On solving v =  6.0 cm 3. Light rays are coming from glass to air m m m m 2 1 2 1 v u R  =  1 4 3 10 1 4 3 15 v   =   ( ) / ( ) 1 4 30 1 45 v + = on solving v =  9 cm 4. Applying u v u R 2 1 2 1  =  m m m Þ 1.44 0.44 1.25 v  ¥ = 1 On solving v = 0.795 cm 5. m m m m 2 1 2 1 v u R  =  1.635 0.635 2.50) v   =  1 9 ( ) ( on solving v = 6.993 cm Lateral magnification m v u =  Þ =  =  6.993 9 0.777 Introductory Exercise 27.3 1. We have 1 1 1 1 1 1 1 2 v u f R R  = =   é ë ê ù û ú ( ) m Þ  + =    é ë ê ù û ú 1 20 1 60 1 1 1 ( ) 1.65 R R Þ  + = ´  3 1 60 2 0.65 R Þ R = ´ = 60 39 0.65 cm 2. Using 1 1 1 v u f  = , we get 1 50 1 1 30   = x Þ  = + 1 1 30 1 50 x On solving x =  18.75 cm m v u =  = 50 18.75 Height of filament image = ´ 2 50 18.75 = 5.3 cm 3. 1 1 1 1 f R R =  + é ë ê ù û ú ( ) m If lens faces becomes opposite three is no change in radius of curvature hence focal length does not change. 4. Using for mula 1 1 1 v u f  = when u ® 0 1 v ® 0 when u f v ® ® ¥ , hence im age moves from sur face to ¥ . 5. 1 1 1 1 1 2 f R R =   é ë ê ù û ú ( ) m Þ 1 1 1 1 2 20 f R R =    é ë ê ù û ú = ´  ( ) 1.3 0.3 Þ f =  100 3 cm (a) When immersed in a liquid of 1.8 refractive index 1 1 2 2 20 1 f R =  æ è ç ö ø ÷  é ë ê ù û ú =  ´  1.3 1.8 0.5 1.8 f ¢ = 36 cm (b) The minimum distance is equal to the focal length = 36 cm 20 O 10 cm Page 3 27 Refraction of Light Introductory Exercise 27.1 1. Let real depth of dust par ti cle is x and thick ness of slab is t From Ist surface m = Real depth App.depth 1.5 cm = x 6 Þ x = 9 cm …(i) From other face m =  t x 4 Þ t x  = ´ 4 1.5 Þ t x = + = + = 6 9 6 16 cm 2. 1 2 4 3 m = Þ m m 2 1 4 3 = …(i) 2 3 3 2 m = Þ m m 3 2 3 2 = …(ii) From Eqs. (i) and (ii), we get m m 3 1 4 3 3 2 2 = ´ = 3. Frequency re main same. Let v 1 is velocity in medium (1) and v 2 in Medium (2) We have m 1 1 = c v and m 2 2 = c v Þ m m 1 2 2 1 = v v Þ v v 2 1 2 1 = m m Similarly, wavelength l m m l 2 1 2 1 = 4. From v n a a = l Þ l a a a v n = = ´ ´ = ´ +  3 10 6 10 5 10 8 14 7 m = 50 nm m l l = = = = a m 500 300 5 3 1.67 Introductory Exercise 27.2 1. Since light rays are com ing from glass to air ap pl y ing m m m m 2 1 2 1 v u R  =  Þ 1 10 1 v  =   1.5 1.5 1.5 Þ 1 1 30 10 1 4 30 v =  + =  + 1.5 .5 Þ v = = 30 3.5 8.57 cm 2. m m m m 2 1 2 1 v u R  =  (a) 1.5 0.5 m   = 1 20 6 ( ) On solving v = 45 cm (b) 1.5 0.5 v   = 1 10 6 ( ) On solving we get v =  90 cm t x 2nd face 1st face t – x (c) 1.5 0.5 v   = 1 3 6 ( ) On solving v =  6.0 cm 3. Light rays are coming from glass to air m m m m 2 1 2 1 v u R  =  1 4 3 10 1 4 3 15 v   =   ( ) / ( ) 1 4 30 1 45 v + = on solving v =  9 cm 4. Applying u v u R 2 1 2 1  =  m m m Þ 1.44 0.44 1.25 v  ¥ = 1 On solving v = 0.795 cm 5. m m m m 2 1 2 1 v u R  =  1.635 0.635 2.50) v   =  1 9 ( ) ( on solving v = 6.993 cm Lateral magnification m v u =  Þ =  =  6.993 9 0.777 Introductory Exercise 27.3 1. We have 1 1 1 1 1 1 1 2 v u f R R  = =   é ë ê ù û ú ( ) m Þ  + =    é ë ê ù û ú 1 20 1 60 1 1 1 ( ) 1.65 R R Þ  + = ´  3 1 60 2 0.65 R Þ R = ´ = 60 39 0.65 cm 2. Using 1 1 1 v u f  = , we get 1 50 1 1 30   = x Þ  = + 1 1 30 1 50 x On solving x =  18.75 cm m v u =  = 50 18.75 Height of filament image = ´ 2 50 18.75 = 5.3 cm 3. 1 1 1 1 f R R =  + é ë ê ù û ú ( ) m If lens faces becomes opposite three is no change in radius of curvature hence focal length does not change. 4. Using for mula 1 1 1 v u f  = when u ® 0 1 v ® 0 when u f v ® ® ¥ , hence im age moves from sur face to ¥ . 5. 1 1 1 1 1 2 f R R =   é ë ê ù û ú ( ) m Þ 1 1 1 1 2 20 f R R =    é ë ê ù û ú = ´  ( ) 1.3 0.3 Þ f =  100 3 cm (a) When immersed in a liquid of 1.8 refractive index 1 1 2 2 20 1 f R =  æ è ç ö ø ÷  é ë ê ù û ú =  ´  1.3 1.8 0.5 1.8 f ¢ = 36 cm (b) The minimum distance is equal to the focal length = 36 cm 20 O 10 cm 6. Using 1 1 1 v u f  = 1 1 20 12 10 v   = ( ) On solving v = 20 cm Magnification =  =  v u 1 Hence the image of same size and inverted. Let the distance between second lens is x Since magnification is unity image distance also x using again 1 1 1 v u f  = we get 1 1 1 1 10 x x f   = = ( ) Þ x = 20 cm Hence the distance between two lenses = + = 20 20 40 cm cm cm 7. 1 1 1 v f u = + …(i) 1 1 1 1 v f u du = + + …(ii) 1 1 v v u du u u duu ¢  = +  + ( ) ( ) on solving, we get v v v du u u du  ¢ ¢ = + ( ) thickness dv v u du =  2 2 10. Size of im age = ´ = 6 2 3 2 cm. 11. Let im age distance is u   m = 3 Þ v u = 3 1 3 1 1 12 u u + = Þ v = 16 cm 12. Since im age is up right and diminished hence lens is concave. Now u v  = 20 …(i) m v u = = 1 2 Þ 1 2 20 =  u u Þ u = 40 cm and v = 20 cm Þ Using 1 1 1 v u f  =  + = 1 20 1 40 1 f Þ f =  40 cm 13. The im age co in cide it self if light falls nor mally on plane mir ror hence ob ject must be on focus i e . . + 10 cm. 8. 1 1 2 2 1 2 1 2 2 1 1 v u R R + =   ( ) ( ) m /m m /m 1 1 2 4 3 2 4 3 1 v + =   0.2 0.4 0.4 ( / ) ( / ) On solving v = 12 cm 9. Since shift in po si tion D t = 0.1 m Hence real depth = + ( ) 0.1 0.2 m = 0.3 m and apparent depth= 0.2 m m = real depth apparent depth = = 0.3 0.2 1.5 21 Page 4 27 Refraction of Light Introductory Exercise 27.1 1. Let real depth of dust par ti cle is x and thick ness of slab is t From Ist surface m = Real depth App.depth 1.5 cm = x 6 Þ x = 9 cm …(i) From other face m =  t x 4 Þ t x  = ´ 4 1.5 Þ t x = + = + = 6 9 6 16 cm 2. 1 2 4 3 m = Þ m m 2 1 4 3 = …(i) 2 3 3 2 m = Þ m m 3 2 3 2 = …(ii) From Eqs. (i) and (ii), we get m m 3 1 4 3 3 2 2 = ´ = 3. Frequency re main same. Let v 1 is velocity in medium (1) and v 2 in Medium (2) We have m 1 1 = c v and m 2 2 = c v Þ m m 1 2 2 1 = v v Þ v v 2 1 2 1 = m m Similarly, wavelength l m m l 2 1 2 1 = 4. From v n a a = l Þ l a a a v n = = ´ ´ = ´ +  3 10 6 10 5 10 8 14 7 m = 50 nm m l l = = = = a m 500 300 5 3 1.67 Introductory Exercise 27.2 1. Since light rays are com ing from glass to air ap pl y ing m m m m 2 1 2 1 v u R  =  Þ 1 10 1 v  =   1.5 1.5 1.5 Þ 1 1 30 10 1 4 30 v =  + =  + 1.5 .5 Þ v = = 30 3.5 8.57 cm 2. m m m m 2 1 2 1 v u R  =  (a) 1.5 0.5 m   = 1 20 6 ( ) On solving v = 45 cm (b) 1.5 0.5 v   = 1 10 6 ( ) On solving we get v =  90 cm t x 2nd face 1st face t – x (c) 1.5 0.5 v   = 1 3 6 ( ) On solving v =  6.0 cm 3. Light rays are coming from glass to air m m m m 2 1 2 1 v u R  =  1 4 3 10 1 4 3 15 v   =   ( ) / ( ) 1 4 30 1 45 v + = on solving v =  9 cm 4. Applying u v u R 2 1 2 1  =  m m m Þ 1.44 0.44 1.25 v  ¥ = 1 On solving v = 0.795 cm 5. m m m m 2 1 2 1 v u R  =  1.635 0.635 2.50) v   =  1 9 ( ) ( on solving v = 6.993 cm Lateral magnification m v u =  Þ =  =  6.993 9 0.777 Introductory Exercise 27.3 1. We have 1 1 1 1 1 1 1 2 v u f R R  = =   é ë ê ù û ú ( ) m Þ  + =    é ë ê ù û ú 1 20 1 60 1 1 1 ( ) 1.65 R R Þ  + = ´  3 1 60 2 0.65 R Þ R = ´ = 60 39 0.65 cm 2. Using 1 1 1 v u f  = , we get 1 50 1 1 30   = x Þ  = + 1 1 30 1 50 x On solving x =  18.75 cm m v u =  = 50 18.75 Height of filament image = ´ 2 50 18.75 = 5.3 cm 3. 1 1 1 1 f R R =  + é ë ê ù û ú ( ) m If lens faces becomes opposite three is no change in radius of curvature hence focal length does not change. 4. Using for mula 1 1 1 v u f  = when u ® 0 1 v ® 0 when u f v ® ® ¥ , hence im age moves from sur face to ¥ . 5. 1 1 1 1 1 2 f R R =   é ë ê ù û ú ( ) m Þ 1 1 1 1 2 20 f R R =    é ë ê ù û ú = ´  ( ) 1.3 0.3 Þ f =  100 3 cm (a) When immersed in a liquid of 1.8 refractive index 1 1 2 2 20 1 f R =  æ è ç ö ø ÷  é ë ê ù û ú =  ´  1.3 1.8 0.5 1.8 f ¢ = 36 cm (b) The minimum distance is equal to the focal length = 36 cm 20 O 10 cm 6. Using 1 1 1 v u f  = 1 1 20 12 10 v   = ( ) On solving v = 20 cm Magnification =  =  v u 1 Hence the image of same size and inverted. Let the distance between second lens is x Since magnification is unity image distance also x using again 1 1 1 v u f  = we get 1 1 1 1 10 x x f   = = ( ) Þ x = 20 cm Hence the distance between two lenses = + = 20 20 40 cm cm cm 7. 1 1 1 v f u = + …(i) 1 1 1 1 v f u du = + + …(ii) 1 1 v v u du u u duu ¢  = +  + ( ) ( ) on solving, we get v v v du u u du  ¢ ¢ = + ( ) thickness dv v u du =  2 2 10. Size of im age = ´ = 6 2 3 2 cm. 11. Let im age distance is u   m = 3 Þ v u = 3 1 3 1 1 12 u u + = Þ v = 16 cm 12. Since im age is up right and diminished hence lens is concave. Now u v  = 20 …(i) m v u = = 1 2 Þ 1 2 20 =  u u Þ u = 40 cm and v = 20 cm Þ Using 1 1 1 v u f  =  + = 1 20 1 40 1 f Þ f =  40 cm 13. The im age co in cide it self if light falls nor mally on plane mir ror hence ob ject must be on focus i e . . + 10 cm. 8. 1 1 2 2 1 2 1 2 2 1 1 v u R R + =   ( ) ( ) m /m m /m 1 1 2 4 3 2 4 3 1 v + =   0.2 0.4 0.4 ( / ) ( / ) On solving v = 12 cm 9. Since shift in po si tion D t = 0.1 m Hence real depth = + ( ) 0.1 0.2 m = 0.3 m and apparent depth= 0.2 m m = real depth apparent depth = = 0.3 0.2 1.5 21 AIEEE Corner 1. We have r i + ° + = ° 90 180 Þ r i =  90 From Snell’s law 1.5 = =  sin sin sin sin ( ) i r i i 90 Þ tan i = 1.5 Þ i =  tan ( ) 1 1.5 2. n v u w w = = = air 0.229 343 1498 Critical angle q = = °  sin ( . ) 1 0229 13.2 3. Sped in glycrine v c n g g = = ´ 3 10 8 1.47 t v g 1 8 8 20 20 3 10 10 = = ´ ´ = ´  1.47 9.8 s Speed in glycrine v c n g g = = ´ 3 10 8 1.63 t v c 2 8 8 20 20 3 10 10 = = ´ ´  ´  1.63 10.8 ~ t t 2 1 8 1086 10  =  ´  ( . ) 9.8 = ´  1.67 s 10 8 4. (a) t v 1 6 1 6 8 1 10 1 10 3 10 = ´ = ´ ´   m 1.2 m / = ´ ´  1.2 10 3 10 6 8 Þ t 1 14 10 = ´  0.4 s t 2 6 8 14 1 0 3 1 0 1 0 = ´ ´ = ´   1.5 0.5 s t 3 6 8 6 8 14 1 10 3 10 10 3 10 10 = ´ ´ = ´ ´ = ´    /1.8 1.8 0.6 Hence t 1 is least and t 1 14 10 = ´  0.4 s (b) Total number of wavelengths = + + 1 1 1 2 3 m l m l m l m 1.5 m m / / / n n n = ´ + ´ 1000 600 100 600 1.2 nm nm 1.5 nm nm + ´ ´ 1 1000 600 1.8 nm nm = = 4500 600 7.5 5. The given wave equation is E y t x ( , ) = ´  ´ ´ é ë ê ù û ú  E y t ax sin 2 5 10 3 10 2 7 14 p l Comparing with standard equation E y t E ky t x ( , ) sin [ ] =  0 w k = ´ = ´ ´  2 5 10 2 3 10 7 14 p , w p v k = = ´ ´ ´ = ´  w p p 2 3 10 2 5 10 10 14 7 8 / 1.5 m/s Refractive index n c v = = ´ ´ = 3 10 10 2 8 8 1.5 Wavelength in this way l p n k = 2 Þ l p n = ´ = ´  2 25 5 10 5 10 7 7 / m Þ l n = 500 nm If vacuum, wavelength is l then l l n n = Þ l l = = ´ = n n 2 500 1000nm 6. Refraction from plane and spher i cal surfaces 22 Reflected ray Reflected ray Incident ray 90–i 90° i i r Page 5 27 Refraction of Light Introductory Exercise 27.1 1. Let real depth of dust par ti cle is x and thick ness of slab is t From Ist surface m = Real depth App.depth 1.5 cm = x 6 Þ x = 9 cm …(i) From other face m =  t x 4 Þ t x  = ´ 4 1.5 Þ t x = + = + = 6 9 6 16 cm 2. 1 2 4 3 m = Þ m m 2 1 4 3 = …(i) 2 3 3 2 m = Þ m m 3 2 3 2 = …(ii) From Eqs. (i) and (ii), we get m m 3 1 4 3 3 2 2 = ´ = 3. Frequency re main same. Let v 1 is velocity in medium (1) and v 2 in Medium (2) We have m 1 1 = c v and m 2 2 = c v Þ m m 1 2 2 1 = v v Þ v v 2 1 2 1 = m m Similarly, wavelength l m m l 2 1 2 1 = 4. From v n a a = l Þ l a a a v n = = ´ ´ = ´ +  3 10 6 10 5 10 8 14 7 m = 50 nm m l l = = = = a m 500 300 5 3 1.67 Introductory Exercise 27.2 1. Since light rays are com ing from glass to air ap pl y ing m m m m 2 1 2 1 v u R  =  Þ 1 10 1 v  =   1.5 1.5 1.5 Þ 1 1 30 10 1 4 30 v =  + =  + 1.5 .5 Þ v = = 30 3.5 8.57 cm 2. m m m m 2 1 2 1 v u R  =  (a) 1.5 0.5 m   = 1 20 6 ( ) On solving v = 45 cm (b) 1.5 0.5 v   = 1 10 6 ( ) On solving we get v =  90 cm t x 2nd face 1st face t – x (c) 1.5 0.5 v   = 1 3 6 ( ) On solving v =  6.0 cm 3. Light rays are coming from glass to air m m m m 2 1 2 1 v u R  =  1 4 3 10 1 4 3 15 v   =   ( ) / ( ) 1 4 30 1 45 v + = on solving v =  9 cm 4. Applying u v u R 2 1 2 1  =  m m m Þ 1.44 0.44 1.25 v  ¥ = 1 On solving v = 0.795 cm 5. m m m m 2 1 2 1 v u R  =  1.635 0.635 2.50) v   =  1 9 ( ) ( on solving v = 6.993 cm Lateral magnification m v u =  Þ =  =  6.993 9 0.777 Introductory Exercise 27.3 1. We have 1 1 1 1 1 1 1 2 v u f R R  = =   é ë ê ù û ú ( ) m Þ  + =    é ë ê ù û ú 1 20 1 60 1 1 1 ( ) 1.65 R R Þ  + = ´  3 1 60 2 0.65 R Þ R = ´ = 60 39 0.65 cm 2. Using 1 1 1 v u f  = , we get 1 50 1 1 30   = x Þ  = + 1 1 30 1 50 x On solving x =  18.75 cm m v u =  = 50 18.75 Height of filament image = ´ 2 50 18.75 = 5.3 cm 3. 1 1 1 1 f R R =  + é ë ê ù û ú ( ) m If lens faces becomes opposite three is no change in radius of curvature hence focal length does not change. 4. Using for mula 1 1 1 v u f  = when u ® 0 1 v ® 0 when u f v ® ® ¥ , hence im age moves from sur face to ¥ . 5. 1 1 1 1 1 2 f R R =   é ë ê ù û ú ( ) m Þ 1 1 1 1 2 20 f R R =    é ë ê ù û ú = ´  ( ) 1.3 0.3 Þ f =  100 3 cm (a) When immersed in a liquid of 1.8 refractive index 1 1 2 2 20 1 f R =  æ è ç ö ø ÷  é ë ê ù û ú =  ´  1.3 1.8 0.5 1.8 f ¢ = 36 cm (b) The minimum distance is equal to the focal length = 36 cm 20 O 10 cm 6. Using 1 1 1 v u f  = 1 1 20 12 10 v   = ( ) On solving v = 20 cm Magnification =  =  v u 1 Hence the image of same size and inverted. Let the distance between second lens is x Since magnification is unity image distance also x using again 1 1 1 v u f  = we get 1 1 1 1 10 x x f   = = ( ) Þ x = 20 cm Hence the distance between two lenses = + = 20 20 40 cm cm cm 7. 1 1 1 v f u = + …(i) 1 1 1 1 v f u du = + + …(ii) 1 1 v v u du u u duu ¢  = +  + ( ) ( ) on solving, we get v v v du u u du  ¢ ¢ = + ( ) thickness dv v u du =  2 2 10. Size of im age = ´ = 6 2 3 2 cm. 11. Let im age distance is u   m = 3 Þ v u = 3 1 3 1 1 12 u u + = Þ v = 16 cm 12. Since im age is up right and diminished hence lens is concave. Now u v  = 20 …(i) m v u = = 1 2 Þ 1 2 20 =  u u Þ u = 40 cm and v = 20 cm Þ Using 1 1 1 v u f  =  + = 1 20 1 40 1 f Þ f =  40 cm 13. The im age co in cide it self if light falls nor mally on plane mir ror hence ob ject must be on focus i e . . + 10 cm. 8. 1 1 2 2 1 2 1 2 2 1 1 v u R R + =   ( ) ( ) m /m m /m 1 1 2 4 3 2 4 3 1 v + =   0.2 0.4 0.4 ( / ) ( / ) On solving v = 12 cm 9. Since shift in po si tion D t = 0.1 m Hence real depth = + ( ) 0.1 0.2 m = 0.3 m and apparent depth= 0.2 m m = real depth apparent depth = = 0.3 0.2 1.5 21 AIEEE Corner 1. We have r i + ° + = ° 90 180 Þ r i =  90 From Snell’s law 1.5 = =  sin sin sin sin ( ) i r i i 90 Þ tan i = 1.5 Þ i =  tan ( ) 1 1.5 2. n v u w w = = = air 0.229 343 1498 Critical angle q = = °  sin ( . ) 1 0229 13.2 3. Sped in glycrine v c n g g = = ´ 3 10 8 1.47 t v g 1 8 8 20 20 3 10 10 = = ´ ´ = ´  1.47 9.8 s Speed in glycrine v c n g g = = ´ 3 10 8 1.63 t v c 2 8 8 20 20 3 10 10 = = ´ ´  ´  1.63 10.8 ~ t t 2 1 8 1086 10  =  ´  ( . ) 9.8 = ´  1.67 s 10 8 4. (a) t v 1 6 1 6 8 1 10 1 10 3 10 = ´ = ´ ´   m 1.2 m / = ´ ´  1.2 10 3 10 6 8 Þ t 1 14 10 = ´  0.4 s t 2 6 8 14 1 0 3 1 0 1 0 = ´ ´ = ´   1.5 0.5 s t 3 6 8 6 8 14 1 10 3 10 10 3 10 10 = ´ ´ = ´ ´ = ´    /1.8 1.8 0.6 Hence t 1 is least and t 1 14 10 = ´  0.4 s (b) Total number of wavelengths = + + 1 1 1 2 3 m l m l m l m 1.5 m m / / / n n n = ´ + ´ 1000 600 100 600 1.2 nm nm 1.5 nm nm + ´ ´ 1 1000 600 1.8 nm nm = = 4500 600 7.5 5. The given wave equation is E y t x ( , ) = ´  ´ ´ é ë ê ù û ú  E y t ax sin 2 5 10 3 10 2 7 14 p l Comparing with standard equation E y t E ky t x ( , ) sin [ ] =  0 w k = ´ = ´ ´  2 5 10 2 3 10 7 14 p , w p v k = = ´ ´ ´ = ´  w p p 2 3 10 2 5 10 10 14 7 8 / 1.5 m/s Refractive index n c v = = ´ ´ = 3 10 10 2 8 8 1.5 Wavelength in this way l p n k = 2 Þ l p n = ´ = ´  2 25 5 10 5 10 7 7 / m Þ l n = 500 nm If vacuum, wavelength is l then l l n n = Þ l l = = ´ = n n 2 500 1000nm 6. Refraction from plane and spher i cal surfaces 22 Reflected ray Reflected ray Incident ray 90–i 90° i i r We have sin sin 60° = r 1.8 Þ sin sin r = ° 60 1.8 Þ sin r = ´ = 3 2 1.8 0.48 Þ r =  sin ( ) 1 0.48 Þ r ~  ° 28.7 Now MO r 6 = tan Þ MO r = 6 tan Similarly ON = 6 tan r Þ MN MO ON r = + = = ° 12 12 tan tan( ) 28.7 Þ MN = 6.6 cm 7. From Snell’s law 4 3 45 = ° sin sin r Solving we get r = ° 32 EF DE r = = ° tan tan 3 32 = 1.88 m Total length of shadow = + 1 1.88 = 2.88 m 8. The sit u a tion is shown in figure For first surface m m m m 2 1 2 1 v u R  =  Þ 1.5 2.5) 0.5 v   = 1 10 ( Þ 1.5 2.5 v =  =  1 20 1 7 20 Þ v =  30 7 cm This image acts as a virtual object for 2nd surface u 2 20 30 7 170 7 =  + æ è ç ö ø ÷ =  cm and R =  10 cm m m m m 2 1 2 1 v u r  =  Þ 1 170 7 10 v + =   1.5 0.5 / Þ 1 1 20 170 v =  10.5 Þ v =  85 cm Hence final image will produced at 65cm from Ist surface. 9. Here v =  1 cm R =  2 cm Applying m m m m 2 1 2 1 v u R  =  Þ 1 1 1 2 2 1 4   =   =   = 1.5 1.5 0.5 x Þ  = 1.5 x 5 4 Þ x =  =  6 5 1.2 cm 23 2 cm 0.25 cm 10 cm 2 1 45° 45° D B 1 m A 3 m C E F r N N 2 N 1 60° r r r 6 cm N 3 M ORead More
1. What is refraction of light? 
2. How does refraction of light occur? 
3. What are some examples of refraction of light in everyday life? 
4. How does the refractive index affect the refraction of light? 
5. What are some applications of refraction of light? 

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