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 â€¦(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 = â€¦(i) 2 3 3 2 m = Þ m m 3 2 3 2 = â€¦(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 â€“ 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 â€¦(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 = â€¦(i) 2 3 3 2 m = Þ m m 3 2 3 2 = â€¦(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 â€“ 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 â€¦(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 = â€¦(i) 2 3 3 2 m = Þ m m 3 2 3 2 = â€¦(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 â€“ 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 = + â€¦(i) 1 1 1 1 v f u du = + + â€¦(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 â€¦(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 â€¦(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 = â€¦(i) 2 3 3 2 m = Þ m m 3 2 3 2 = â€¦(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 â€“ 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 = + â€¦(i) 1 1 1 1 v f u du = + + â€¦(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 â€¦(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 Snellâ€™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 90â€“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 â€¦(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 = â€¦(i) 2 3 3 2 m = Þ m m 3 2 3 2 = â€¦(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 â€“ 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 = + â€¦(i) 1 1 1 1 v f u du = + + â€¦(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 â€¦(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 Snellâ€™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 90â€“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 Snellâ€™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

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