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Page 1 8. F o cal length of com bi na tion 1 1 1 1 1 1 2 F F F f f = + =  convex convex if f f 1 2 > Þ 1 1 1 2 f f < Þ F =  negative Hence assertion is true. Since power is a measure of converging or divergence of a lens. Hence reason is not true. Correct option is (c). 9. Since glass slab pro duced a net shift. Hence v is in creased. Thus mag ni fied im age is ob tained but im age may be real or vir tual de pend ing on the po si tion of slab. Correct option is (b) 10. In this case im age dis tance of O O 1 2 and are same from the lens. Q 1 1 1 v u f  = and reason is trure. Hence correct option is (d). 11. As ser tion is false since only ray energe if refractve in dex of the col our less than the prism and an gle of in ci dence is less than crit i cal an gle but rea son is true. Cor rect op tion is (a). 12. If two ob ject is placed be tween pole and fo cus im age is real hence as ser tion is true. Also rea son is correct. Hence correct option is (b) 13. Since both as ser tion and rea son are true and rea son e x pla n a tion is cor rect. Hence correct option is (a). Objective Questions (Level 2) ¢ Single option correct 1. We have m = Real depth App.depth Þ 4 3 1 = App.depth Þ App. depth = 3 4 Hence the distance between bird and mirror = + = 2 3 4 11 4 m Since plane mirror form image behind the mirror (for real object) at same distance as object hence the distance between bird and its image = + = 11 4 11 4 11 2 m Correct option is (d). 2. m m m m 2 1 2 1 v u R  =  Here m =  ¥ u 1 1 = and m 2 = 1.5 Þ 1.5 0.5 v R + ¥ = 1 Þ v R = 3 Hence correct option is (b). 3. From figure, r = ° 30 Q r = +  = °  ° q a 90 120 90 37 60° r r 60° 60° Normal O Page 2 8. F o cal length of com bi na tion 1 1 1 1 1 1 2 F F F f f = + =  convex convex if f f 1 2 > Þ 1 1 1 2 f f < Þ F =  negative Hence assertion is true. Since power is a measure of converging or divergence of a lens. Hence reason is not true. Correct option is (c). 9. Since glass slab pro duced a net shift. Hence v is in creased. Thus mag ni fied im age is ob tained but im age may be real or vir tual de pend ing on the po si tion of slab. Correct option is (b) 10. In this case im age dis tance of O O 1 2 and are same from the lens. Q 1 1 1 v u f  = and reason is trure. Hence correct option is (d). 11. As ser tion is false since only ray energe if refractve in dex of the col our less than the prism and an gle of in ci dence is less than crit i cal an gle but rea son is true. Cor rect op tion is (a). 12. If two ob ject is placed be tween pole and fo cus im age is real hence as ser tion is true. Also rea son is correct. Hence correct option is (b) 13. Since both as ser tion and rea son are true and rea son e x pla n a tion is cor rect. Hence correct option is (a). Objective Questions (Level 2) ¢ Single option correct 1. We have m = Real depth App.depth Þ 4 3 1 = App.depth Þ App. depth = 3 4 Hence the distance between bird and mirror = + = 2 3 4 11 4 m Since plane mirror form image behind the mirror (for real object) at same distance as object hence the distance between bird and its image = + = 11 4 11 4 11 2 m Correct option is (d). 2. m m m m 2 1 2 1 v u R  =  Here m =  ¥ u 1 1 = and m 2 = 1.5 Þ 1.5 0.5 v R + ¥ = 1 Þ v R = 3 Hence correct option is (b). 3. From figure, r = ° 30 Q r = +  = °  ° q a 90 120 90 37 60° r r 60° 60° Normal O Hence m = ° ° = sin sin 60 30 3 The correct option is (d). 4. The lens be come di verg ing if m m m m 1 2 2 3  >  Þ m m m 1 3 2 2 + > or 2 2 1 3 m m m < + Hence correct option is (b). 5. 1 1 16 1 1 v f + = Þ v f f 1 16 16 =  m v f f f f 1 1 16 16 16 16 16 =  =   =   ( ) ( ) …(i) and 1 1 6 1 2 v f + = Þ v f f 2 6 6 =  m 2 2 6 = v ( Q image is virtual) Þ m f f f f 2 6 6 6 6 =  =  ( ) …(ii) But m m 1 2 = Þ   =  f f f f 16 6 Þ  + =  6 16 f f Þ 2 22 f = Þ f = 11 cm Hence correct option is (d). 6. Let real depth at any in stant t of the wa ter is h then volume of wa ter V R h = p 2 Þ dV dt R dh dt = p 2 …(i) Let apparent depth at this instant is h¢ Q m = Real depth Apparent depth Þ n n h h 2 1 1 = Þ h n n h 1 1 2 = Now dh dt x n n dh dt ¢ = = 1 2 Þ dh dt xn n = 2 1 …(ii) From Eq. (i) and (ii), we get dV dt R xn n = p 2 2 1 Hence correct option is (b). 7. Q q = ° 37 Þ i =   = = ° [ ( )] 90 90 37 q q Applying Snell’s law on face BC. sin sin 37 3 5 2 1 ° = = r m m Þ sin sin r = ° = ´ = 4 3 37 5 3 3 5 1 Þ r = ° 90 Hence deviation d = ° + ° = ° 90 37 127 Correct option is (b). 8. Let refractive in dex of liq uid is m For position of fish w.r.t. bird is m = Real depth App.depth Þ m = x h 1 …(i) For position of bird w.r.t. fish is 1 2 m = y h …(2) From Eq. (i) and (ii) we get u h h = 2 1 Hence correct option is (a). 9. Þ ( ) R t R  + = 2 2 2 3 Þ 2 9 Rt ~  Q R T >> 38 90–q i r 90 m = 5/3 A 90° air C B air (R – t ) R t 3 Page 3 8. F o cal length of com bi na tion 1 1 1 1 1 1 2 F F F f f = + =  convex convex if f f 1 2 > Þ 1 1 1 2 f f < Þ F =  negative Hence assertion is true. Since power is a measure of converging or divergence of a lens. Hence reason is not true. Correct option is (c). 9. Since glass slab pro duced a net shift. Hence v is in creased. Thus mag ni fied im age is ob tained but im age may be real or vir tual de pend ing on the po si tion of slab. Correct option is (b) 10. In this case im age dis tance of O O 1 2 and are same from the lens. Q 1 1 1 v u f  = and reason is trure. Hence correct option is (d). 11. As ser tion is false since only ray energe if refractve in dex of the col our less than the prism and an gle of in ci dence is less than crit i cal an gle but rea son is true. Cor rect op tion is (a). 12. If two ob ject is placed be tween pole and fo cus im age is real hence as ser tion is true. Also rea son is correct. Hence correct option is (b) 13. Since both as ser tion and rea son are true and rea son e x pla n a tion is cor rect. Hence correct option is (a). Objective Questions (Level 2) ¢ Single option correct 1. We have m = Real depth App.depth Þ 4 3 1 = App.depth Þ App. depth = 3 4 Hence the distance between bird and mirror = + = 2 3 4 11 4 m Since plane mirror form image behind the mirror (for real object) at same distance as object hence the distance between bird and its image = + = 11 4 11 4 11 2 m Correct option is (d). 2. m m m m 2 1 2 1 v u R  =  Here m =  ¥ u 1 1 = and m 2 = 1.5 Þ 1.5 0.5 v R + ¥ = 1 Þ v R = 3 Hence correct option is (b). 3. From figure, r = ° 30 Q r = +  = °  ° q a 90 120 90 37 60° r r 60° 60° Normal O Hence m = ° ° = sin sin 60 30 3 The correct option is (d). 4. The lens be come di verg ing if m m m m 1 2 2 3  >  Þ m m m 1 3 2 2 + > or 2 2 1 3 m m m < + Hence correct option is (b). 5. 1 1 16 1 1 v f + = Þ v f f 1 16 16 =  m v f f f f 1 1 16 16 16 16 16 =  =   =   ( ) ( ) …(i) and 1 1 6 1 2 v f + = Þ v f f 2 6 6 =  m 2 2 6 = v ( Q image is virtual) Þ m f f f f 2 6 6 6 6 =  =  ( ) …(ii) But m m 1 2 = Þ   =  f f f f 16 6 Þ  + =  6 16 f f Þ 2 22 f = Þ f = 11 cm Hence correct option is (d). 6. Let real depth at any in stant t of the wa ter is h then volume of wa ter V R h = p 2 Þ dV dt R dh dt = p 2 …(i) Let apparent depth at this instant is h¢ Q m = Real depth Apparent depth Þ n n h h 2 1 1 = Þ h n n h 1 1 2 = Now dh dt x n n dh dt ¢ = = 1 2 Þ dh dt xn n = 2 1 …(ii) From Eq. (i) and (ii), we get dV dt R xn n = p 2 2 1 Hence correct option is (b). 7. Q q = ° 37 Þ i =   = = ° [ ( )] 90 90 37 q q Applying Snell’s law on face BC. sin sin 37 3 5 2 1 ° = = r m m Þ sin sin r = ° = ´ = 4 3 37 5 3 3 5 1 Þ r = ° 90 Hence deviation d = ° + ° = ° 90 37 127 Correct option is (b). 8. Let refractive in dex of liq uid is m For position of fish w.r.t. bird is m = Real depth App.depth Þ m = x h 1 …(i) For position of bird w.r.t. fish is 1 2 m = y h …(2) From Eq. (i) and (ii) we get u h h = 2 1 Hence correct option is (a). 9. Þ ( ) R t R  + = 2 2 2 3 Þ 2 9 Rt ~  Q R T >> 38 90–q i r 90 m = 5/3 A 90° air C B air (R – t ) R t 3 Þ 2 900 3 300 R = = mm mm mm Þ R = = 150 15 mm cm Hence correct option is (a). 10. 1 1 1 1 1 v v f + = Þ v fv v f 1 1 1 =  m v v f v f 1 1 1 1 =  =   ( ) …(i) 1 1 1 2 2 v v f + = Þ v fv v f 2 2 2 =  m v v f f 2 2 2 2 = =  m …(ii) Q m m 1 2 = Þ   = f v f f v f 1 2 Þ f v v = + 1 2 2 Hence the correct option is (d). 11. 1 1 1 1 2 1 2 F f f d f f = +  Þ F f f f f d = +  1 2 1 2 as d increases f f d 1 2 +  decreases hence F increases. Hence image move to right. Correct option is (b). 12. In this case, minimum de vi a tion of ray 1 is same as ray 2. Hence correct option is (c). 13. For critical an gle at glass air sur face sin q m c g = = 1 2 3 …(i) Now for glass water surface. m m q w g c r = sin sin Þ 4 3 3 2 2 3 / / sin = ´ r Þ sin r = 4 3 Now for water air surface m m w a r r = ¢ sin sin Þ 4 3 4 3 1 = ´ ¢ sin r Þ sin r¢ = 1 Þ r¢ = ° 90 Hence correct option is (d). 14. F or lim it ing an g le of in ci dent emer gent ra y be come par al lel to the 2nd face Q sin sin r¢ ° = 90 3 7 Þ r¢ =  sin 1 3 7 Now r r + ° + ¢ = ° 30 90 Þ r r = °  ¢ 60 Þ r = °   ( sin / ) 60 3 7 1 Now m = sin sin i r Þ 7 3 60 3 7 1 =   sin sin [ sin / ] i Þ i = ´    sin { / sin ( sin / } 1 1 7 3 60 3 7 = ´ °  ° ì í î ü ý þ  sin sin ( ) 1 7 3 60 21 = ´ °  sin { / sin } 1 7 3 19 =  °  sin { ~ 1 30 0.49} Hence correct option is (a). 15. The im age form the ob ject it self if the rays in ci dent par al lel to op ti cal axis on the mir ror i.e., image of refraction is formed at ¥. It is possible when O is placed at focus i e . ., d=10cm Hence correct option is (c). 16. The dot will ap pear at c for all val ues of m. Since po si tion does not in same me dium. Hence correct option is (b). 39 i 30+r' 60° 90°–r' r' r Page 4 8. F o cal length of com bi na tion 1 1 1 1 1 1 2 F F F f f = + =  convex convex if f f 1 2 > Þ 1 1 1 2 f f < Þ F =  negative Hence assertion is true. Since power is a measure of converging or divergence of a lens. Hence reason is not true. Correct option is (c). 9. Since glass slab pro duced a net shift. Hence v is in creased. Thus mag ni fied im age is ob tained but im age may be real or vir tual de pend ing on the po si tion of slab. Correct option is (b) 10. In this case im age dis tance of O O 1 2 and are same from the lens. Q 1 1 1 v u f  = and reason is trure. Hence correct option is (d). 11. As ser tion is false since only ray energe if refractve in dex of the col our less than the prism and an gle of in ci dence is less than crit i cal an gle but rea son is true. Cor rect op tion is (a). 12. If two ob ject is placed be tween pole and fo cus im age is real hence as ser tion is true. Also rea son is correct. Hence correct option is (b) 13. Since both as ser tion and rea son are true and rea son e x pla n a tion is cor rect. Hence correct option is (a). Objective Questions (Level 2) ¢ Single option correct 1. We have m = Real depth App.depth Þ 4 3 1 = App.depth Þ App. depth = 3 4 Hence the distance between bird and mirror = + = 2 3 4 11 4 m Since plane mirror form image behind the mirror (for real object) at same distance as object hence the distance between bird and its image = + = 11 4 11 4 11 2 m Correct option is (d). 2. m m m m 2 1 2 1 v u R  =  Here m =  ¥ u 1 1 = and m 2 = 1.5 Þ 1.5 0.5 v R + ¥ = 1 Þ v R = 3 Hence correct option is (b). 3. From figure, r = ° 30 Q r = +  = °  ° q a 90 120 90 37 60° r r 60° 60° Normal O Hence m = ° ° = sin sin 60 30 3 The correct option is (d). 4. The lens be come di verg ing if m m m m 1 2 2 3  >  Þ m m m 1 3 2 2 + > or 2 2 1 3 m m m < + Hence correct option is (b). 5. 1 1 16 1 1 v f + = Þ v f f 1 16 16 =  m v f f f f 1 1 16 16 16 16 16 =  =   =   ( ) ( ) …(i) and 1 1 6 1 2 v f + = Þ v f f 2 6 6 =  m 2 2 6 = v ( Q image is virtual) Þ m f f f f 2 6 6 6 6 =  =  ( ) …(ii) But m m 1 2 = Þ   =  f f f f 16 6 Þ  + =  6 16 f f Þ 2 22 f = Þ f = 11 cm Hence correct option is (d). 6. Let real depth at any in stant t of the wa ter is h then volume of wa ter V R h = p 2 Þ dV dt R dh dt = p 2 …(i) Let apparent depth at this instant is h¢ Q m = Real depth Apparent depth Þ n n h h 2 1 1 = Þ h n n h 1 1 2 = Now dh dt x n n dh dt ¢ = = 1 2 Þ dh dt xn n = 2 1 …(ii) From Eq. (i) and (ii), we get dV dt R xn n = p 2 2 1 Hence correct option is (b). 7. Q q = ° 37 Þ i =   = = ° [ ( )] 90 90 37 q q Applying Snell’s law on face BC. sin sin 37 3 5 2 1 ° = = r m m Þ sin sin r = ° = ´ = 4 3 37 5 3 3 5 1 Þ r = ° 90 Hence deviation d = ° + ° = ° 90 37 127 Correct option is (b). 8. Let refractive in dex of liq uid is m For position of fish w.r.t. bird is m = Real depth App.depth Þ m = x h 1 …(i) For position of bird w.r.t. fish is 1 2 m = y h …(2) From Eq. (i) and (ii) we get u h h = 2 1 Hence correct option is (a). 9. Þ ( ) R t R  + = 2 2 2 3 Þ 2 9 Rt ~  Q R T >> 38 90–q i r 90 m = 5/3 A 90° air C B air (R – t ) R t 3 Þ 2 900 3 300 R = = mm mm mm Þ R = = 150 15 mm cm Hence correct option is (a). 10. 1 1 1 1 1 v v f + = Þ v fv v f 1 1 1 =  m v v f v f 1 1 1 1 =  =   ( ) …(i) 1 1 1 2 2 v v f + = Þ v fv v f 2 2 2 =  m v v f f 2 2 2 2 = =  m …(ii) Q m m 1 2 = Þ   = f v f f v f 1 2 Þ f v v = + 1 2 2 Hence the correct option is (d). 11. 1 1 1 1 2 1 2 F f f d f f = +  Þ F f f f f d = +  1 2 1 2 as d increases f f d 1 2 +  decreases hence F increases. Hence image move to right. Correct option is (b). 12. In this case, minimum de vi a tion of ray 1 is same as ray 2. Hence correct option is (c). 13. For critical an gle at glass air sur face sin q m c g = = 1 2 3 …(i) Now for glass water surface. m m q w g c r = sin sin Þ 4 3 3 2 2 3 / / sin = ´ r Þ sin r = 4 3 Now for water air surface m m w a r r = ¢ sin sin Þ 4 3 4 3 1 = ´ ¢ sin r Þ sin r¢ = 1 Þ r¢ = ° 90 Hence correct option is (d). 14. F or lim it ing an g le of in ci dent emer gent ra y be come par al lel to the 2nd face Q sin sin r¢ ° = 90 3 7 Þ r¢ =  sin 1 3 7 Now r r + ° + ¢ = ° 30 90 Þ r r = °  ¢ 60 Þ r = °   ( sin / ) 60 3 7 1 Now m = sin sin i r Þ 7 3 60 3 7 1 =   sin sin [ sin / ] i Þ i = ´    sin { / sin ( sin / } 1 1 7 3 60 3 7 = ´ °  ° ì í î ü ý þ  sin sin ( ) 1 7 3 60 21 = ´ °  sin { / sin } 1 7 3 19 =  °  sin { ~ 1 30 0.49} Hence correct option is (a). 15. The im age form the ob ject it self if the rays in ci dent par al lel to op ti cal axis on the mir ror i.e., image of refraction is formed at ¥. It is possible when O is placed at focus i e . ., d=10cm Hence correct option is (c). 16. The dot will ap pear at c for all val ues of m. Since po si tion does not in same me dium. Hence correct option is (b). 39 i 30+r' 60° 90°–r' r' r 17. We have for to tal internal reflection sin sin / / i 90 6 5 3 2 = Þ sin i = 4 5 Þ i = æ è ç ö ø ÷ = °  sin 1 4 5 53 Hence the ray will not cross BC if i > ° 53 Þ 90 180 + + = ° q i q =  90 i Q i > ° 53 Þ q < ° 37 Hence correct option is (a). 18. For reflection at curved sur face 1 1 3 2 1 1 10 v x   =  æ è ç ö ø ÷ ´ ( ) Þ v x x =  20 20 This image act as virtual object for planeglasswater surface Þ m m m m g w g w x x ¥   =  ¥ ( ) 20 20 Þ x = 20 cm Hence answer is (c). 19. The ra tio of fo cal length in the sit u a tion II and III is 1 : 1. Hence correct option is (c). 20. We have 1 1 1 OB OA f   = ( ) Þ 1 1 1 OB OA f + = Þ f OB OA OA OB = + . Q OB OA AB + = Þ f OB OA AB = . …(i) Now AB AC BC 2 2 2 = + ( ) OA OB OC OA OB OC + = + + + 2 2 2 2 2 Þ OA OB OAOB C OA OB 2 2 2 2 2 2 20 + + = + + Þ OC OA OB 2 = …(ii) Putting this value in Eq. (i), we get Þ f OC AB = 2 Hence correct option is (c). 21. The shift pro duce Dt t w g =  é ë ê ê ù û ú ú 1 2 m =  é ë ê ù û ú 36 1 1 9 8 / Q w g m = = 3 2 4 3 9 8 / / = ´ = 36 1 9 4 cm Hence correct option is (b). 22. m = Real depth App. depth Þ 4 3 = real depth 10.5 cm Þ Real depth = ´ 4 3 10.5 cm = 14 cm Hence correct option is (d). 23. Q y 0 1 = + cm and y i =  2 cm m v u =  Þ 2 = v u Now let x be the position of lens then v x =  50 and v x = + ( ) 40 . Þ 2 50 40 =  + x x Þ 80 2 50 + =  x x Þ  = 3 30 x Þ x =  10 cm Hence correct option is (c). 40 3/2 B E C D q q i 90° 6/5 A Page 5 8. F o cal length of com bi na tion 1 1 1 1 1 1 2 F F F f f = + =  convex convex if f f 1 2 > Þ 1 1 1 2 f f < Þ F =  negative Hence assertion is true. Since power is a measure of converging or divergence of a lens. Hence reason is not true. Correct option is (c). 9. Since glass slab pro duced a net shift. Hence v is in creased. Thus mag ni fied im age is ob tained but im age may be real or vir tual de pend ing on the po si tion of slab. Correct option is (b) 10. In this case im age dis tance of O O 1 2 and are same from the lens. Q 1 1 1 v u f  = and reason is trure. Hence correct option is (d). 11. As ser tion is false since only ray energe if refractve in dex of the col our less than the prism and an gle of in ci dence is less than crit i cal an gle but rea son is true. Cor rect op tion is (a). 12. If two ob ject is placed be tween pole and fo cus im age is real hence as ser tion is true. Also rea son is correct. Hence correct option is (b) 13. Since both as ser tion and rea son are true and rea son e x pla n a tion is cor rect. Hence correct option is (a). Objective Questions (Level 2) ¢ Single option correct 1. We have m = Real depth App.depth Þ 4 3 1 = App.depth Þ App. depth = 3 4 Hence the distance between bird and mirror = + = 2 3 4 11 4 m Since plane mirror form image behind the mirror (for real object) at same distance as object hence the distance between bird and its image = + = 11 4 11 4 11 2 m Correct option is (d). 2. m m m m 2 1 2 1 v u R  =  Here m =  ¥ u 1 1 = and m 2 = 1.5 Þ 1.5 0.5 v R + ¥ = 1 Þ v R = 3 Hence correct option is (b). 3. From figure, r = ° 30 Q r = +  = °  ° q a 90 120 90 37 60° r r 60° 60° Normal O Hence m = ° ° = sin sin 60 30 3 The correct option is (d). 4. The lens be come di verg ing if m m m m 1 2 2 3  >  Þ m m m 1 3 2 2 + > or 2 2 1 3 m m m < + Hence correct option is (b). 5. 1 1 16 1 1 v f + = Þ v f f 1 16 16 =  m v f f f f 1 1 16 16 16 16 16 =  =   =   ( ) ( ) …(i) and 1 1 6 1 2 v f + = Þ v f f 2 6 6 =  m 2 2 6 = v ( Q image is virtual) Þ m f f f f 2 6 6 6 6 =  =  ( ) …(ii) But m m 1 2 = Þ   =  f f f f 16 6 Þ  + =  6 16 f f Þ 2 22 f = Þ f = 11 cm Hence correct option is (d). 6. Let real depth at any in stant t of the wa ter is h then volume of wa ter V R h = p 2 Þ dV dt R dh dt = p 2 …(i) Let apparent depth at this instant is h¢ Q m = Real depth Apparent depth Þ n n h h 2 1 1 = Þ h n n h 1 1 2 = Now dh dt x n n dh dt ¢ = = 1 2 Þ dh dt xn n = 2 1 …(ii) From Eq. (i) and (ii), we get dV dt R xn n = p 2 2 1 Hence correct option is (b). 7. Q q = ° 37 Þ i =   = = ° [ ( )] 90 90 37 q q Applying Snell’s law on face BC. sin sin 37 3 5 2 1 ° = = r m m Þ sin sin r = ° = ´ = 4 3 37 5 3 3 5 1 Þ r = ° 90 Hence deviation d = ° + ° = ° 90 37 127 Correct option is (b). 8. Let refractive in dex of liq uid is m For position of fish w.r.t. bird is m = Real depth App.depth Þ m = x h 1 …(i) For position of bird w.r.t. fish is 1 2 m = y h …(2) From Eq. (i) and (ii) we get u h h = 2 1 Hence correct option is (a). 9. Þ ( ) R t R  + = 2 2 2 3 Þ 2 9 Rt ~  Q R T >> 38 90–q i r 90 m = 5/3 A 90° air C B air (R – t ) R t 3 Þ 2 900 3 300 R = = mm mm mm Þ R = = 150 15 mm cm Hence correct option is (a). 10. 1 1 1 1 1 v v f + = Þ v fv v f 1 1 1 =  m v v f v f 1 1 1 1 =  =   ( ) …(i) 1 1 1 2 2 v v f + = Þ v fv v f 2 2 2 =  m v v f f 2 2 2 2 = =  m …(ii) Q m m 1 2 = Þ   = f v f f v f 1 2 Þ f v v = + 1 2 2 Hence the correct option is (d). 11. 1 1 1 1 2 1 2 F f f d f f = +  Þ F f f f f d = +  1 2 1 2 as d increases f f d 1 2 +  decreases hence F increases. Hence image move to right. Correct option is (b). 12. In this case, minimum de vi a tion of ray 1 is same as ray 2. Hence correct option is (c). 13. For critical an gle at glass air sur face sin q m c g = = 1 2 3 …(i) Now for glass water surface. m m q w g c r = sin sin Þ 4 3 3 2 2 3 / / sin = ´ r Þ sin r = 4 3 Now for water air surface m m w a r r = ¢ sin sin Þ 4 3 4 3 1 = ´ ¢ sin r Þ sin r¢ = 1 Þ r¢ = ° 90 Hence correct option is (d). 14. F or lim it ing an g le of in ci dent emer gent ra y be come par al lel to the 2nd face Q sin sin r¢ ° = 90 3 7 Þ r¢ =  sin 1 3 7 Now r r + ° + ¢ = ° 30 90 Þ r r = °  ¢ 60 Þ r = °   ( sin / ) 60 3 7 1 Now m = sin sin i r Þ 7 3 60 3 7 1 =   sin sin [ sin / ] i Þ i = ´    sin { / sin ( sin / } 1 1 7 3 60 3 7 = ´ °  ° ì í î ü ý þ  sin sin ( ) 1 7 3 60 21 = ´ °  sin { / sin } 1 7 3 19 =  °  sin { ~ 1 30 0.49} Hence correct option is (a). 15. The im age form the ob ject it self if the rays in ci dent par al lel to op ti cal axis on the mir ror i.e., image of refraction is formed at ¥. It is possible when O is placed at focus i e . ., d=10cm Hence correct option is (c). 16. The dot will ap pear at c for all val ues of m. Since po si tion does not in same me dium. Hence correct option is (b). 39 i 30+r' 60° 90°–r' r' r 17. We have for to tal internal reflection sin sin / / i 90 6 5 3 2 = Þ sin i = 4 5 Þ i = æ è ç ö ø ÷ = °  sin 1 4 5 53 Hence the ray will not cross BC if i > ° 53 Þ 90 180 + + = ° q i q =  90 i Q i > ° 53 Þ q < ° 37 Hence correct option is (a). 18. For reflection at curved sur face 1 1 3 2 1 1 10 v x   =  æ è ç ö ø ÷ ´ ( ) Þ v x x =  20 20 This image act as virtual object for planeglasswater surface Þ m m m m g w g w x x ¥   =  ¥ ( ) 20 20 Þ x = 20 cm Hence answer is (c). 19. The ra tio of fo cal length in the sit u a tion II and III is 1 : 1. Hence correct option is (c). 20. We have 1 1 1 OB OA f   = ( ) Þ 1 1 1 OB OA f + = Þ f OB OA OA OB = + . Q OB OA AB + = Þ f OB OA AB = . …(i) Now AB AC BC 2 2 2 = + ( ) OA OB OC OA OB OC + = + + + 2 2 2 2 2 Þ OA OB OAOB C OA OB 2 2 2 2 2 2 20 + + = + + Þ OC OA OB 2 = …(ii) Putting this value in Eq. (i), we get Þ f OC AB = 2 Hence correct option is (c). 21. The shift pro duce Dt t w g =  é ë ê ê ù û ú ú 1 2 m =  é ë ê ù û ú 36 1 1 9 8 / Q w g m = = 3 2 4 3 9 8 / / = ´ = 36 1 9 4 cm Hence correct option is (b). 22. m = Real depth App. depth Þ 4 3 = real depth 10.5 cm Þ Real depth = ´ 4 3 10.5 cm = 14 cm Hence correct option is (d). 23. Q y 0 1 = + cm and y i =  2 cm m v u =  Þ 2 = v u Now let x be the position of lens then v x =  50 and v x = + ( ) 40 . Þ 2 50 40 =  + x x Þ 80 2 50 + =  x x Þ  = 3 30 x Þ x =  10 cm Hence correct option is (c). 40 3/2 B E C D q q i 90° 6/5 A 24. If the plane surface of planocon v e x lens is sil vered it be have the con cave mir ror of fo cal length f m /2 Q f m = 10 cm Þ f e = 5 cm hence R = 10 cm Correct option is (c). 25. Since lens made real and mag ni fied im age, hence it is a con vex lens when lens dipped in wa ter its fo cal length. 1 1 1 1 1 2 f R R w g =  æ è ç ç ö ø ÷ ÷ + æ è ç ç ö ø ÷ ÷ m m =  æ è ç ç ö ø ÷ ÷ + æ è ç ç ö ø ÷ ÷ =  + æ è ç ç ö ø ÷ 4 3 3 2 1 1 1 1 9 1 1 1 2 1 2 / / R R R R ÷ Q f is  ve lens behave as concave, hence the image is virtual and magnified. Correct option is (c). 26. The prism trans mit the light for which an gle of in ci dence( c), 2 9 0 c £ ° Þ c £ ° 4 5 Hence m = = ° = = 1 1 4 5 2 s i n s i n C 1 . 4 1 4 Correct option is (b). 27. 1 1 16 1 v f   = ( ) Q   m v u = = 3 For convex lens v u = 3 1 48 1 16 1 + = f Þ f = 12 cm (for real image) Similarly when distance is 6 cm, 3 times virtual image is formed hence mirror is convex with focal length 12 cm. Correct option is (c). 28. v n g g = l Þ c n g g m l = …(i) and c n w w m l = …(ii) Dividing (i) and (ii) m m l l w g g w = Þ 4 3 4 5 mg = Þ m g = 5 3 Hence correct option is (a). 29. x f f d f d f f d = +  +  1 2 1 1 2 ( ) and y f d f f d =  +  ( ) 1 1 2 D Here f f d 1 2 20 30 = = = cm cm , and D = = 55m 0.5 cm Putting these values we get x = 25 cm and y = 0.25 cm Correct option is (b). 30. Since for each q an gle of in ci dence at glassair boundry re mains 0° hence there will never be to tal in ter nal reflection. Correct option is (d). 31. Diameter = ´ m Orig i nal di am e ter = ´ =  4 3 1 4 3 cm cm Hence correct option is (a). 32. 1 1 1 20 1 20 1 f =  + é ë ê ù û ú ( ) 1.5 Þ f 1 20 = cm Here u 1 30 =  cm 1 1 1 1 1 v u f  = Þ 1 1 20 1 30 1 v =  Þ v 1 60 = cm Magnification   m v u 1 60 30 2 =  = = + (Inverted image) 41 30 cm f = – 20 f = 20 cm 5 mm 90°Read More

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