GEOMETRICAL OPTICS
OBJECTIVE-II
1. If the light moving in a straight line bends by a small but fixed angle, it may be a case of
(a) reflection
(b) refraction
(c) diffraction
(d) dispersion.
ANSWER: (a), (b)
EXPLANATION: If the light moving is incident on a reflective surface such that the angle between the light and the surface is small the reflected ray will also make a small but fixed angle. So it may be a case of reflection. Hence the option (a).
Suppose the light ray falls on a transparent medium surface (with different µ) with a small angle of incidence it will bend towards or away from the normal by a small but fixed angle. So it may be a case of refraction. Hence the option (b).
2. Mark the correct options,
(a) If the incident rays are converging, we have a real object.
(b) If the final rays are converging, we have a real image.
(c) The image of a virtual object is called a virtual image.
(d) If the image is virtual, the corresponding object is called a virtual object.
ANSWER: (b)
EXPLANATION: The definitions in the option (c) and (d) are not true.
The situation in the option (a) is not necessary.
The converging final rays make a real image, hence the option (b) is true.
3. Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or nor?
(a) Pole
(b) Focus
(c) Radius of curvature
(d) Principal axis.
ANSWER: (a), (c), (d)
EXPLANATION: The rays near the optical axis are called paraxial rays. The pole, radius of curvature and principal axis are dependent on the geometry of the mirror only not on the paraxial rays. But the focus is dependent on the paraxial rays because only these converge nearly to a single point. Hence the options (a), (c), (d).
4. The image of an extended object placed perpendicular to the principal axis of a mirror, will be erect if
(a) the object and the image are both real
(b) the object and the image are both virtual
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real.
ANSWER: (c), (d)
EXPLANATION: When the object is erect, the image will be erect only for the conditions mentioned in options (c) and (d).
5. A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black,
(a) the image will be shifted downward
(b) the image will be shifted upward
(c) the image will not be shifted
(d) the intensity of the image will decrease.
ANSWER: (c), (d)
EXPLANATION: In this condition, only the rays from the upper half of the lens is blocked. Hence the intensity of the image decreases but remains at the same position. Hence the option (c) and (d).
Diagram for Q-5
6. Consider three converging lenses L₁, L₂ and L₃ having identical geometrical construction, the index of refraction of L₁ and L₂ are µ₁ and µ₂ respectively. The upper half of the lens L₃ has a refractive index µ₁ and the lower half has µ₂ (figure 18-Q4). A point object O is imaged at O₁ by the lens L₁ and at O₂ by the lens L₂ placed in the same position. If L₃ is placed at the same place,
(a) there will be an image at O₁
(b) there will be an image at O₂
(c) the only image will form somewhere between O₁ and O₂
(d) the only image will form away from O₂.
Figure for Q-6
ANSWER: (a), (b)
EXPLANATION: In this case, the upper part of the lens L₃ will act as lens L₁ and the lower half as lens L₂. These two halves will form two images at O₁ and O₂, though their intensities will be reduced. Hence the options (a) and (b).
7. A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(a) must be less than 10 cm
(b) must be greater than 20 cm
(c) must not be greater than 20 cm
(d) must not be less than 10 cm.
ANSWER: (?) Must be greater than 10 cm
EXPLANATION: Let the distance of the image (if formed on screen) be x from the lens. Then the distance of the object will be (40-x) from the screen. If the distance towards right from the lens is considered positive then,
u = -(40-x), v = x. from the lens formula
1/f = 1/v - 1/u
→1/f = 1/x +1/(40-x) ={40-x+x}/x(40-x)
→1/f = 40/x(40-x)
→f = x(40-x)/40
For max value of f
df/dx = 0
→(40-2x)/40 = 0
→1-x/20 = 0
x = 20 cm
since d²f/dx² = -1/20 which is negative hence for x = 20 cm f is maximum.
so f = 20(40-20)/40 cm =20*20/40 cm = 10 cm.
So the focal length of the lens is more than 10 cm that is why the image failed to form on the screen.
1. If the light moving in a straight line bends by a small but fixed angle, it may be a case of
(a) reflection
(b) refraction
(c) diffraction
(d) dispersion.
ANSWER: (a), (b)
EXPLANATION: If the light moving is incident on a reflective surface such that the angle between the light and the surface is small the reflected ray will also make a small but fixed angle. So it may be a case of reflection. Hence the option (a).
Suppose the light ray falls on a transparent medium surface (with different µ) with a small angle of incidence it will bend towards or away from the normal by a small but fixed angle. So it may be a case of refraction. Hence the option (b).
2. Mark the correct options,
(a) If the incident rays are converging, we have a real object.
(b) If the final rays are converging, we have a real image.
(c) The image of a virtual object is called a virtual image.
(d) If the image is virtual, the corresponding object is called a virtual object.
ANSWER: (b)
EXPLANATION: The definitions in the option (c) and (d) are not true.
The situation in the option (a) is not necessary.
The converging final rays make a real image, hence the option (b) is true.
3. Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or nor?
(a) Pole
(b) Focus
(c) Radius of curvature
(d) Principal axis.
ANSWER: (a), (c), (d)
EXPLANATION: The rays near the optical axis are called paraxial rays. The pole, radius of curvature and principal axis are dependent on the geometry of the mirror only not on the paraxial rays. But the focus is dependent on the paraxial rays because only these converge nearly to a single point. Hence the options (a), (c), (d).
4. The image of an extended object placed perpendicular to the principal axis of a mirror, will be erect if
(a) the object and the image are both real
(b) the object and the image are both virtual
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real.
ANSWER: (c), (d)
EXPLANATION: When the object is erect, the image will be erect only for the conditions mentioned in options (c) and (d).
5. A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black,
(a) the image will be shifted downward
(b) the image will be shifted upward
(c) the image will not be shifted
(d) the intensity of the image will decrease.
ANSWER: (c), (d)
EXPLANATION: In this condition, only the rays from the upper half of the lens is blocked. Hence the intensity of the image decreases but remains at the same position. Hence the option (c) and (d).
6. Consider three converging lenses L₁, L₂ and L₃ having identical geometrical construction, the index of refraction of L₁ and L₂ are µ₁ and µ₂ respectively. The upper half of the lens L₃ has a refractive index µ₁ and the lower half has µ₂ (figure 18-Q4). A point object O is imaged at O₁ by the lens L₁ and at O₂ by the lens L₂ placed in the same position. If L₃ is placed at the same place,
(a) there will be an image at O₁
(b) there will be an image at O₂
(c) the only image will form somewhere between O₁ and O₂
(d) the only image will form away from O₂.
ANSWER: (a), (b)
EXPLANATION: In this case, the upper part of the lens L₃ will act as lens L₁ and the lower half as lens L₂. These two halves will form two images at O₁ and O₂, though their intensities will be reduced. Hence the options (a) and (b).
7. A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(a) must be less than 10 cm
(b) must be greater than 20 cm
(c) must not be greater than 20 cm
(d) must not be less than 10 cm.
ANSWER: (?) Must be greater than 10 cm
EXPLANATION: Let the distance of the image (if formed on screen) be x from the lens. Then the distance of the object will be (40-x) from the screen. If the distance towards right from the lens is considered positive then,
u = -(40-x), v = x. from the lens formula
1/f = 1/v - 1/u
→1/f = 1/x +1/(40-x) ={40-x+x}/x(40-x)
→1/f = 40/x(40-x)
→f = x(40-x)/40
For max value of f
df/dx = 0
→(40-2x)/40 = 0
→1-x/20 = 0
x = 20 cm
since d²f/dx² = -1/20 which is negative hence for x = 20 cm f is maximum.
so f = 20(40-20)/40 cm =20*20/40 cm = 10 cm.
So the focal length of the lens is more than 10 cm that is why the image failed to form on the screen.
(a) reflection
(b) refraction
(c) diffraction
(d) dispersion.
ANSWER: (a), (b)
EXPLANATION: If the light moving is incident on a reflective surface such that the angle between the light and the surface is small the reflected ray will also make a small but fixed angle. So it may be a case of reflection. Hence the option (a).
Suppose the light ray falls on a transparent medium surface (with different µ) with a small angle of incidence it will bend towards or away from the normal by a small but fixed angle. So it may be a case of refraction. Hence the option (b).
2. Mark the correct options,
(a) If the incident rays are converging, we have a real object.
(b) If the final rays are converging, we have a real image.
(c) The image of a virtual object is called a virtual image.
(d) If the image is virtual, the corresponding object is called a virtual object.
ANSWER: (b)
EXPLANATION: The definitions in the option (c) and (d) are not true.
The situation in the option (a) is not necessary.
The converging final rays make a real image, hence the option (b) is true.
3. Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or nor?
(a) Pole
(b) Focus
(c) Radius of curvature
(d) Principal axis.
ANSWER: (a), (c), (d)
EXPLANATION: The rays near the optical axis are called paraxial rays. The pole, radius of curvature and principal axis are dependent on the geometry of the mirror only not on the paraxial rays. But the focus is dependent on the paraxial rays because only these converge nearly to a single point. Hence the options (a), (c), (d).
4. The image of an extended object placed perpendicular to the principal axis of a mirror, will be erect if
(a) the object and the image are both real
(b) the object and the image are both virtual
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real.
ANSWER: (c), (d)
EXPLANATION: When the object is erect, the image will be erect only for the conditions mentioned in options (c) and (d).
5. A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black,
(a) the image will be shifted downward
(b) the image will be shifted upward
(c) the image will not be shifted
(d) the intensity of the image will decrease.
ANSWER: (c), (d)
EXPLANATION: In this condition, only the rays from the upper half of the lens is blocked. Hence the intensity of the image decreases but remains at the same position. Hence the option (c) and (d).
 |
| Diagram for Q-5 |
6. Consider three converging lenses L₁, L₂ and L₃ having identical geometrical construction, the index of refraction of L₁ and L₂ are µ₁ and µ₂ respectively. The upper half of the lens L₃ has a refractive index µ₁ and the lower half has µ₂ (figure 18-Q4). A point object O is imaged at O₁ by the lens L₁ and at O₂ by the lens L₂ placed in the same position. If L₃ is placed at the same place,
(a) there will be an image at O₁
(b) there will be an image at O₂
(c) the only image will form somewhere between O₁ and O₂
(d) the only image will form away from O₂.
 |
| Figure for Q-6 |
ANSWER: (a), (b)
EXPLANATION: In this case, the upper part of the lens L₃ will act as lens L₁ and the lower half as lens L₂. These two halves will form two images at O₁ and O₂, though their intensities will be reduced. Hence the options (a) and (b).
7. A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(a) must be less than 10 cm
(b) must be greater than 20 cm
(c) must not be greater than 20 cm
(d) must not be less than 10 cm.
ANSWER: (?) Must be greater than 10 cm
EXPLANATION: Let the distance of the image (if formed on screen) be x from the lens. Then the distance of the object will be (40-x) from the screen. If the distance towards right from the lens is considered positive then,
u = -(40-x), v = x. from the lens formula
1/f = 1/v - 1/u
→1/f = 1/x +1/(40-x) ={40-x+x}/x(40-x)
→1/f = 40/x(40-x)
→f = x(40-x)/40
For max value of f
df/dx = 0
→(40-2x)/40 = 0
→1-x/20 = 0
x = 20 cm
since d²f/dx² = -1/20 which is negative hence for x = 20 cm f is maximum.
so f = 20(40-20)/40 cm =20*20/40 cm = 10 cm.
So the focal length of the lens is more than 10 cm that is why the image failed to form on the screen.
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Links to the Chapters
Links to the Chapters
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 18 - Geometrical Optics
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
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Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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Click here for → EXERCISES (1-10)
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CHAPTER- 6 - Friction
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Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
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For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
