OPTICAL INSTRUMENTS
OBJECTIVE-I
1. The size of an object as perceived by an eye depends primarily on
(a) the actual size of the object
(b) the distance of the object from the eye
(c) the aperture of the pupil
(d) size of the image formed on the retina.
ANSWER: (d)
EXPLANATION: The size of an object as perceived by an eye depends on the angle subtended by the object on the eye. Wider the angle larger is the image formed on the retina and bigger image is perceived by the eye.
2. The muscles of a normal eye are least strained when the eye is focused on an object
(a) far away from the eye
(b) very close to the eye
(c) at about 25 cm from the eye
(d) at about 1 m from the eye.
ANSWER: (a)
EXPLANATION: When the eyes are focused far away from the eye the image of the object is formed on the retina. As the object comes closer, the muscles are contracted to reduce the focal length of the eye lens so that again the image is formed on the retina to see the object clearly. Hence the muscles are least strained when eyes focus at objects far away from it.
3. A normal eye is not able to see objects closer than 25 cm because
(a) the focal length of the eye is 25 cm
(b) the distance of the retina from the eye-lens is 25 cm
(c) the eye is not able to decrease the distance between the eye lens and the retina beyond a limit
(d) the eye is not able to decrease the focal length beyond a limit.
ANSWER: (d)
EXPLANATION: The focal length of the lens in the eyes is variable. It is maximum when the muscles in the eyes are relaxed for seeing the faraway objects. For nearer objects the muscles contract to reduce the focal length of the lens. But there is a limit to this contraction of the muscles and hence the limit for the focal length. This limit allows the normal eyes to see objects up to 25 cm closer.
4. When objects at different distances are seen by the eyes, which of the following remain constant?
(a) The focal length of the eye lens.
(b) The object-distance from the eye lens.
(c) The radii of curvature of the eye lens.
(d) The image-distance from the eye lens.
ANSWER: (d)
EXPLANATION: The focal length and the radii of curvature of the eye lens varies when the objects at different distances are seen. Hence (a) and (c) are not true. Since the object distance varies in the problem hence (b) is not true. The image in each case forms at the retina and the distance of the retina and the eye lens is fixed hence the option (d) is true.
5. A person A can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person B having muscle stronger than A, but all other parameters of eye identical to that of A?
(a) 25 cm to 200 cm
(b) 18 cm to 200 cm
(c) 25 cm to 300 cm
(d) 18 cm to 300 cm
ANSWER: (b)
EXPLANATION: For both the persons when the muscles are relaxed the distant vision will be 200 cm. Since muscles of B are stronger, he will be able to reduce the focal length of the eye more than person B. Hence B may see objects clearly nearer than 25 cm. Hence the option (b).
6. The focal length of a normal eye lens is about
(a) 1 mm
(b) 2 cm
(c) 25 cm
(d) 1 m
ANSWER: (b)
EXPLANATION: The normal eye is when the muscles are relaxed. In this stage, the eye is focused at a faraway distance and the light rays are nearly parallel. The image is formed at the retina which is the focus of the eye lens. The focal length is the distance between the eye lens and the retina which is about 2 cm.
Diagram for Q-6
7. The distance of the eye lens from the retina is x. For a normal eye, the maximum focal length of the eye lens
(a) = x
(b) < x
(c) > x
(d) = 2x.
ANSWER: (a)
EXPLANATION: The maximum focal length of the eye lens is when the muscles are relaxed i.e. when the eye is focused at a far away distance from the eye. In this case, the incoming rays are nearly parallel and they are focused at the retina. Hence the focal length of the eye lens is the distance between the eye lens and the retina which in this case is =x. Hence the option (a).
8. A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m
(a) if he is farsighted
(b) if he is nearsighted
(c) if his vision is normal
(d) in each of these cases.
ANSWER: (d)
EXPLANATION: For a farsighted person, who can not see objects clearly for a distance of 25 cm, the gasses in question will correct his vision for the near distances and he may see the objects from 25 cm to 1m. Without the glasses, he was not able to sufficiently contract the muscles to focus at the retina.
For a nearsighted person who can not see objects clearly at a far away distances, if such glasses are given he may get his distant vision even reduced to 1 m only. Similar may be a case for a person with normal vision. Such glasses may reduce his distant vision from infinity to 1 m only.
Hence the option (d).
9. An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
(a) on f but not on u
(b) on u but not on f
(c) on f as well as u
(d) neither on f nor on u
ANSWER: (c)
EXPLANATION: For a simple microscope, the angular magnification is given as D/f for the normal adjustment and for the object within the focal length it is =1+D/f. Hence the angular magnification depends on both the f and u.
10. To increase the angular magnification of a simple microscope, one should increase
(a) the focal length of the lens
(b) the power of the lens
(c) the aperture of the lens
(d) the object size.
ANSWER: (b)
EXPLANATION: The angular magnification is given as
m = D/f, for normal adjustment and
=1+D/f for the object within f.
→m =D*P or 1+D*P, where P=1/f =power of the lens.
It is clear that m depends upon P, power of the lens.
11. A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5X before his eyes. The angular magnification achieved is
(a) 5
(b) 2.5
(c) 1
(d) 0.2
ANSWER: (c)
EXPLANATION: With the simple microscope of angular magnification 5X, the final image will appear to be at D. Initially the object was also at the near point =D. Let the height of the object = h and the height of the final image = h'.
The lateral magnification
=h'/h=v/u =(-D)/(-D) =1
→h' = h
Now the angular magnification
m =Angle subtended by the final image/the angle subtended by the object.
→m =(h'/D)/(h/D)
→m =h'/h = 1
Hence the option (c).
1. The size of an object as perceived by an eye depends primarily on
(a) the actual size of the object
(b) the distance of the object from the eye
(c) the aperture of the pupil
(d) size of the image formed on the retina.
ANSWER: (d)
EXPLANATION: The size of an object as perceived by an eye depends on the angle subtended by the object on the eye. Wider the angle larger is the image formed on the retina and bigger image is perceived by the eye.
2. The muscles of a normal eye are least strained when the eye is focused on an object
(a) far away from the eye
(b) very close to the eye
(c) at about 25 cm from the eye
(d) at about 1 m from the eye.
ANSWER: (a)
EXPLANATION: When the eyes are focused far away from the eye the image of the object is formed on the retina. As the object comes closer, the muscles are contracted to reduce the focal length of the eye lens so that again the image is formed on the retina to see the object clearly. Hence the muscles are least strained when eyes focus at objects far away from it.
3. A normal eye is not able to see objects closer than 25 cm because
(a) the focal length of the eye is 25 cm
(b) the distance of the retina from the eye-lens is 25 cm
(c) the eye is not able to decrease the distance between the eye lens and the retina beyond a limit
(d) the eye is not able to decrease the focal length beyond a limit.
ANSWER: (d)
EXPLANATION: The focal length of the lens in the eyes is variable. It is maximum when the muscles in the eyes are relaxed for seeing the faraway objects. For nearer objects the muscles contract to reduce the focal length of the lens. But there is a limit to this contraction of the muscles and hence the limit for the focal length. This limit allows the normal eyes to see objects up to 25 cm closer.
4. When objects at different distances are seen by the eyes, which of the following remain constant?
(a) The focal length of the eye lens.
(b) The object-distance from the eye lens.
(c) The radii of curvature of the eye lens.
(d) The image-distance from the eye lens.
ANSWER: (d)
EXPLANATION: The focal length and the radii of curvature of the eye lens varies when the objects at different distances are seen. Hence (a) and (c) are not true. Since the object distance varies in the problem hence (b) is not true. The image in each case forms at the retina and the distance of the retina and the eye lens is fixed hence the option (d) is true.
5. A person A can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person B having muscle stronger than A, but all other parameters of eye identical to that of A?
(a) 25 cm to 200 cm
(b) 18 cm to 200 cm
(c) 25 cm to 300 cm
(d) 18 cm to 300 cm
ANSWER: (b)
EXPLANATION: For both the persons when the muscles are relaxed the distant vision will be 200 cm. Since muscles of B are stronger, he will be able to reduce the focal length of the eye more than person B. Hence B may see objects clearly nearer than 25 cm. Hence the option (b).
6. The focal length of a normal eye lens is about
(a) 1 mm
(b) 2 cm
(c) 25 cm
(d) 1 m
ANSWER: (b)
EXPLANATION: The normal eye is when the muscles are relaxed. In this stage, the eye is focused at a faraway distance and the light rays are nearly parallel. The image is formed at the retina which is the focus of the eye lens. The focal length is the distance between the eye lens and the retina which is about 2 cm.
7. The distance of the eye lens from the retina is x. For a normal eye, the maximum focal length of the eye lens
(a) = x
(b) < x
(c) > x
(d) = 2x.
ANSWER: (a)
EXPLANATION: The maximum focal length of the eye lens is when the muscles are relaxed i.e. when the eye is focused at a far away distance from the eye. In this case, the incoming rays are nearly parallel and they are focused at the retina. Hence the focal length of the eye lens is the distance between the eye lens and the retina which in this case is =x. Hence the option (a).
8. A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m
(a) if he is farsighted
(b) if he is nearsighted
(c) if his vision is normal
(d) in each of these cases.
ANSWER: (d)
EXPLANATION: For a farsighted person, who can not see objects clearly for a distance of 25 cm, the gasses in question will correct his vision for the near distances and he may see the objects from 25 cm to 1m. Without the glasses, he was not able to sufficiently contract the muscles to focus at the retina.
For a nearsighted person who can not see objects clearly at a far away distances, if such glasses are given he may get his distant vision even reduced to 1 m only. Similar may be a case for a person with normal vision. Such glasses may reduce his distant vision from infinity to 1 m only.
Hence the option (d).
9. An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
(a) on f but not on u
(b) on u but not on f
(c) on f as well as u
(d) neither on f nor on u
ANSWER: (c)
EXPLANATION: For a simple microscope, the angular magnification is given as D/f for the normal adjustment and for the object within the focal length it is =1+D/f. Hence the angular magnification depends on both the f and u.
10. To increase the angular magnification of a simple microscope, one should increase
(a) the focal length of the lens
(b) the power of the lens
(c) the aperture of the lens
(d) the object size.
ANSWER: (b)
EXPLANATION: The angular magnification is given as
m = D/f, for normal adjustment and
=1+D/f for the object within f.
→m =D*P or 1+D*P, where P=1/f =power of the lens.
It is clear that m depends upon P, power of the lens.
11. A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5X before his eyes. The angular magnification achieved is
(a) 5
(b) 2.5
(c) 1
(d) 0.2
ANSWER: (c)
EXPLANATION: With the simple microscope of angular magnification 5X, the final image will appear to be at D. Initially the object was also at the near point =D. Let the height of the object = h and the height of the final image = h'.
The lateral magnification
=h'/h=v/u =(-D)/(-D) =1
→h' = h
Now the angular magnification
m =Angle subtended by the final image/the angle subtended by the object.
→m =(h'/D)/(h/D)
→m =h'/h = 1
Hence the option (c).
(a) the actual size of the object
(b) the distance of the object from the eye
(c) the aperture of the pupil
(d) size of the image formed on the retina.
ANSWER: (d)
EXPLANATION: The size of an object as perceived by an eye depends on the angle subtended by the object on the eye. Wider the angle larger is the image formed on the retina and bigger image is perceived by the eye.
2. The muscles of a normal eye are least strained when the eye is focused on an object
(a) far away from the eye
(b) very close to the eye
(c) at about 25 cm from the eye
(d) at about 1 m from the eye.
ANSWER: (a)
EXPLANATION: When the eyes are focused far away from the eye the image of the object is formed on the retina. As the object comes closer, the muscles are contracted to reduce the focal length of the eye lens so that again the image is formed on the retina to see the object clearly. Hence the muscles are least strained when eyes focus at objects far away from it.
3. A normal eye is not able to see objects closer than 25 cm because
(a) the focal length of the eye is 25 cm
(b) the distance of the retina from the eye-lens is 25 cm
(c) the eye is not able to decrease the distance between the eye lens and the retina beyond a limit
(d) the eye is not able to decrease the focal length beyond a limit.
ANSWER: (d)
EXPLANATION: The focal length of the lens in the eyes is variable. It is maximum when the muscles in the eyes are relaxed for seeing the faraway objects. For nearer objects the muscles contract to reduce the focal length of the lens. But there is a limit to this contraction of the muscles and hence the limit for the focal length. This limit allows the normal eyes to see objects up to 25 cm closer.
4. When objects at different distances are seen by the eyes, which of the following remain constant?
(a) The focal length of the eye lens.
(b) The object-distance from the eye lens.
(c) The radii of curvature of the eye lens.
(d) The image-distance from the eye lens.
ANSWER: (d)
EXPLANATION: The focal length and the radii of curvature of the eye lens varies when the objects at different distances are seen. Hence (a) and (c) are not true. Since the object distance varies in the problem hence (b) is not true. The image in each case forms at the retina and the distance of the retina and the eye lens is fixed hence the option (d) is true.
5. A person A can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person B having muscle stronger than A, but all other parameters of eye identical to that of A?
(a) 25 cm to 200 cm
(b) 18 cm to 200 cm
(c) 25 cm to 300 cm
(d) 18 cm to 300 cm
ANSWER: (b)
EXPLANATION: For both the persons when the muscles are relaxed the distant vision will be 200 cm. Since muscles of B are stronger, he will be able to reduce the focal length of the eye more than person B. Hence B may see objects clearly nearer than 25 cm. Hence the option (b).
6. The focal length of a normal eye lens is about
(a) 1 mm
(b) 2 cm
(c) 25 cm
(d) 1 m
ANSWER: (b)
EXPLANATION: The normal eye is when the muscles are relaxed. In this stage, the eye is focused at a faraway distance and the light rays are nearly parallel. The image is formed at the retina which is the focus of the eye lens. The focal length is the distance between the eye lens and the retina which is about 2 cm.
 |
| Diagram for Q-6 |
7. The distance of the eye lens from the retina is x. For a normal eye, the maximum focal length of the eye lens
(a) = x
(b) < x
(c) > x
(d) = 2x.
ANSWER: (a)
EXPLANATION: The maximum focal length of the eye lens is when the muscles are relaxed i.e. when the eye is focused at a far away distance from the eye. In this case, the incoming rays are nearly parallel and they are focused at the retina. Hence the focal length of the eye lens is the distance between the eye lens and the retina which in this case is =x. Hence the option (a).
8. A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m
(a) if he is farsighted
(b) if he is nearsighted
(c) if his vision is normal
(d) in each of these cases.
ANSWER: (d)
EXPLANATION: For a farsighted person, who can not see objects clearly for a distance of 25 cm, the gasses in question will correct his vision for the near distances and he may see the objects from 25 cm to 1m. Without the glasses, he was not able to sufficiently contract the muscles to focus at the retina.
For a nearsighted person who can not see objects clearly at a far away distances, if such glasses are given he may get his distant vision even reduced to 1 m only. Similar may be a case for a person with normal vision. Such glasses may reduce his distant vision from infinity to 1 m only.
Hence the option (d).
9. An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
(a) on f but not on u
(b) on u but not on f
(c) on f as well as u
(d) neither on f nor on u
ANSWER: (c)
EXPLANATION: For a simple microscope, the angular magnification is given as D/f for the normal adjustment and for the object within the focal length it is =1+D/f. Hence the angular magnification depends on both the f and u.
10. To increase the angular magnification of a simple microscope, one should increase
(a) the focal length of the lens
(b) the power of the lens
(c) the aperture of the lens
(d) the object size.
ANSWER: (b)
EXPLANATION: The angular magnification is given as
m = D/f, for normal adjustment and
=1+D/f for the object within f.
→m =D*P or 1+D*P, where P=1/f =power of the lens.
It is clear that m depends upon P, power of the lens.
11. A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5X before his eyes. The angular magnification achieved is
(a) 5
(b) 2.5
(c) 1
(d) 0.2
ANSWER: (c)
EXPLANATION: With the simple microscope of angular magnification 5X, the final image will appear to be at D. Initially the object was also at the near point =D. Let the height of the object = h and the height of the final image = h'.
The lateral magnification
=h'/h=v/u =(-D)/(-D) =1
→h' = h
Now the angular magnification
m =Angle subtended by the final image/the angle subtended by the object.
→m =(h'/D)/(h/D)
→m =h'/h = 1
Hence the option (c).
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Links to the Chapters
Links to the Chapters
CHAPTER- 19 - Optical Instruments
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
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 19 - Optical Instruments
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
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
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Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
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Click here for → Exercise(43-54)
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
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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"
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Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for EXERCISES (Question number 1 to 10)
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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"
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