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Physics M.C.Q (1 Marks)

Optical Instruments · Jharkhand Board (JAC) STD 12 Science (Jharkhand - English Medium). 12 questions.

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M.C.Q (1 Marks)

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12 questions · auto-graded multiple-choice test.

MCQ 11 Mark
The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is:
  • ✓
    Always strained in looking at an object.
  • B
    Strained for objects at large distances only.
  • C
    Strained for objects at short distances only.
  • D
    Unstrained for all distances.
Answer
Correct option: A.
Always strained in looking at an object.
Explanation:
The maximum focal length of a normal eye is equal to the distance of the lens from the retina. In case it is greater than the distance, the eye will be strained while focusing the objects on the retina that is at a fixed distance from the eye lens.
 
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MCQ 21 Mark
A person A can clearly see objects between 25cm and 200cm. Which of the following may represent the range of clear vision for a person B having muscles stronger than A, but all other parameters of eye identical to that of A?
  • A
    25cm to 200cm.
  • ✓
    18cm to 200cm.
  • C
    25cm to 300cm.
  • D
    18cm to 300cm.
Answer
Correct option: B.
18cm to 200cm.
Explanation:
Person B has stronger ciliary muscles than person A. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person A. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.
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MCQ 31 Mark
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
  • ✓
    1
  • D
    0.2
Answer
Correct option: C.
1
Explanation:
We have,
h = Object height
u = Object distance = 25cm
D = Near point = 25cm
Now,
$\text{m}=\frac{\frac{\text{h}}{\text{u}}}{\frac{\text{h}}{\text{D}}}$
$\Rightarrow\text{m}=\frac{\frac{\text{h}}{25}}{\frac{\text{h}}{25}}$
$\Rightarrow \text{m}=1$
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MCQ 41 Mark
The size of an object as perceived by an eye depends primarily on:
  • A
    Actual size of the object.
  • B
    Distance of the object from the eye.
  • C
    Aperture of the pupil.
  • ✓
    Size of the image formed on the retina.
Answer
Correct option: D.
Size of the image formed on the retina.
Explanation:
An eye consists of a lens that works on the principle on which a glass lens works. It forms the image on the screen called retina. The magnification, in this case, depends on the ratio of the image to the object height.
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MCQ 51 Mark
To increase the angular magnification of a simple microscope, one should increase:
  • A
    The focal length of the lens.
  • ✓
    The power of the lens.
  • C
    The aperture of the lens.
  • D
    The object size.
Answer
Correct option: B.
The power of the lens.
Explanation:
For a simple microscope in normal adjustment, the object is placed at a distance equal to f (the ​focal length) from the lens, And the angular magnification is given by the relation
$\text{m}=\frac{\text{D}}{\text{f}}$
for $\text{u}<\text{f},\text{m}=\frac{\text{D}}{\text{f}}+1$
power of lens $=\frac{1}{\text{f}}$
Angular magnification depends on power.
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MCQ 61 Mark
The muscles of a normal eye are least strained when the eye is focused on an object:
  • ✓
    Far away from the eye.
  • B
    Very close to the eye.
  • C
    At about 25cm from the eye.
  • D
    At about 1m from the eye.
Answer
Correct option: A.
Far away from the eye.
Explanation:
A normal eye can see from 25cm to infinity, it faces least difficulty and strain focusing on the object as far as it could be.
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MCQ 71 Mark
When objects at different distances are seen by the eye, 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.
  • ✓
    The image-distance from the eye-lens.
Answer
Correct option: D.
The image-distance from the eye-lens.
Explanation:
In the human eye, the image is formed on the retina, which is at a fixed distance from the eye lens.
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MCQ 81 Mark
The distance of the eye-lens from the retina is x. For a normal eye, the maximum focal length of the eye-lens:
  • ✓
    = x.
  • B
    < x.
  • C
    > x.
  • D
    = 2x.
Answer
Correct option: A.
= x.
Explanantion:
For a normal eye, we have:
Far point at which the object can be placed, $\text{u}=\infty$
Distance between the eye lens and the retina, $\text{v}=\text{x}$
Thus, we have:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{\text{x}}-\frac{1}{\infty}$
$\Rightarrow\text{f}=\text{x}$
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MCQ 91 Mark
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.
  • ✓
    On f as well as u.
  • D
    Neither on f nor on u.
Answer
Correct option: C.
On f as well as u.
Explanation:
​The angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object on an unaided eye.
In a simple microscope,
$\text{m}=\frac{\frac{\text{h}}{\text{x}}}{\frac{\text{h}}{\text{D}}}$
Here,
u = Object distance from the lems
D = Image distance form the lens
h = Height of the object
In normal adjustment, the object is placed at a distance equal to focal length (f) from the lens and then magnification is given by m
$=\frac{\text{D}}{\text{f}}$
for $\text{u}<\text{f},\text{ m}=\frac{\text{D}}{\text{f}}+1$
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MCQ 101 Mark
The focal length of a normal eye-lens is about:
  • A
    1mm.
  • ✓
    2cm.
  • C
    25cm.
  • D
    1m.
Answer
Correct option: B.
2cm.
Explanation:
Given:
Near point of the human eye, u = -25cm
Distance between the retina and the eye lens, v = 2cm (approximately)
thus, we have the focal length, f.
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{1}{2}-\frac{1}{-25}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{27}{50}$
$\Rightarrow\text{x}\cong2\text{cm}$
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MCQ 111 Mark
A man wearing glasses of focal length +1m cannot clearly see beyond 1m:
  • A
    If he is farsighted.
  • B
    If he is nearsighted.
  • C
    If his vision is normal.
  • ✓
    In each of these cases.
Answer
Correct option: D.
In each of these cases.
Explanation:
The man is wearing glasses of positive power (converging lens). Hence, he cannot see nearby objects clearly. In other words, he is farsighted. Since he cannot see beyond 1m, he is nearsighted. If a person with normal vision wears glasses of focal length +1m, then the person will not be able to see beyond 1m.
 
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MCQ 121 Mark
A normal eye is not able to see objects closer than 25cm because:
  • A
    The focal length of the eye is 25cm.
  • B
    The distance of the retina from the eye-lens is 25cm.
  • C
    The eye is not able to decrease the distance between the eye-lens and the retina beyond a limit.
  • ✓
    The eye is not able to decrease the focal length beyond a limit.
Answer
Correct option: D.
The eye is not able to decrease the focal length beyond a limit.
Explanation:
The ciliary muscles adjust the focal length to form an image on the retina, but the muscles cannot be strained beyond a limit. Hence, if the object is brought too close to the eye, the focal length cannot be adjusted to form the image on the retina.
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