Download App

Physics M.C.Q (1 Marks)

Optical Instruments · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 16 questions.

Questions

M.C.Q (1 Marks)

Take a timed test

16 questions · 1 auto-graded MCQ + 15 self-marked written.

Question 11 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?
  1. 25cm to 200cm.
  2. 18cm to 200cm.
  3. 25cm to 300cm.
  4. 18cm to 300cm.
Answer
  1. 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.
View full question & answer→
Question 21 Mark
The muscles of a normal eye are least strained when the eye is focused on an object:
  1. Far away from the eye.
  2. Very close to the eye.
  3. At about 25cm from the eye.
  4. At about 1m from the eye.
Answer
  1. 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.
View full question & answer→
Question 31 Mark
When objects at different distances are seen by the eye, which of the following remain constant?
  1. The focal length of the eye-lens.
  2. The object-distance from the eye-lens.
  3. The radii of curvature of the eye-lens.
  4. The image-distance from the eye-lens.
Answer
  1. 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.
View full question & answer→
Question 41 Mark
A man wearing glasses of focal length +1m cannot clearly see beyond 1m:
  1. If he is farsighted.
  2. If he is nearsighted.
  3. If his vision is normal.
  4. In each of these cases.
Answer
  1. 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.
View full question & answer→
Question 51 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:
  1. 5
  2. 2.5
  3. 1
  4. 0.2
Answer
  1. 1
Explanation:
We have,
h = Object height
u = Object distance = 25cm
D = Near point = 25cm
Now,
$\text{m}=\frac{\frac{\text{h}}{\text{u}}}{\frac{\text{h}}{\text{D}}}$
$\Rightarrow\text{m}=\frac{\frac{\text{h}}{25}}{\frac{\text{h}}{25}}$
$\Rightarrow \text{m}=1$
View full question & answer→
MCQ 61 Mark
The focal length of the objective of a compound microscope if $f_o$ and its distance from the eyepiece is $L.$ The object is placed at a distance $u$ from the objective. For proper working of the instrument:
  • A
    $L < u$
  • ✓
    $L > u$
  • C
    $f_o < L < 2f_o$
  • D
    $L > 2f_o$
Answer
Correct option: B.
$L > u$
In a compound microscope, the objective lens of a short focal length, $f_o$ is used.
The focal length of the objective lens is less than the focal length of the eyepiece, $f_{​e},.$
The object is placed at a distance slightly greater than its focal length.
The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length.
This image acts as its object and the final image forms in between length, $L$ of the microscope.
$\therefore L > f_o​ + f_{e }> 2f_o​$
Also, $L > u$
View full question & answer→
Question 71 Mark
To increase the angular magnification of a simple microscope, one should increase:
  1. The focal length of the lens.
  2. The power of the lens.
  3. The aperture of the lens.
  4. The object size.
Answer
  1. 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.
View full question & answer→
Question 81 Mark
Mark the correct options:
  1. If the far point goes ahead, the power of the divergent lens should be reduced.
  2. If the near point goes ahead, the power of the convergent lens should be reduced.
  3. If the far point is 1 m away from the eye, divergent lens should be used.
  4. If the near point is 1 m away from the eye, divergent lens should be used.
Answer
  1. If the far point goes ahead, the power of the divergent lens should be reduced.
  1. If the far point is 1m away from the eye, the divergent lens should be used.
Explanation:
As the far point (x) is shifited ahead, the focal length (f) will be increased.
thus. we have
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}-\frac{1}{-\infty}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}$
$\Rightarrow\text{f}=-\text{x}$
As power (p) is equal to reciprocal of the focal length, it will be reduced, Also, because the focal length is negative, the lens used will be divergent when the far point is 1m away.
View full question & answer→
Question 91 Mark
An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends:
  1. On f but not on u.
  2. On u but not on f.
  3. On f as well as u.
  4. Neither on f nor on u.
Answer
  1. 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 $\text{u}<\text{f},\text{ m}=\frac{\text{D}}{\text{f}}+1$
View full question & answer→
Question 101 Mark
The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is:
  1. Always strained in looking at an object.
  2. Strained for objects at large distances only.
  3. Strained for objects at short distances only.
  4. Unstrained for all distances.
Answer
  1. Aways 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.
View full question & answer→
Question 111 Mark
When we see an object, the image formed on the retina is:
  1. Real.
  2. Virtual.
  3. Erect.
  4. Inverted.
Answer
  1. Ral.
  2. Ierted.
Explanation:
The retina acts as a screen in the eye; only real, inverted images can be obtained on the screen.
View full question & answer→
Question 121 Mark
The size of an object as perceived by an eye depends primarily on:
  1. Actual size of the object.
  2. Distance of the object from the eye.
  3. Aperture of the pupil.
  4. Size of the image formed on the retina.
Answer
  1. 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.
View full question & answer→
Question 131 Mark
 A normal eye is not able to see objects closer than 25cm because:
  1. The focal length of the eye is 25cm.
  2. The distance of the retina from the eye-lens is 25cm.
  3. The eye is not able to decrease the distance between the eye-lens and the retina beyond a limit.
  4. The eye is not able to decrease the focal length beyond a limit.
Answer
  1. 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.
View full question & answer→
Question 141 Mark
In which of the following the final image is erect?
  1. Simple microscope.
  2. Compound microscope.
  3. Astronomical telescope.
  4. Galilean telescope.
Answer
  1. Simple microscope.
  2. Galilean telescope.
Explanation:
In a simple microscope, the image formed is virtual and above the axis on the same side of the object. Similarly, in a Galilean telescope, the image formed is virtual, erect and magnified and between the objective lens and the eyepiece. A compound microscope and an astronomical telescope produce inverted images.
View full question & answer→
Question 151 Mark
The distance of the eye-lens from the retina is x. For a normal eye, the maximum focal length of the eye-lens:
  1. = x.
  2. < x.
  3. > x.
  4. = 2x.
Answer
  1. = 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:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{\text{x}}-\frac{1}{\infty}$
$\Rightarrow\text{f}=\text{x}$
View full question & answer→
Question 161 Mark
The focal length of a normal eye-lens is about:
  1. 1mm.
  2. 2cm.
  3. 25cm.
  4. 1m.
Answer
  1. 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.
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{1}{2}-\frac{1}{-25}$
$\Rightarrow\frac{1}{\text{f}}\cong\frac{27}{50}$
$\Rightarrow\text{x}\cong2\text{cm}$
View full question & answer→
Generate Paper FreeAll Optical Instruments topics