Physics

Optical Instruments and Human Eye

428 Questions

Optical instruments and the human eye explore how lenses and mirrors are used to magnify and resolve images. Key topics include the magnifying power of astronomical telescopes, compound microscope configurations, and hyperfocal distance calculations. These physics concepts are vital for general science competitive exams.

Telescope magnifying powerCompound microscope lensesHyperfocal distance calculationHuman eye resolutionSpherical mirror magnification

Optical Instruments and Human Eye Questions

Multiple choice

Should you start on high power objective on a microscope? 

  1. yes

  2. no
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

You should always start focusing with the lowest power objective lens because it has a wider field of view, making it easier to locate the specimen. Starting on high power increases the risk of crashing the lens into the slide and makes focusing very difficult. Once the specimen is centered and focused, you can safely rotate to higher powers.

Multiple choice

What is the total magnification of a microscope if the eyepiece is 10x and the other is 40x? 

  1. 50x

  2. 400x

  3. 4000x

  4. 40x
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the total magnification of a microscope, you multiply the magnification of the eyepiece by the magnification of the objective lens. Multiplying 10 by 40 gives a total magnification of 400x. Adding the magnifications together is a common mistake that leads to the incorrect answer of 50x.

Multiple choice

With a telescope, you can see

  1. the same mumber of stars you can see with just your eyes.

  2. more stars than than you can see with just your eyes.

  3. fewer stars than you can see with just your eyes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Telescopes gather light from objects that are too faint or too far away for the naked eye to detect. This allows observers to see many more stars than are visible without assistance.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

How can resolving power of the instrument be increased?

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The ability of an optical instruments to show the images of two adjacent point objects as separate is called :

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Two lenses of focal lengths $+ 100 cm$ and $+ 5 cm$ are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at $\displaystyle \infty $)

  1. $95 cm$
  2. $100 cm$
  3. $105 cm$
  4. $500 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

In an astronomical microscope, the focal length of the objective is made :

  1. shorter than that of the eye piece

  2. greater than that of the eye piece

  3. half of the eye piece

  4. equal to that of the eye piece
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an Astronomical telescope, the objective lens has a greater radius than the eyepiece.

Thus the objective lens has a greater focal length than the eyepiece.

Option B is correct.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

If the apertature of a telescope is decreased the resolving power will

  1. increases

  2. decreases

  3. remain same

  4. zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resolving power of a telescope=$\dfrac{a}{1.22\lambda}$

where $a$ is the aperture of the telescope.
Thus resolving power$\propto $ aperture.
Hence, if the aperture of telescope decreases, the resolving power decreases.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The resolving power of a telescope depends on :

  1. length of telescope

  2. focal length of objective

  3. diameter of the objective

  4. focal length of eyepiece
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Resolving power of telescope $R=\dfrac{1}{\Delta \theta}=\dfrac{a}{1.22 \lambda}$
where, $\Delta \theta$ is angular separation between two objects.
            $a$ is the diameter of the objective.
            $\lambda$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The diameter of the objective of a telescope is $a$, its magnifying power is $m$ and wavelength of light $\lambda $ . The resolving power of the telescope is :

  1. $\dfrac{(1.22\lambda )}{a}$
  2. $\dfrac{1.22a}{\lambda} $
  3. $\lambda (1.22a)$
  4. $\dfrac {a} {1.22\lambda} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power of telescope:
$R=\dfrac{1}{ \theta}=\dfrac{a}{1.22 \lambda}$


where $\theta$ is angular resolution, a is diameter of the objective and $\lambda$ is wavelength of light.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The resolving power of human eye is :

  1. $\approx 1'$
  2. $\approx 1^{0}$
  3. $\approx 10"$
  4. $\approx 5"$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The normal pupil size of a human eye is 4 mm.which sets a minimum resolution approximately 1' to 2'.we want to pull small objects as close to our eyes as possible to be able to see them, but there is a minimum distance of comfortable viewing which is roughly at 25 cm. Hence, correct option is A.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

An electron microscope is superior to an optical microscope in terms of:

  1. having better resolving power

  2. being easy to handle

  3. low cost

  4. quickness of observation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The biggest advantage of an electron microscope over optical microscope is that they have a higher resolution and are therefore capable of a higher magnification ( up to $2 \ million$ times ). 

However, optical microscopes show a useful magnification up to $1000-2000 $ times. This is a limit imposed by the wavelength of light. Electron microscopes, therefore, allow for the visualization of structures that would normally be not visible by optical microscopy.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Resolving power of a telescope increases with :

  1. increase in focal length of eyepiece

  2. increase in focal length of objective

  3. increase in aperture of eyepiece

  4. increase in aperture of objective
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power of a telescope:
$R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of the objective
so, $R$ increases when a is increased and $a$  increases when aperture of objective is increased