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 digital painting tux paint physics

What is the maximum size to which you magnify the picture?

  1. $2x$
  2. $4x$
  3. $6x$
  4. $8x$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Maximum size to which you magnify the picture is 8x.

Minimum size to which you magnify the picture is 2x
Magnify the picture is the process of enlarging the apparent size like zoom in the picture.

Multiple choice physics motion and measurement measuring length measurement of small and large distances measurement of distance

Which of the following is the best method to measure microscopic distances?

  1. Optical microscope

  2. Meter scale

  3. Screw gauge

  4. Diffraction pattern
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Optical microscopes are enabled with objective fitted with lenses with different magnifications. This helps to magnify a small objects and so necessary measurements.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A reflecting telescope has a large mirror for its objective with radius of curvature equal to 80 cm. The magnifying power of this telescope if eye piece used has a focal length of 1.6 cm is:

  1. 100

  2. 50

  3. 25

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The focal length of objective mirror $f _0 = \dfrac{R}{2} = \dfrac{80}{2} = 40 \,cm$
and focal length of eye piece = 1.6 cm
$\therefore $ magnifying power, $m = \dfrac{f _0}{f _e} = \dfrac{40}{1.6 } = 25$

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A telescope is used to resolve two stars separated by $4.6 \, \times \, 10^{-6}$ rad. If the wavelength of light used is 5460 $\overset{0}{A}$, what should be the aperture of the objective of the telescope? 

  1. 0.1488 m

  2. 0.567 m

  3. 1 m

  4. 2 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Aperature of the telescope, $D = \dfrac{1.22\lambda}{d\theta}$ 


Here, $\lambda = 5460\ \overset{0}{A} = 5460 \times 10^{-10}m$


$ d\theta= 4.6 \times10^{-6}\ rad$

$\therefore D = \dfrac{1.22\times5460\times10^{-10}}{4.6\times 10^{-6}}$
$ = 0.1488m $

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

In an astronomical telescope of refracting type:

  1. Eyepiece has greater focal length

  2. Objective has greater focal length

  3. Objective and eyepiece have equal focal length

  4. Eyepiece has greater aperture than the objective
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

An astronomical telescope of refracting type has objective of greater focal length to have more surface area for incoming light from celestial bodies. 

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

Four lenses of focal length $+15cm,+20cm,+150cm$ and $+250cm$ are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be

  1. $+15cm$
  2. $+20cm$
  3. $+150cm$
  4. $+250cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In astronomical telescope, magnification is inversely proportional to focal length of eye piece. Hence, it will be minimum i.e. +15cm as in question.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The objective of a telescope A has diameter 3 times that of the objective of telescope B, How much greater amount of light is gathered by A as compared to B?

  1. $\dfrac { 1 }{ 9 } $
  2. $\dfrac { 1 }{ 3 } $

  3. 3

  4. 9
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The amount of light gathered by a telescope objective is proportional to its area, which scales with the square of its diameter (A proportional to D^2). Since the diameter of telescope A is 3 times that of telescope B, the light gathering power is 3^2 = 9 times greater.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

For an electron microscope, which of the following is false ?

  1. It uses magnetic lens to converge electron beam.

  2. Its resolving power is directly proportional to accelerating potential of electron.

  3. Its resolving power is inversely proportional to wavelength of electrons
  4. Magnification attained with the help of of it of the order of $10^6$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an electron microscope: (A) is correct - magnetic lenses are used to focus the electron beam. (C) is correct - resolving power is inversely proportional to electron wavelength (R ∝ 1/λ). (D) is correct - electron microscopes can achieve magnifications of ~10⁶. (B) is FALSE - resolving power is NOT directly proportional to accelerating potential; it increases with potential because higher potential reduces electron wavelength (λ = h/√(2meV)), but this is an inverse relationship, not direct proportionality.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

 If an object subtend angle of $2^o$ at eye when seen through telescope having objective and eyepiece of focal length $f = 60\, cm $and $f = 5\, cm$ respectively than angle subtend by image at eye piece will be 

  1. $16^o$
  2. $50^o$
  3. $10^o$
  4. $24^o$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The magnifying power of a telescope in normal adjustment is given by m = f_o / f_e = 60 / 5 = 12. The angle subtended by the image at the eyepiece is equal to the magnification times the angle subtended by the object at the objective, so beta = m * alpha = 12 * 2 degrees = 24 degrees.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A telescope has an objective of focal length $50 cm$ and an eye-piece of focal length $5 m$ THe least distance of distinct vision is $25 cm$. The telescope is focused for distinct vission a scale $200 cm$ away from the objective.The separation between the two lenses is nearly 

  1. $71 cm$
  2. $61 cm$
  3. $81 cm$
  4. $51 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to : ( final image is at $\infty$)

  1. The length of the telescope

  2. Half the length of the telescope

  3. Double the length of the telescope

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The astronomical telescope makes use of two positive lenses: the objective, which forms the image of a distant object at its focal length, and the eyepiece, which acts as a simple magnifier with which to view the image formed by the objective. Its length is equal to the sum of the focal lengths of the objective and eyepiece, and its angular magnification is -fo/fe , giving an inverted image.
Hence, the sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to the length of the telescope.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eyepiece, find the separation between the lenses required for distinct vision

  1. 32.4 cm

  2. 42.3 cm

  3. 24.3 cm

  4. 30.24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f _{0}=20cm$


$f _{e}=5cm$

$V _{e}=30cm = D$

$L=?$

$L _{D}=f _{0}+ \dfrac{Df _e}{D+f _e}$ $=20+\dfrac{5\times 30}{35}=24.3 cm$

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment.Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective,then identify the correct choice

  1. eye piece should be displaced by 2 cm away from the objective

  2. eye piece should be displaced by 2 cm towards the objective

  3. eye piece should be displaced by 5 cm towards the objective

  4. eye piece should be displaced by 5 cm away from the objective
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know, $ \dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1} {u} $


We seeing heavenly body,  $u =  \infty$
$v = f = 1 m$

When seeing 21 m far
$u = - 21 m $
$f = 1 m$
$v = \dfrac {f u} {(f+u)} = 21 / 20 = 1.05 m$

So, eye piece need to move $1.05-1 = 0.05 m$ further away from objective

Answer. D) eye piece should be displaced by $5 cm$ away from the objective