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

To increase both the resolving power and magnifying power of a telescope

  1. Both the focal length and aperture of the objective has to be increased.

  2. The focal length of the objective has to be increased.

  3. The aperture of the objective has to be increased.

  4. The wavelength of light has to be decreased.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power, $R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of objective $\lambda$ is wavelength of light
magnifying power $m=\dfrac{-f _{0}}{f _{e}}\left ( 1+\dfrac{f _{e}}{D} \right )$
so, decreasing the wavelength of light increases the resolving power and magnifying power of telescope.

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

If accelerating potential increases from $20\ KV$ to $80\ KV$ in an electron microscope, its resolving power $R$ would change to

  1. $\dfrac{R}{4}$
  2. $4R$
  3. $2R$
  4. $\dfrac{R}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{1}{2}mv^{2}= eV$

$mv= \sqrt{2eVm}$

And $\lambda = \dfrac{h}{mV}$

$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV _{1}m}}{\sqrt{eV _{2}m}}$

$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$

$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$

$R\ \propto \dfrac{1}{\lambda}$

so $R$ would change to $2R$.

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

The least resolvable angle by a telescope using objective of aperture 5 m is nearly                ($\lambda = 4000A^{\circ}$)

  1. $\dfrac{1}{50^{\circ}}$
  2. $\dfrac{1}{50}$ minute
  3. $\dfrac{1}{50}$sec
  4. $\dfrac{1}{500}$sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

     $R= \dfrac{9}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{5}{1.22\times 4000\times 10^{-10}}$

  $\Delta \theta = \dfrac{1}{50}sec$

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

The angular resolution of a telescope of 10 cm diameter at a wavelength of 5000Å is of the order of:

  1. 10$^{6}$ rad
  2. $10^{-2}$ rad
  3. $10^{-4}$ rad
  4. $10^{-5}$ rad
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$R= \dfrac{1}{\Delta \theta }= \dfrac{a}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{0.10}{1.22\times 5000\times 10^{-10}}$

$\Delta \theta = 6.1\times 10^{-6}\ rad$

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

If the wavelength of light used is $6000\mathring { A } $. The angular resolution of telescope of objective lens having diameter $10cm$ is ______ rad

  1. $7.52\times { 10 }^{ -6 }$
  2. $6.10\times { 10 }^{ -6 }$
  3. $6.55\times { 10 }^{ -6 }$
  4. $7.32\times { 10 }^{ -6 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Limit of resolution $\sin { \theta  } =\theta =\cfrac { 1.22\lambda  }{ D } $
putting the values

$\theta=\dfrac{1.22\times6000\times10^{-10}}{0.1}$

$\theta=7.32\times10^{-6}$

Option (D) is correct.

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

The ratio of resolving power of telescope, when lights of wavelength $4000\overset{o}{A}$ and $5000\overset{o}{A}$ are used, is _________.

  1. $6 : 5$
  2. $5 : 4$
  3. $4 : 5$
  4. $9 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resolving power(R.P.) $\propto \lambda^{-1}$

Therefore $\dfrac{R.P. _1}{R.P. _2}=\dfrac{\lambda _2}{\lambda _1}$
Given:
$\lambda _1=4000\overset{o}{A}$
$\lambda _2=5000\overset{o}{A}$
Hence $\dfrac{R.P. _1}{R.P. _2}=\dfrac{5000\overset{o}{A}}{4000\overset{o}{A}}$
$\dfrac{R.P. _1}{R.P. _2}=\dfrac{5}{4}$
Therefore the correct option is (B).

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

ASSERTION: Resolving power of telescope is more if the diameter of the objective lens is more.
REASON:Objective lens of large diameter collects more light.

  1. both A and R are correct and R is correct explanation of A

  2. A and R both are correct but R is not correct explanation of A

  3. A is true but R is false

  4. both A and R is false
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,
$ RP=\dfrac { D }{ 1.22  \lambda}$


Hence $R$ is correct but for larger resolution objects making small angle be distinguished or very close objects  should be  distinguished.

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

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil of diameter 3 mm. Approximately what is the maximum distance up to which these dots can be resolved by the eye.

  1. 5 m

  2. 6 m

  3. 1 m

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

$ \dfrac { 1.22\lambda  }{ 3mm } =\dfrac { 1mm }{ d } $


$ or\quad d=\dfrac { 3\times { 10 }^{ -6 } }{ 1.22\times 5\times { 10 }^{ -7 } } =5m$

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

Since the objective lens merely forms an enlarged real image that is viewed by the eyepiece, the overall angular magnification M of the compound microscope is the product of the lateral magnification $ { m } _{ 1 }$ of the objective and the angular magnification $ { M } _{ 2 }$ of the eyepiece. The former is given by
$ { m } _{ 1 }=\dfrac { { S } _{ 1 }^{ ' } }{ { S } _{ 1 } } $
Where $ { S } _{ 1 }and{ S } _{ 1 }^{ ' }$ are the object and image distance for the objective lens. Ordinarily the object is very close to the focus, resulting in an image whose distance from the  objective is much larger than the focal length $ { f } _{ 1 }$. Thus $ { S } _{ 1 }$ is approximately equal to $ { f } _{ 1 }$ and $ { m } _{ 1 }$ =$ -\dfrac { { S } _{ 1 }^{ ' } }{ { f } _{ 1 } } $, approximately. The angular magnification of the eyepiece from $ { M }=-\dfrac { { u }^{ ' } }{ u } =\dfrac { { y }/{ f } }{ { y }/{ 25 } } =\dfrac { 25 }{ f } $ (f in centimeters) is $ { M } _{ 2 }=25cm/{ f } _{ 2 },$ Where $ { f } _{ 2 }$ is the focal length of the eyepiece, considered as a simple lens. Hence the overall magnification M of the compound microscope is, apart from a negative sign, which is customarily ignored,
$ { M }={ m } _{ 1 }{ M } _{ 2 }=\dfrac { \left( 25cm \right) { S } _{ 1 }^{ ' } }{ f } $
1. What is the resolving power of the instrument whose magnifying power is given in the passage?

  1. $ \dfrac { \mu \sin { \theta } }{0 .61\lambda } $
  2. $ \dfrac { \mu \sin { \theta } }{ 1.22\lambda } $
  3. $ \dfrac { \mu \sin { \theta } }{ \lambda } $
  4. $ \dfrac { \sin { \theta } }{ 1.22\lambda } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mentioned instrument is compound microscope and its resolving power is $ R.P=\dfrac { 2\mu \sin { \theta  }  }{ 1.22\lambda  } =\dfrac { \mu \sin { \theta  }  }{ 0.61\lambda  } $


where, $ \mu$ is refractive index of medium, $ \theta$ is the semi-vertical angle of the cone of the rays received by the objective.

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

A person wishes to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between these pillars (resolving power of normal human eye is 1')?

  1. 1 m

  2. 3.2 m

  3. 0.5 m

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

Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.

Hence,$\theta=\dfrac{d}{D}$ 
Hence,$d=\theta D=\dfrac{1}{60}\times \dfrac{\pi}{180}\times 110000=3.2m$

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 eye-piece

  2. increase in focal length of objective

  3. increase in aperture of eye piece

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

Resolving power $= \displaystyle \dfrac{\lambda}{d \lambda}$ plane transmission granting 


Resolving power for telescope

$= \displaystyle \frac{1}{\text{limit of resolution}} = \dfrac{d}{1.22 \lambda} = \dfrac{d _0}{d _1}$

by increasing the aperture of objective resolving power can be increased.

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

The limit of resolution of eye is approximately

  1. $1^0$
  2. $1'$
  3. $1 mm$
  4. $1 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The resolution of the human eye is the smallest object our  eye can see. This is limited by the diffraction limit, which is approximated by the angular size ratio of the object's size versus the distance to the object.
The normal pupil size of a human eye is 4 mm, which sets a minimum angular resolution of the eye  and to able to see the small objects we bring them as close to our eyes as possible, but there is a minimum distance for comfortable viewing which is roughly at 25 cm.But quoted figure for the smallest resolvable size is 0.1 mm, showing that the diffraction limit is a crucial factor in visual resolving power.

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

Aperture of the human eye is 2 mm. Assuming the mean wavelength of light to be 5000 $\overset{o}{A}$, the angular resolution limit of the eye is nearly:

  1. 2 minute

  2. 1 minute

  3. 0.5 minute

  4. 1.5 minute
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the angular limit of resolution of human eye is R then
R = $\displaystyle\frac{1.22\lambda}{a}$ = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}}$ rad
         
      = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}} \times \frac{180}{\pi} \times 60$ minute = 1 minute