Physics

Optical Instruments and Human Eye

416 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

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
  

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

The magnifying power of an astronomical telescope is $8$, then the ratio of the focal length of the objective to the focal length of the eyepiece is : (final image is at $\displaystyle \infty $)

  1. $8$
  2. $\displaystyle \frac { 1 }{ 8 } $
  3. $0.45$

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

magnifying power of telescope  (M P)$=8$
The magnifying power of an astronomical telescope is $=\cfrac{focal  \ length \ of \ object (-f _0)  }{focal \ length \ of \ eyepiece(f _e)}=8$
hence, the ratio of the focal length of the objective to the focal length of the eyepiece is $8$.

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 diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wave length of light =500 nm]

  1. 10 m

  2. 5 m

  3. 15 m

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

We have, $\dfrac{y}{D}\ge 1.22 \dfrac{\lambda}{d}$


$\Rightarrow D \le \dfrac{yd}{(1.22)\lambda} = \dfrac{10^{-3}\times 3\times 10^{-3}}{(1.22)\times 5\times 10^{-7}}$

$=\dfrac{30}{6.1}\approx 5m$

$\therefore  D _{max} = 5m$

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

An astronaut is looking down on earth's surface from a space shuttle an altitude of 400 km Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of the size of about :

  1. 0.5m

  2. 5m

  3. 50m

  4. 500m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The resolving power of an instrument is given by the formula, $R.P=1.22\times \frac{\lambda D}{d}$
Here, d is aperture of the instruments, D is distance of satellite from the earth. Here eye is the optical instruments.
$\displaystyle R.P=\frac{1.22\times 500\times 10^{-9}}{5\times 10^{-3}}\times 400\times 1000$
          $\displaystyle =1.22\times \frac{10^{-2}}{10^{-3}}\times 4 = 1.22\times 40=50 m$

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

Assertion: The resolving power of a telescope is more if the diameter of the objective lens is more.
Reason: Objective lens of large diameter collects more light.

  1. Both assertion and reason are true but the reason is the correct explanation of assertion

  2. Both assertion and reason are true but the reason is not the correct explanation of assertion

  3. Assertion is true but reason is false

  4. Both the assertion and reason are false

  5. Reason is true but assertion is false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resolving power of a telescope increases as diameter of objective lens increases.
Resolving Power =D1.22λD1.22λ
where D is diameter of objective and λλ is wavelength of light used.

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

The limit of resolution of microscope, if the numerical aperture of microscope is 0.12, and the wavelength of light used is 600 nm, is 

  1. 0.3$\mu $m
  2. 1.2 $\mu $m
  3. 2.5$\mu $m
  4. 3$\mu $m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a microscope, the limit of resolution is given by,
 $X =  \dfrac{\lambda}{2A} $
where $ \lambda $ is the wavelength of light used, and A is the numerical aperture.
Hence, substituting the values,  $X =  \dfrac{600}{2 \times 0.12} $,
which gives, $X= 2.5  \mu m $

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

A telescope has an objective lens of 10 cm diameter and is situated at a distance of $1km$ for two objects. The minimum distance between these two objects, which can be resolved by the telesope, when the mean wavelength of light is 5000Å is of the order of

  1. 5 cm

  2. 0.5 mm

  3. 5 m

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

Resolution power $= \dfrac {d\lambda}{D} = \dfrac {1000 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = 5mm$