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

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

The optical instrument which is used in every cricket match is.:

  1. Simple microscope

  2. Compound microscope

  3. Astronomical telescope

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

Binocular is used in every cricket match for the purpose of zooming.
Because, binoculars are a pair of mirror symmetrical telescopes that allow a user to view distant objects using both eyes. To allow both eyes to view distant objects symmetrically, the binoculars require two separate telescopes, one for each eye, held together in the device called binoculars allowing binocular vision. The telescopes are mounted symmetrically side-by-side and aligned to point accurately in the same direction, providing the user with undistorted vision of distant objects. Unlike a monocular (telescope) which only makes use of one telescope to view objects, binoculars are able to provide the user with three-dimensional viewing of distant objects, whilst promoting visual clarity and acuity.

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

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

An astronomical telescope has a large aperture to

  1. reduce spherical aberration

  2. have high resolution

  3. increase span of observation

  4. have low dispersion
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The aperture of an astronomical telescope is defined as the diameter of the objective lens.
In telescopes since, the stars are very far away from away us and emit light internsities, we need a large aperture to increase the amount of light entering the telescope thereby increasing the resolution

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

The limit of resolution of an optical instrument is the smallest angle that two points on an object have to subtend at the eye so that they are.

  1. Unresolved

  2. Well resolved

  3. Just resolved

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

Limit of resolution of an optical instrument is the minimum angle that two points on an object have to subtend at the eye so that they are just resolved. (C)

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

To increase the magnification of a telescope 

  1. the objective lens should be of large focal length and eyepiece should be of small focal length.

  2. the objective and eyepiece both should be of large focal length.

  3. both the objective and eyepiece should be of smaller focal lengths

  4. the objective should be of small focal length and eyepiece should be of large focal length
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$m=\dfrac { { f } _{ o } }{ { f } _{ e } } $
from the above relation we can see that magnification is directly proportional to the focal length of objective lens.
Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Magnification of an object ($m$), is equal to

  1. $\cfrac {v+f}{f}$
  2. $\cfrac {vf}{v-f}$
  3. $\cfrac {f}{v+f}$

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

we know,mirror formula
$\cfrac {1}{f}=\cfrac{1}{v}+\cfrac{1}{u}, magnification(m)=\cfrac{-v}{u}$
$\cfrac{1}{u}=\cfrac{1}{f}-\cfrac{1}{v}$
$\cfrac{1}{u}=\cfrac{v-f}{fv}$
${u}=\cfrac{fv}{v-f}$
$(m)=\cfrac{-v}{u}$,substituting $u$.
$m=\cfrac{-v}{1}\times\cfrac{v-f}{fv}$
$m=\cfrac{f-v}{f}$

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

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star ; -

  1. $305 \times 10^{-9} $ radian
  2. $152.5 \times 10^{-9} $ radian
  3. $610 \times 10^{-9} $ radian
  4. $457.5 \times 10^{-9} $ radian
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Limit of resolution of telescope = $\dfrac{1.22 \lambda}{D}$
$\theta = \dfrac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}} = 305 \times 10^{-9}$ radian

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

In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power of the microscope will change from R to

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

Electron microscope is a microscope that can magnify very small details with high resolving power due to the use of electrons as the source of illumination. Since the wavelength of electrons are 100,000 times shorter than visible light the electron microscopes have greater resolving power
We have the Abbe's formula as resolution limit $d=\dfrac{0.61\lambda}{NA}$(NA is the numerical aperture)


The resolving power increases when d, the minimum distance that can be seen between two points in the image, decreases. Thus, according to the formula the resolving power is inversely proportional to the wavelength.

Resolving Power  $\propto \dfrac{1}{\lambda}$

A higher voltage will give the electrons a higher speed. Thus the electrons will have a smaller de Broglie wavelength according to the equation,  $\lambda=h/mv$

$\lambda\propto\dfrac{1}{\sqrt V}$

Thus we get Resolving power $\propto \sqrt{V}$

$ \implies\dfrac{R}{R'} = \sqrt{\dfrac{20}{80}} $

Thus, $R' = 2R$

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

An astronomical telescope, consists of two thin lenses set $36 cm$ a part and has a magnifying power $8$. Calculate the focal length of the lenses.

  1. $32 cm$
  2. $18 cm$
  3. $25 cm$
  4. $36 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $f _o+f _e=36 cm$
$M=-8$ (magnifying power is negative)
Now, $M=\cfrac {f _o}{f _e}$
$\therefore -8=-8=-\cfrac {f _o}{f _e}$
$\Rightarrow f _o=8f _e$

From the equations, we have
$8f _e+f _e=36$ or $f _e=4\ cm$

Again, $f _o=8f _e=8\times 4=32\ cm$

Multiple choice polarisation of light polarisation wave optics optics physics

The diameter of an objective of a telescope, which can just resolve two stars situated at an angular displacement of ${10^{ - 4}}$ degree, should be $\left( {\lambda  = 5000\,{A^0}} \right)$ 

  1. 35 mm

  2. 35 cm

  3. 35 m

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

$\Delta \theta  = \left( {\dfrac{{1.22\lambda }}{d}} \right)$
$d = \left( {\dfrac{{1.22 \times \lambda }}{{\Delta \theta }}} \right)$
$ \Rightarrow \,\,{10^{ - 4}}\, \to \,\dfrac{\lambda }{{180}} \times {10^{ - 4}}$
$ = 1.74 \times {10^{ - 6}}\,rad.$
$d = \left( {\dfrac{{1.22 \times 5000 \times {{10}^{ - 10}}}}{{1.74 \times {{10}^{ - 6}}}}} \right)$
$ = 0.35\,m\,\,or\,\,35\,cm$