Tag: resolving power of optical instruments

Questions Related to resolving power of optical instruments

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

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

  1. Simple microscope

  2. Compound microscope

  3. Astronomical telescope

  4. Binocular

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Correct Option: D
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.

A person wants to see two pillars from a distance of 11 km, separately. The distance between the pillars must be approximately

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

  1. 3.2m

  2. 1m

  3. 0.25 m

  4. 0.5 m

Show answer
Correct Option: A
Explanation:

Resolving power of eye = $ 1' = \dfrac {\pi} {60 \times 180} radians$ (approximately)


Let the separation between pillars $= x m$
Distance $= 11000 m$

=> $\dfrac {x} {11000} = \dfrac {\pi}{60 \times 180}$

=> $ x = 3.2 m$ (approximately)

Answer. A)$ 3.2m $

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

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

  1. 5 cm

  2. 0.5 mm

  3. 5 m

  4. 5 mm

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Correct Option: D
Explanation:

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

An astronomical telescope has a large aperture to

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

  1. reduce spherical aberration

  2. have high resolution

  3. increase span of observation

  4. have low dispersion

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Correct Option: B
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

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.

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

  1. Unresolved

  2. Well resolved

  3. Just resolved

  4. None of these

Show answer
Correct Option: C
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)

To increase the magnification of a telescope 

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

  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

Show answer
Correct Option: A
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.

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

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

  1. $\cfrac {v+f}{f}$

  2. $\cfrac {vf}{v-f}$

  3. $\cfrac {f}{v+f}$

  4. None of these

Show answer
Correct Option: D
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 $u$.
$m=\cfrac{-v}{1}\times\cfrac{v-f}{fv}$
$m=\cfrac{f-v}{f}$

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 ; -

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

  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

Show answer
Correct Option: A
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

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

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

  1. $2 R$

  2. $\dfrac{R}{2}$

  3. $4R$

  4. $3R$

Show answer
Correct Option: A
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, $R' = 2R$

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.

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

  1. $32 cm$

  2. $18 cm$

  3. $25 cm$

  4. $36 cm$

Show answer
Correct Option: A
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, $f _o=8f _e=8\times 4=32\ cm$