Tag: resolving power of optical instruments

Questions Related to resolving power of optical instruments

 A beam of plane polarised light falls on a polarizer which rotates about axis of ray with angular velocity $\omega $. The energy passing through polrizer in one revolution if incident power is P is :

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

  1. $ \dfrac {\pi P} {2\omega} $

  2. $ \dfrac {\pi P} {\omega} $

  3. $ \dfrac {2 \pi P} {\omega} $

  4. $ \dfrac {3 \pi P} {2\omega} $

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Correct Option: A

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.

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

  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

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Correct Option: B
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.

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.

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

  1. 5 m

  2. 6 m

  3. 1 m

  4. 4 m

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Correct Option: A
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$

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?

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

  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 } $

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Correct Option: A
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.

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')?

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

  1. 1 m

  2. 3.2 m

  3. 0.5 m

  4. 5 m

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Correct Option: B
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,$d=\theta D=\dfrac{1}{60}\times \dfrac{\pi}{180}\times 110000=3.2m$

The resolving power of an electron microscope operated at 16 kV is R. The resolving power of the electron microscope when operated at 4 kV is

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

  1. R/4

  2. R/2

  3. 4R

  4. 2R

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Correct Option: B
Explanation:

$\sqrt{\dfrac{16}{4}}=\dfrac{R}{x}$


$x=\dfrac{R}{2}$

option $B$ is correct 

Wavelength of light used in an optical instrument are $\lambda _1 = 4000 A^o  and \lambda _2 = 5000 A^0$, then ratio of their respective resolving powers (corresponding to $\lambda _1   \ and  \ \lambda _2$) is

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

  1. 16:25

  2. 9:1

  3. 4:5

  4. 5:4

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Correct Option: D
Explanation:

Resolving power of an optical instrument $\displaystyle \propto \dfrac {1}{\lambda}$


$\displaystyle \dfrac{\text{Resolving  power  at} \lambda _1} {\text{Resolving  power  at} \lambda _2}=\dfrac {\lambda _1}{\lambda _2}$

$\displaystyle \left [\text{Limit of resolution}  \propto \dfrac{1}{\text{resolving  power}}\right]$

$\therefore$  Ratio  of  resolving  power $= \displaystyle \dfrac {5000}{4000} = \dfrac {5}{4} = 5 : 4$

Resolving power of a telescope increases with

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

  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

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Correct Option: D
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.

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. increases span of observation

  4. have low dispersion

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Correct Option: B
Explanation:

Larger aperture help in high (increased) resolution of telescope and large amount of light collection.

The limit of resolution of eye is approximately

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

  1. $1^0$

  2. $1'$

  3. $1 mm$

  4. $1 cm$

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Correct Option: B
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.