Tag: diffraction

Questions Related to 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?

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

Show answer
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

Show answer
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

Show answer
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

Show answer
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. $1 cm$

Show answer
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.

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:

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

  1. 2 minute

  2. 1 minute

  3. 0.5 minute

  4. 1.5 minute

Show answer
Correct Option: B
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
  

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 $)

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

  1. $8$

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

  3. $0.45$

  4. None of these

Show answer
Correct Option: A
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 $8$.

A photographer changes the aperture of his camera and reduces it to half of the original aperture. The exposure time now should be:

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

  1. same as before

  2. double than before

  3. four times than before

  4. half than before

Show answer
Correct Option: D
Explanation:

Answer is D.

The amount of light captured while taking a photo is known as the exposure, and it's affected by three things - the shutter speed, the aperture diameter, and the ISO or film speed. 
Aperture is measured using the "f-number", sometimes called the "f-stop", which describes the diameter of the aperture. A lower f-number relates to a wider aperture (one that lets in more light), while a higher f-number means a narrower aperture (less light).
Because of the way f-numbers are calculated, a stop doesn't relate to a doubling or halving of the value, but to a multiplying or dividing by 1.41 (the square root of 2). For example, going from f/2.8 to f/4 is a decrease of 1 stop because 4 = 2.8 * 1.41. Changing from f/16 to f/11 is an increase of 1 stop because 11 = 16 / 1.41.
Hence, when the aperture of the camera reduces to half of the original aperture, the exposure time should be half than before.