Tag: resolution of optical instruments

Questions Related to resolution of optical instruments

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.

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

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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

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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

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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.

Wavelength of light used in an optical instrument  are $\lambda _1 = 4000 \mathring { A } $ and $ \lambda _2 = 5000 \mathring { A} $ 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:
We know, 
$ Resolving\ Power\ \propto \dfrac{1}{\lambda} $

Given, 
$ \lambda _1 = 4000 \mathring{A} $
$ \lambda _2 = 5000 \mathring{A} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{ \lambda _2}{ \lambda _1} $

$ \Rightarrow \dfrac{RP _{\lambda _1}}{RP _{\lambda _2}} = \dfrac{5000}{4000} = \dfrac{5}{4} $

$ \Rightarrow RP _{\lambda _1} : RP _{\lambda _2} = 5:4 $

Hence, the correct answer is OPTION D. 

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]

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

  1. 10 m

  2. 5 m

  3. 15 m

  4. None of these

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Correct Option: B
Explanation:

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


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

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

$\therefore  D _{max} = 5m$

doctor prescribes spectacles to a patient with a combination of a convex lens of focal length 40 em, and concave lens of focal length 25 em then the power of spectacles will be.

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

  1. -6.5 D

  2. 1.5 D

  3. -1.5 D

  4. -8.5 D

Show answer
Correct Option: C
Explanation:

For combination of lenses, power
$P=P _1+P _2=\frac {100}{40}-\frac{100}{25}=-1.5 D$

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 :

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

  1. 0.5m

  2. 5m

  3. 50m

  4. 500m

Show answer
Correct Option: C
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.
$\displaystyle R.P=\frac{1.22\times 500\times 10^{-9}}{5\times 10^{-3}}\times 400\times 1000$
          $\displaystyle =1.22\times \frac{10^{-2}}{10^{-3}}\times 4 = 1.22\times 40=50 m$

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.

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

  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

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Correct Option: A
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.

High quality lens system for optical instrument is made by using ..............

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

  1. Concave and convex lens

  2. Convex lens

  3. Concave lens

  4. Concave and convex mirror

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Correct Option: A