Tag: resolution of optical instruments

Questions Related to resolution of optical instruments

The resolving power of human eye is :

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

  1. $\approx 1'$

  2. $\approx 1^{0}$

  3. $\approx 10"$

  4. $\approx 5"$

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Correct Option: A
Explanation:

The normal pupil size of a human eye is 4 mm.which sets a minimum resolution approximately 1' to 2'.we want to pull small objects as close to our eyes as possible to be able to see them, but there is a minimum distance of comfortable viewing which is roughly at 25 cm. Hence, correct option is A.

An electron microscope is superior to an optical microscope in terms of:

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

  1. having better resolving power

  2. being easy to handle

  3. low cost

  4. quickness of observation

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Correct Option: A
Explanation:

The biggest advantage of an electron microscope over optical microscope is that they have a higher resolution and are therefore capable of a higher magnification ( up to $2 \ million$ times ). 

However, optical microscopes show a useful magnification up to $1000-2000 $ times. This is a limit imposed by the wavelength of light. Electron microscopes, therefore, allow for the visualization of structures that would normally be not visible by optical microscopy.

Assertion : 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 correct and Reason is correct explanation of Assertion

  2. Assertion and Reason both are correct but Reason is not correct explanation of Assertion.

  3. Assertion is true but Reason is false.

  4. Both Assertion and Reason are false.

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Correct Option: B
Explanation:

Resolving power of a telescope is more if the diameter of the objective lens is more because $R=\dfrac{a}{1.22 \lambda}$
where, a is diameter of the objective. objective lens of large diameter collects more light but does not increase the resolving power of the telescope because resolving power increases when angular separation increases.

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 eyepiece

  2. increase in focal length of objective

  3. increase in aperture of eyepiece

  4. increase in aperture of objective

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Correct Option: D
Explanation:

Resolving power of a telescope:
$R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of the objective
so, $R$ increases when a is increased and $a$  increases when aperture of objective is increased

To increase both the resolving power and magnifying power of a telescope

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

  1. Both the focal length and aperture of the objective has to be increased.

  2. The focal length of the objective has to be increased.

  3. The aperture of the objective has to be increased.

  4. The wavelength of light has to be decreased.

Show answer
Correct Option: D
Explanation:

Resolving power, $R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of objective $\lambda$ is wavelength of light
magnifying power $m=\dfrac{-f _{0}}{f _{e}}\left ( 1+\dfrac{f _{e}}{D} \right )$
so, decreasing the wavelength of light increases the resolving power and magnifying power of telescope.

If accelerating potential increases from $20\ KV$ to $80\ KV$ in an electron microscope, its resolving power $R$ would change to

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

  1. $\dfrac{R}{4}$

  2. $4R$

  3. $2R$

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

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Correct Option: C
Explanation:

$\dfrac{1}{2}mv^{2}= eV$

$mv= \sqrt{2eVm}$

And $\lambda = \dfrac{h}{mV}$

$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV _{1}m}}{\sqrt{eV _{2}m}}$

$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$

$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$

$R\ \propto \dfrac{1}{\lambda}$

so $R$ would change to $2R$.

The least resolvable angle by a telescope using objective of aperture 5 m is nearly                ($\lambda = 4000A^{\circ}$)

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

  1. $\dfrac{1}{50^{\circ}}$

  2. $\dfrac{1}{50}$  minute

  3. $\dfrac{1}{50}$sec

  4. $\dfrac{1}{500}$sec

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Correct Option: C
Explanation:

     $R= \dfrac{9}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{5}{1.22\times 4000\times 10^{-10}}$

  $\Delta \theta = \dfrac{1}{50}sec$

The angular resolution of a telescope of 10 cm diameter at a wavelength of 5000Å is of the order of:

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

  1. 10$^{6}$ rad

  2. $10^{-2}$ rad

  3. $10^{-4}$ rad

  4. $10^{-5}$ rad

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Correct Option: D
Explanation:

$R= \dfrac{1}{\Delta \theta }= \dfrac{a}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{0.10}{1.22\times 5000\times 10^{-10}}$

$\Delta \theta = 6.1\times 10^{-6}\ rad$

If the wavelength of light used is $6000\mathring { A } $. The angular resolution of telescope of objective lens having diameter $10cm$ is ______ rad

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

  1. $7.52\times { 10 }^{ -6 }$

  2. $6.10\times { 10 }^{ -6 }$

  3. $6.55\times { 10 }^{ -6 }$

  4. $7.32\times { 10 }^{ -6 }$

Show answer
Correct Option: D
Explanation:

Limit of resolution $\sin { \theta  } =\theta =\cfrac { 1.22\lambda  }{ D } $
putting the values

$\theta=\dfrac{1.22\times6000\times10^{-10}}{0.1}$

$\theta=7.32\times10^{-6}$

Option (D) is correct.

The ratio of resolving power of telescope, when lights of wavelength $4000\overset{o}{A}$ and $5000\overset{o}{A}$ are used, is _________.

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

  1. $6 : 5$

  2. $5 : 4$

  3. $4 : 5$

  4. $9 : 1$

Show answer
Correct Option: B
Explanation:

Resolving power(R.P.) $\propto \lambda^{-1}$

Therefore $\dfrac{R.P. _1}{R.P. _2}=\dfrac{\lambda _2}{\lambda _1}$
Given:
$\lambda _1=4000\overset{o}{A}$
$\lambda _2=5000\overset{o}{A}$
Hence $\dfrac{R.P. _1}{R.P. _2}=\dfrac{5000\overset{o}{A}}{4000\overset{o}{A}}$
$\dfrac{R.P. _1}{R.P. _2}=\dfrac{5}{4}$
Therefore the correct option is (B).