Tag: resolving power of optical instruments

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

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

  1. having better resolving power

  2. being easy to handle

  3. low cost

  4. quickness of observation
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

Resolving power of a telescope increases with :

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

  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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

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

  2. $4R$

  3. $2R$

  4. $\dfrac{R}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $2R$.

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

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

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

  2. $\dfrac{1}{50}$  minute

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

  4. $\dfrac{1}{500}$sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

     $R= \dfrac{9}{1.22\lambda }$

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

  $\Delta \theta = \dfrac{1}{50}sec$

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

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

  1. 10$^{6}$ rad

  2. $10^{-2}$ rad

  3. $10^{-4}$ rad

  4. $10^{-5}$ rad
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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

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

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

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

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

  4. $7.32\times { 10 }^{ -6 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $6 : 5$

  2. $5 : 4$

  3. $4 : 5$

  4. $9 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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).

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

A photograph of the moon was taken with telescope. Later on, it was found that a housefly was siting on the objective lens of the telescope. In photograph

  1. the image of the housefly will be reduced

  2. there is a reduction in the intensity of the image

  3. there is an increase in the intensity of the image

  4. the image of the housefly will be enlarged
Reveal answer Fill a bubble to check yourself
D Correct answer