Tag: physics

Questions Related to physics

In an astronomical microscope, the focal length of the objective is made :

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

  1. shorter than that of the eye piece

  2. greater than that of the eye piece

  3. half of the eye piece

  4. equal to that of the eye piece

Show answer
Correct Option: B
Explanation:

In an Astronomical telescope, the objective lens has a greater radius than the eyepiece.

Thus the objective lens has a greater focal length than the eyepiece.

Option B is correct.

If the apertature of a telescope is decreased the resolving power will

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

  1. increases

  2. decreases

  3. remain same

  4. zero

Show answer
Correct Option: B
Explanation:

Resolving power of a telescope=$\dfrac{a}{1.22\lambda}$

where $a$ is the aperture of the telescope.
Thus resolving power$\propto $ aperture.
Hence, if the aperture of telescope decreases, the resolving power decreases.

The resolving power of a telescope depends on :

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

  1. length of telescope

  2. focal length of objective

  3. diameter of the objective

  4. focal length of eyepiece

Show answer
Correct Option: C
Explanation:

Resolving power of telescope $R=\dfrac{1}{\Delta \theta}=\dfrac{a}{1.22 \lambda}$
where, $\Delta \theta$ is angular separation between two objects.
            $a$ is the diameter of the objective.
            $\lambda$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.

The diameter of the objective of a telescope is $a$, its magnifying power is $m$ and wavelength of light $\lambda $ . The resolving power of the telescope is :

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

  1. $\dfrac{(1.22\lambda )}{a}$

  2. $\dfrac{1.22a}{\lambda} $

  3. $\lambda (1.22a)$

  4. $\dfrac {a} {1.22\lambda} $

Show answer
Correct Option: D
Explanation:

Resolving power of telescope:
$R=\dfrac{1}{ \theta}=\dfrac{a}{1.22 \lambda}$


where $\theta$ is angular resolution, a is diameter of the objective and $\lambda$ is wavelength of light.

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. $\approx 5"$

Show answer
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

Show answer
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.

Show answer
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

Show answer
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. $\dfrac{R}{2}$

Show answer
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 $2R$.