Tag: resolving power of optical instruments

Questions Related to resolving power of optical instruments

How can resolving power of the instrument be increased?

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

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.

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Correct Option: A,B
Explanation:

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

The ability of an optical instruments to show the images of two adjacent point objects as separate is called :

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

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these

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Correct Option: C
Explanation:

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

Two lenses of focal lengths $+ 100 cm$ and $+ 5 cm$ are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at $\displaystyle \infty $)

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

  1. $95 cm$

  2. $100 cm$

  3. $105 cm$

  4. $500 cm$

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Correct Option: C
Explanation:

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$

In optical instruments, the lenses are used to form images by :

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

  1. Reflection

  2. Refraction

  3. Dispersion

  4. Scattering

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Correct Option: B
Explanation:

In optical instruments, the lenses are used to form images by Refraction.

In which of the following the final image is erect? 

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

  1. Compound microscope

  2. Astronomical telescope

  3. Simple microscope

  4. All of the above

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Correct Option: C
Explanation:

The image formed by the Compound microscope and Astronomical telescope is inverted,but in case of Simple microscope it form erect image.

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

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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

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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.