Tag: resolution of optical instruments

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Questions Related to resolution of optical instruments

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

If the refractive indices of a prism for red, yellow and violet colours be 1.61, 1.63 and 1.65 respectively, then the dispersive power of the prism will be:

  1. $\dfrac { 1.65-1.62 }{ 1.61-1 }$

  2. $\dfrac { 1.62-1.61 }{ 1.65-1 }$

  3. $\dfrac { 1.65-1.61 }{ 1.63-1 }$

  4. $\dfrac { 1.65-1.63 }{ 1.61-1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Dispersive power of the prism=$\dfrac{\mu _v-\mu _r}{\mu _y-1}=\dfrac{1.65-1.61}{1.63-1}$

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

How can resolving power of the instrument be increased?

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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.

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

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

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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 $)

  1. $95 cm$

  2. $100 cm$

  3. $105 cm$

  4. $500 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

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

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

  1. increases

  2. decreases

  3. remain same

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

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

The resolving power of a telescope depends on :

  1. length of telescope

  2. focal length of objective

  3. diameter of the objective

  4. focal length of eyepiece
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 :

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

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

  3. $\lambda (1.22a)$

  4. $\dfrac {a} {1.22\lambda} $
Reveal answer Fill a bubble to check yourself
D Correct answer
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.