Tag: diffraction

Questions Related to diffraction

An astronomical telescope has a large aperture to

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

  1. reduce spherical aberration

  2. have high resolution

  3. increase span of observation

  4. have low dispersion

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Correct Option: B
Explanation:

The aperture of an astronomical telescope is defined as the diameter of the objective lens.
In telescopes since, the stars are very far away from away us and emit light internsities, we need a large aperture to increase the amount of light entering the telescope thereby increasing the resolution

The limit of resolution of an optical instrument is the smallest angle that two points on an object have to subtend at the eye so that they are.

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

  1. Unresolved

  2. Well resolved

  3. Just resolved

  4. None of these

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Correct Option: C
Explanation:

Limit of resolution of an optical instrument is the minimum angle that two points on an object have to subtend at the eye so that they are just resolved. (C)

To increase the magnification of a telescope 

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

  1. the objective lens should be of large focal length and eyepiece should be of small focal length.

  2. the objective and eyepiece both should be of large focal length.

  3. both the objective and eyepiece should be of smaller focal lengths

  4. the objective should be of small focal length and eyepiece should be of large focal length

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Correct Option: A
Explanation:
$m=\dfrac { { f } _{ o } }{ { f } _{ e } } $
from the above relation we can see that magnification is directly proportional to the focal length of objective lens.

Magnification of an object ($m$), is equal to

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

  1. $\cfrac {v+f}{f}$

  2. $\cfrac {vf}{v-f}$

  3. $\cfrac {f}{v+f}$

  4. None of these

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Correct Option: D
Explanation:

we know,mirror formula
$\cfrac {1}{f}=\cfrac{1}{v}+\cfrac{1}{u}, magnification(m)=\cfrac{-v}{u}$
$\cfrac{1}{u}=\cfrac{1}{f}-\cfrac{1}{v}$
$\cfrac{1}{u}=\cfrac{v-f}{fv}$
${u}=\cfrac{fv}{v-f}$
$(m)=\cfrac{-v}{u}$,substituting $u$.
$m=\cfrac{-v}{1}\times\cfrac{v-f}{fv}$
$m=\cfrac{f-v}{f}$

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star ; -

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

  1. $305 \times 10^{-9} $ radian

  2. $152.5 \times 10^{-9} $ radian

  3. $610 \times 10^{-9} $ radian

  4. $457.5 \times 10^{-9} $ radian

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Correct Option: A
Explanation:

Limit of resolution of telescope = $\dfrac{1.22 \lambda}{D}$
$\theta = \dfrac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}} = 305 \times 10^{-9}$ radian

In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power of the microscope will change from R to

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

  1. $2 R$

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

  3. $4R$

  4. $3R$

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Correct Option: A
Explanation:

Electron microscope is a microscope that can magnify very small details with high resolving power due to the use of electrons as the source of illumination. Since the wavelength of electrons are 100,000 times shorter than visible light the electron microscopes have greater resolving power
We have the Abbe's formula as resolution limit $d=\dfrac{0.61\lambda}{NA}$(NA is the numerical aperture)


The resolving power increases when d, the minimum distance that can be seen between two points in the image, decreases. Thus, according to the formula the resolving power is inversely proportional to the wavelength.

Resolving Power  $\propto \dfrac{1}{\lambda}$

A higher voltage will give the electrons a higher speed. Thus the electrons will have a smaller de Broglie wavelength according to the equation,  $\lambda=h/mv$

$\lambda\propto\dfrac{1}{\sqrt V}$

Thus we get Resolving power $\propto \sqrt{V}$

$ \implies\dfrac{R}{R'} = \sqrt{\dfrac{20}{80}} $

Thus, $R' = 2R$

An astronomical telescope, consists of two thin lenses set $36 cm$ a part and has a magnifying power $8$. Calculate the focal length of the lenses.

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

  1. $32 cm$

  2. $18 cm$

  3. $25 cm$

  4. $36 cm$

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Correct Option: A
Explanation:

Here, $f _o+f _e=36 cm$
$M=-8$ (magnifying power is negative)
Now, $M=\cfrac {f _o}{f _e}$
$\therefore -8=-8=-\cfrac {f _o}{f _e}$
$\Rightarrow f _o=8f _e$

From the equations, we have
$8f _e+f _e=36$ or $f _e=4\ cm$

Again, $f _o=8f _e=8\times 4=32\ cm$