Tag: wave optics

Questions Related to wave optics

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 $2R$.

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

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

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

  2. $\dfrac{1}{50}$  minute

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

  4. $\dfrac{1}{500}$sec

Show answer
Correct Option: C
Explanation:

     $R= \dfrac{9}{1.22\lambda }$

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

  $\Delta \theta = \dfrac{1}{50}sec$

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

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

  1. 10$^{6}$ rad

  2. $10^{-2}$ rad

  3. $10^{-4}$ rad

  4. $10^{-5}$ rad

Show answer
Correct Option: D
Explanation:

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

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

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

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

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

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

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

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

  4. $7.32\times { 10 }^{ -6 }$

Show answer
Correct Option: D
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.

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

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

  1. $6 : 5$

  2. $5 : 4$

  3. $4 : 5$

  4. $9 : 1$

Show answer
Correct Option: B
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).

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

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

  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

Show answer
Correct Option: D

 A beam of plane polarised light falls on a polarizer which rotates about axis of ray with angular velocity $\omega $. The energy passing through polrizer in one revolution if incident power is P is :

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

  1. $ \dfrac {\pi P} {2\omega} $

  2. $ \dfrac {\pi P} {\omega} $

  3. $ \dfrac {2 \pi P} {\omega} $

  4. $ \dfrac {3 \pi P} {2\omega} $

Show answer
Correct Option: A

ASSERTION: Resolving power of 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 A and R are correct and R is correct explanation of A

  2. A and R both are correct but R is not correct explanation of A

  3. A is true but R is false

  4. both A and R is false

Show answer
Correct Option: B
Explanation:

We have,
$ RP=\dfrac { D }{ 1.22  \lambda}$


Hence $R$ is correct but for larger resolution objects making small angle be distinguished or very close objects  should be  distinguished.

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil of diameter 3 mm. Approximately what is the maximum distance up to which these dots can be resolved by the eye.

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

  1. 5 m

  2. 6 m

  3. 1 m

  4. 4 m

Show answer
Correct Option: A
Explanation:

$ \dfrac { 1.22\lambda  }{ 3mm } =\dfrac { 1mm }{ d } $


$ or\quad d=\dfrac { 3\times { 10 }^{ -6 } }{ 1.22\times 5\times { 10 }^{ -7 } } =5m$

Since the objective lens merely forms an enlarged real image that is viewed by the eyepiece, the overall angular magnification M of the compound microscope is the product of the lateral magnification $ { m } _{ 1 }$ of the objective and the angular magnification $ { M } _{ 2 }$ of the eyepiece. The former is given by
$ { m } _{ 1 }=\dfrac { { S } _{ 1 }^{ ' } }{ { S } _{ 1 } } $
Where $ { S } _{ 1 }and{ S } _{ 1 }^{ ' }$ are the object and image distance for the objective lens. Ordinarily the object is very close to the focus, resulting in an image whose distance from the  objective is much larger than the focal length $ { f } _{ 1 }$. Thus $ { S } _{ 1 }$ is approximately equal to $ { f } _{ 1 }$ and $ { m } _{ 1 }$ =$ -\dfrac { { S } _{ 1 }^{ ' } }{ { f } _{ 1 } } $, approximately. The angular magnification of the eyepiece from $ { M }=-\dfrac { { u }^{ ' } }{ u } =\dfrac { { y }/{ f } }{ { y }/{ 25 } } =\dfrac { 25 }{ f } $ (f in centimeters) is $ { M } _{ 2 }=25cm/{ f } _{ 2 },$ Where $ { f } _{ 2 }$ is the focal length of the eyepiece, considered as a simple lens. Hence the overall magnification M of the compound microscope is, apart from a negative sign, which is customarily ignored,
$ { M }={ m } _{ 1 }{ M } _{ 2 }=\dfrac { \left( 25cm \right) { S } _{ 1 }^{ ' } }{ f } $
1. What is the resolving power of the instrument whose magnifying power is given in the passage?

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

  1. $ \dfrac { \mu \sin { \theta } }{0 .61\lambda } $

  2. $ \dfrac { \mu \sin { \theta } }{ 1.22\lambda } $

  3. $ \dfrac { \mu \sin { \theta } }{ \lambda } $

  4. $ \dfrac { \sin { \theta } }{ 1.22\lambda } $

Show answer
Correct Option: A
Explanation:

The mentioned instrument is compound microscope and its resolving power is $ R.P=\dfrac { 2\mu \sin { \theta  }  }{ 1.22\lambda  } =\dfrac { \mu \sin { \theta  }  }{ 0.61\lambda  } $


where, $ \mu$ is refractive index of medium, $ \theta$ is the semi-vertical angle of the cone of the rays received by the objective.