Tag: wave optics

Questions Related to wave optics

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

In case of linearly polarised light, the magnitude of the electric field vector.

polarisation of light polarisation wave optics optics physics

  1. Does not change with time

  2. Varies periodically with time

  3. Increases and decreases linearly with time

  4. Is parallel to the direction of propagation

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Correct Option: B
Explanation:

In any type of light whether polarised or unpolarised, the magnitude of electric field vector always varies periodically with time.
Actually the change in electric field vector gives rise to periodically changing magnetic field.

If the angle between the pass axis of the polarizer and the analyzer is 45, the ratio of the intensities of original light and the transmitted light after passing through the analyzer is 

polarisation of light polarisation wave optics optics physics

  1. $\dfrac{I}{2}$

  2. $\dfrac{I}{3}$

  3. I

  4. $\dfrac{I}{4}$

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Correct Option: D
Explanation:

The intensity of the light after passing through the polariser
$I \, = \, I _0 \, cos^2\phi \, = \, I _0 \, cos^245$
$= \, I _0 \left ( \dfrac{I}{\sqrt2} \right )^2 \, = \, \dfrac{I}{2} \, \times \, \, \dfrac{I}{2} \, = \, \, \dfrac{I}{4} \,\, \left ( \because \, I _0 \, = \, \, \dfrac{I}{2} \right )$

The angle between the pass axis of polarizer and analyzer is $45^{\circ}$. The percentage of polarised light passing through analyzer is:

polarisation of light polarisation wave optics optics physics

  1. 75%

  2. 25%

  3. 50%

  4. 100%

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Correct Option: C
Explanation:

Given,

$\theta=45^0$

By malus law,
$I\propto cos^2\theta$
$I\propto (cos45^0)^2$

$I\propto \left(\dfrac{1}{\sqrt{2}}\right)^2$

$I\propto (0.5)$
The percentage of polarized light passing through analyzer is $50$%.
The correct option is C. 

When ordinary light is made incident on a quarter wave plate, the emergent light is:

polarisation of light polarisation wave optics optics physics

  1. linearly polarised

  2. circularly polarised

  3. unpolarised

  4. elliptically polarised

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Correct Option: D
Explanation:

A quarter-wave plate is basically used for the elliptical polarization of the unpolarized light incident over it. The quarter-wave plate produces the electric field at the various angle about the axis of the light and thereby making the light elliptically polarized.


In case the electric field are at an angle of $90^\circ$ from one another, it is a special case of the circularly polarized light.