chapter 10 wave optics question answer

Dear students class 12th physics chapter 10 wave optics question answer important short questions with answers pdf have been collected at one place to get you all prepared for your 12th Bihar board exam, as well as for UP board exams ,Jharkhand board exam ,MP board exams ,Rajsthan board exams ,cbse board exams and other board exams. Visit my nawendu classes youtube channel for more help.

12th physics WAVE OPTICS -short Question & Answer

Q (1)- What is necessary condition for interference?

Answer- Necessary conditions for interference of light are

(i) Both light sources must be coherent i.e. both sources are obtained form same source.

(ii) Source slits should be narrow.

(iii) Both light sources should be monochromatic.

(iv) Both coherent sources should be at short distance so that fringe width should be large and clear.

(v)The distance between the coherent light sources and the screen should be large so that the fringes are wider ($\beta =\frac{\lambda D}{d}$) and clear.

(vi) The amplitudes of both waves should be equal or nearly equal.

Q (2)- Write two essential conditions of constructive interference.

Bihar Board 2021

Answer- There following two essential conditions for constructive interference.

(i) superposing waves must be in same phase.

(ii) Both light sources must be coherent.

Q (3)- Describe young’s double slit experiment.

Answer- Read in nawendu classes note.

Q (4)- Why does a soup bubble of colourless solution appear coloured in the presence of white light?

Answer- There are two layers in soap bubble. Some part of incident light is reflected by upper surface and rest by lower surface. So light waves reflected by both surfaces superpose while entering in our eyes that is why colourless soap bubbles appear coloured due to coloured interference fringes of white light.

Q (5)- Give some examples of interference of light in nature.

Answer- (i) soap bubbles appear coloured due to interference of light
(ii) When oil floats on the surface of water then it appears coloured due to interference of light.

Q (6)- If the incident ray in young’s experiment be white. Then how will the interference fringes be seen?
Answer- Since wavelengths of seven colours of white light are different. Distance of interference fringes is directly proportional to wavelength ($x=\frac{n\lambda D}{d}$) so interference fringes formed by different colours of white light will be at different distances therefore interference fringes will be of different colours (colourful).

Q (7)- Interference fringes are not formed by two independent sources. Why?
Answer- Interference fringes are also formed by two independent sources but they do not appear constant and clearly because phase difference (or path difference) between two super posing waves does not remain constant. When two sources are obtained from a single source then any change in original source will be equally applied to both sources so their path-difference remains constant. Therefore interference fringes are seen constant and clear.

Q (8)- Write the expression for the width of the interference fringe.

Bihar Board 2014

Answer - Fringe width =$\beta =\frac{\lambda D}{d}$

Where $\lambda =$ wavelength of light

D = distance between the light sources and the curtain

d = distance between two light sources

Q (9)- Difference between interference and diffraction of light.

Answer- Interference of light
(i) This phenomena occurs due to superposition of two waves coming from two coherent sources.

(ii) All bright fringes are of equal Intensity.

(iii) All fringes are of equal width.
(iv) Its dark fringes are almost dark.

Diffraction of light

(i) This phenomena occurs due to superposition of secondary wavelets Coming from the same wave front.

(ii) All bright fringes are of unequal intensity.

(iii) All fringes are of unequal width.

(iv) Its dark fringes are of different intensity.

Q (10)- If young’s double slit experiment is performed by covering double-slit by red and blue transparent paper then what will be change in width of fringe?
Answer- Since we know that width of fringe is directly proportion al to wave-length ($\beta =\frac{\lambda D}{d}$).

If double-slits are covered by blue paper after covering red paper then wavelength of light will decrease so fringe-width will be also decreased.

Q (11)- In the experiments using bi-prism small line (narrow slit) holes are used, Why ?Answer- Because a wide slit can be supposed to be made of many fine slits. Each slit produces separate interference pattern. These interference patterns overlap each-other and fringes are not seen clearly. Hence narrow slit holes are used in bi-prism experiment.
Q (12)- What is mallus law of polarization?

Answer- If angle between plane of vibration of plane polarized light and plane of easy transmission is $\theta $ then intensity of light after transmitting through a polarizer is ${{\cos }^{2}}\theta $ times of the intensity of incident light.

Q (13)-. What is a polaroid ? Describe some of its applications.

Answer- The material which polarises (plane polarized) light is called Polaroid.
Uses of Polaroid – (i) To avoid glare of light.
(ii) To avoid the dazzling light of a car approaching from the apposite side during night driving. (iii) To watch 3-D- pictures (films) spectacles of Polaroid are put on.
((iv) In holography. (v) In calculators and watches letters and numbers are formed by liquid crystal display through polarisation of light.

Q (14)- Write Brewster's law of polarization of light.

Bihar Board 2014 ,2017, 2021

Answer- Brewster's law - The tangent of the angle of polarization is equal to the refractive index of the medium.

$\mu =\tan {{i}_{p}}$

where $\mu $= refractive index of the medium

${{i}_{p}}$= polarization angle

Q (15)- What do you understand by diffraction of light?

Bihar Board 2020

What do you mean by diffraction of light?

Bihar Board 2022

Answer- The phenomenon of bending of light through the edge of an object or an aperture (or a hole) is called diffraction of light.

For diffraction of light, the size of the object or aperture should be of the order of wavelength of light.

Q (16)- How does diffraction put a limit on the resolution power of instruments.

Answer- Due to diffraction of light image of a point object is actually airy disk. Two airy disk of two point objects overlap each-other. So two point objects look distinct only up to a certain distance between them.

Thus diffraction put a limit on the resolution power of optical instruments.

Q (17)- In sound diffraction takes place by simple obstacle than light. Explain.

Answer- Size of simple obstacle is not of the order of wavelength of light. But wavelength of sound wave is of the order of simple obstacle (1meter to 2 meter) that is why diffraction of sound wave takes place by simple obstacles.