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All questions of Wave Optics for NEET Exam

Light of wavelength 500 nm is incident on a slit of width 0.1 mm. The width of the central bright line on the screen is 2m. What is the distance of the screen?​
a)1 m
b)200 m
c)0.75 m
d)0.5 m
Correct answer is option 'B'. Can you explain this answer?

Swara Sharma answered
Beta(central Maxima)=2lambda D(distance of the screen)/d(distance between slits).

beta=2m.

wavelength= 500nm=5×10^-7m.

d=0.1mm=1×10^-4m.

2=2×5×10^-7 D/10^-4.

1=5×10^-3 D .

1/5×10^-3=D .

1000/5=D.

200m=D
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If the Young’s apparatus is immersed in water, the effect on fringe width will​
  • a)
    remain same
  • b)
    decrease
  • c)
    increase
  • d)
    Cant say because the experiment cannot be carried in any other medium except air
Correct answer is option 'B'. Can you explain this answer?

Kalyan Joshi answered
Effect of Immersing Young's Apparatus in Water on Fringe Width

When the Young's apparatus is immersed in water, the effect on the fringe width can be explained as follows:

1. Refractive Index of Water: Water has a higher refractive index than air. This means that light passing through water will be bent more than in air.

2. Path Difference: Path difference is the difference in the distance traveled by two waves from the source to the point where they interfere with each other. When the apparatus is immersed in water, the path difference between the two waves will change due to the change in refractive index.

3. Decrease in Fringe Width: As the path difference changes, the interference pattern will also change. The fringe width will decrease because the change in the path difference will cause the interference pattern to become more spread out.

4. Explanation: The fringe width is inversely proportional to the square root of the refractive index. Since the refractive index of water is higher than air, the fringe width will decrease when the apparatus is immersed in water.

Conclusion

In conclusion, when the Young's apparatus is immersed in water, the fringe width will decrease due to the change in refractive index, which leads to a change in the path difference and the interference pattern.

A rotating calcite crystal is placed over an ink dot. On seeing through the crystal one finds:
  • a)
    Two stationary dots
  • b)
    Two dots moving along straight lines
  • c)
    One dot rotating about the other
  • d)
    Both dots rotating about a common axis
Correct answer is option 'C'. Can you explain this answer?

Anaya Patel answered
In calcite crystal there is a phenomenon of double refraction. So, we see one dot as a stationary object whereas the other being refracted at a little displaced position seems to be rotating about the first dot.

An unpolarised beam of intensity Io is incident on a polarizer and analyser placed in contact. The angle between the transmission axes of the polarizer and the analyser is θ. What is the intensity of light emerging out of the analyser?​
  • a)
    (I0 cos 2θ)/2
  • b)
    I0 cos2θ
  • c)
    I0 cosθ
  • d)
    I0
Correct answer is option 'A'. Can you explain this answer?

Nikita Singh answered
 Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E0 is the amplitude of the electric vector transmitted by the polarizer, then intensity I0 of the light incident on the analyser is
I ∞ E02
 
The electric field vector E0 can be resolved into two rectangular components i.e E0 cosθ and E0 sinθ. The analyzer will transmit only the component ( i.e E0 cosθ ) which is parallel to its transmission axis. However, the component E0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
 
I ∞ ( E0 x cosθ )2
 
I / I0 = ( E0 x cosθ )2 / E02 = cos2θ
 
I = I0 x cos2θ
when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,
I = I0 x cos2θ/2
 

Two coherent sources produce a dark fringe when phase difference between the interfering waves is(n integer)​
  • a)
    2π
  • b)
    (2n – 1)π
  • c)
    zero
  • d)
    n
Correct answer is option 'B'. Can you explain this answer?

Shruti Sarkar answered
Dark fringes will be produced when there are destructive interference. The condition for that is the two waves should have a phase difference of an odd integral multiple of π.

A beam of light has a wavelength of 650 nm in vacuum. What is the wavelength of these waves in the liquid whose index of refraction at this wavelength is 1.47?
  • a)
    472nm
  • b)
    442nm
  • c)
    462nm
  • d)
    452nm
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
We know that,
Wavelength of light in a material:
λ= λ0/n
where, λ=wavelength of material, λ0=wavelength of light in vacuum,
n=index of refraction of material.
So, 650x10-9m/1.47=442.177nm.

Diffraction of light gives the information of
  • a)
    Transverse nature
  • b)
    Longitudinal nature
  • c)
    Both transverse and longitudinal
  • d)
    Neither transverse nor longitudina
Correct answer is option 'D'. Can you explain this answer?

Vijay Bansal answered
Transverse and longitudinal waves. Light and other types of electromagnetic radiation are transverse waves. Water waves and S waves (a type of seismic wave) are also transverse waves. In transverse waves, the vibrations are at right angles to the direction of travel.

Diffraction pattern cannot be observed with:
a)one wide slit
b)two narrow slits
c)one narrow slit
d)large number of narrow slits
Correct answer is option 'A'. Can you explain this answer?

Pooja Mehta answered
If one illuminates two parallel slits, the light from the two slits again interferes. Here the interference is a more pronounced pattern with a series of alternating light and dark bands. ... He also proposed (as a thought experiment) that if detectors were placed before each slit, the interference pattern would disappear.

The phenomena diffraction can take place in sound waves.
  • a)
    Yes
  • b)
    No
  • c)
    Only Interference
  • d)
    Under certain conditions only
Correct answer is option 'A'. Can you explain this answer?

Rohan Singh answered
YES, Sound waves, on the other hand, are longitudinal, meaning that they oscillate parallel to the direction of their motion. Since there is no component of a sound wave's oscillation that is perpendicular to its motion, sound waves cannot be polarized

Which of the following undergoes largest diffraction?
  • a)
    Ultraviolet light
  • b)
    Infra red light
  • c)
    Radio waves
  • d)
    Y – rays
Correct answer is option 'C'. Can you explain this answer?

Anjana Sharma answered
Maximum diffraction occurs when size of obstacle is almost equal to wavelength of light wave. Hence maximum diffraction occurs for larger wavelength . As wavelength of radio wave is higher than others, maximum diffraction will occur for it.   

When viewed in white light, a soap bubble shows colours because of
  • a)
    Interference
  • b)
    Scattering
  • c)
    Dispersion
  • d)
    Diffraction
Correct answer is option 'A'. Can you explain this answer?

Nayanika Datta answered
Soap Bubble and Colors

Introduction:
When white light passes through a soap bubble, it undergoes a phenomenon known as dispersion, resulting in the appearance of various colors. This colorful phenomenon is due to interference of light waves and can be explained by the principles of interference and thin film interference.

Interference:
Interference is the interaction of two or more light waves, resulting in either constructive or destructive interference. Constructive interference occurs when the peaks and troughs of the waves align, leading to reinforcement and the formation of bright regions. Destructive interference occurs when the peaks and troughs of the waves cancel each other out, resulting in dark regions.

Thin Film Interference:
A soap bubble can be considered as a thin film of water sandwiched between two layers of air. When light waves hit the surface of the soap bubble, a portion of the light is reflected, and another portion enters the bubble. This incident light can also be seen as a combination of different wavelengths, or colors, of light.

Multiple Reflections and Interference:
As the light waves enter the soap bubble, they undergo multiple reflections from the inner and outer surfaces of the film. These reflections create a path difference between the different waves. If the path difference is an integer multiple of the wavelength of light, constructive interference occurs, resulting in bright colors. If the path difference is a half-integer multiple of the wavelength, destructive interference occurs, leading to dark regions.

Dispersion:
In addition to interference, dispersion also plays a role in the formation of colors in soap bubbles. Dispersion is the phenomenon where different colors of light have different wavelengths and therefore different speeds when passing through a medium. As the light waves enter the soap film, they are separated into their constituent colors due to their different wavelengths. This separation of colors is what gives rise to the vibrant and distinct colors observed in soap bubbles.

Conclusion:
In conclusion, soap bubbles show colors when viewed in white light due to the phenomenon of interference and dispersion. The interaction of light waves in the thin film of the bubble, combined with the separation of colors based on their wavelengths, leads to the formation of various colors. This colorful display is a result of the complex interplay between interference and dispersion.

Shape of the wave front of portion of the wave front of light from a distant star intercepted by the Earth
  • a)
    plane
  • b)
    spherical
  • c)
    hyperboloid
  • d)
    conical
Correct answer is option 'A'. Can you explain this answer?

Aryan Keshri answered
Option A ....Since the light coming from the distance star to the earth will be at Infinity , the rays coming from it will be parallel rays , thus the wavefront will be a plane .

The Brewster’s angle for a transparent medium is 600.The angle of incidence is​
  • a)
    450
  • b)
    900
  • c)
    600
  • d)
    300
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized.
So the angle of incidence would be 60°

The intensity of light transmitted by the analyzer is maximum when
  • a)
    The transmission axes of the analyzer and the polarizer are perpendicular
  • b)
    The reflected ray is fully polarized
  • c)
    The transmission axes of the analyzer and the polarizer are parallel
  • d)
    The reflected ray is partially polarized
Correct answer is option 'C'. Can you explain this answer?

Alfiya I answered
From malu's law, If I0 is intensity of plane polarised light incident on an analyser and (thete) is the angle between axes of polariser and analyser, then intensity of polarised light transmitted from analyser ,I = I0 cos(^2) thete, ie, when (thete)=0, cos(^2) 0=1, then I= I0 ie, the transmission axes of polariser and analyser are paralell.

The change in diffraction pattern of a single slit, when the monochromatic source of light is replaced by a source of white light will be
  • a)
    Diffracted image is not clear
  • b)
    Diffracted image gets dispersed into constituent colours of white light
  • c)
    The image of the slit becomes infinitely wide
  • d)
    Clear colourful fringe pattern
Correct answer is option 'B'. Can you explain this answer?

Dr Manju Sen answered
  • When a source of white light is used instead of a monochromatic source, the diffracted image of the slit gets dispersed into constituent colours of white light.
  • The central maximum will be white and on either side of the central maximum, there will be coloured fringes.

Emission and absorption is best described by,
  • a)
    dual / schizophrenic model
  • b)
    a wave model
  • c)
    None of the above
  • d)
    particle model
Correct answer is option 'D'. Can you explain this answer?

Nandini Iyer answered
Distance between the slits, d=0.28×10−3 m
Distance between the slits and the screen, D=1.4m
Distance between the central fringe and the fourth (n=4) fringe, 
u=1.2×10−2 m
In case of a constructive interference, we have the relation for the distance between the two fringes as:
u=nλD/d
⇒λ=ud/nD=6×10−7m=600nm

In the Young’s double slit experiment, the distance of p th dark fringe from the central maximum is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Neha Sharma answered
Position of the pth dark fringe is at a distance of 
y=(2p−1) (λD/2d)​ from the centre
where 
λ: Wavelength
D: Distance between slits 
d: slit width

In an interference experiment monochromatic light is replaced by white light, we will see:​
  • a)
    uniform darkness on the screen
  • b)
    a few coloured bands and then uniform illumination
  • c)
    equally spaced white and dark bands
  • d)
    uniform illumination on the screen
Correct answer is option 'B'. Can you explain this answer?

Arun Khanna answered
Therefore if monochromatic light in Young's interference experiment is replaced by white light, then the waves of each wavelength form their separate interference patterns. The resultant effect of all these patterns is obtained on the screen. i.e., the waves of all colours reach at mid point M in same phase we will see​.a few coloured bands and then uniform illumination

If the light is completely polarized by reflection, then angle between the reflected and refracted light is
  • a)
    π
  • b)
    2π
  • c)
    π/2
  • d)
    π/4
Correct answer is option 'C'. Can you explain this answer?

Sant Singh answered
This is in accordance with Brewsters law. When light is incident. When light is incident at Brewsters angle then the refracted and reflected ray are perpendicular to reach other.

According to Huygens construction relation between old and new wave fronts is
  • a)
    new wave front is parallel to old wave front
  • b)
    new wave front is the forward envelope of the secondary waves
  • c)
    new wave front is perpendicular to old wave front
  • d)
    new wave front is tangential to old wave front
Correct answer is option 'B'. Can you explain this answer?

Roshni Desai answered
The Huygens-Fresnel principle states that every point on a wavefront is a source of wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. The new wavefront is a line tangent to all of the wavelets.
The correct answer is option B.

A paper, with two marks having separation d, isheld normal to the line of sight of an observer ata distance of 50m. The diameter of the eye-lensof the observer is 2 mm. Which of the followingis the least value of d, so that the marks can beseen as separate ? The mean wavelength ofvisible light may be taken as 5000 Å.     [2002]
  • a)
    1.25 m
  • b)
    13.6 cm
  • c)
    12.5 cm
  • d)
    2.5 mm
Correct answer is option 'C'. Can you explain this answer?

Gowri Nair answered
Angular limit of resolution of eye, 
where, d is diameter of eye lens.
Also, if y is the minimum separation between two objects at distance D from eye then
θ = y/D
Here, wavelength λ = 5000Å = 5x10-7m
D = 50 m
Diameter of eye lens = 2 mm = 2x10-3 m
From eq. (1), minimum separation is
y = 5 * 10-7 * 50 / 2 * 10-3
y = 0.0125 m
y = 1.25 cm

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
  • a)
    0.32 mm
  • b)
    0.2 mm
  • c)
    0.3 mm
  • d)
    0.25 mm
Correct answer is option 'B'. Can you explain this answer?

Poulomi Desai answered
Wavelength of light beam, λ
Distance of the screen from the slit, D=1m
For first minima, n=1
Distance between the slits is d
Distance of the first minimum from the centre of the screen can be obtained as, x = 2.5mm = 2.5×10−3
Now, nλ = xd/D
⇒ d= nλD/x = 0.2mm
Therefore, the width of the slits is 0.2 mm.

Which of the following phenomenon cannot be explained by diffraction?
  • a)
    lack of sharp boundary when a point source of light illuminates a straightedge
  • b)
    The rainbow fringes of an oil slick on water
  • c)
    fringes observed from a large number of parallel slits
  • d)
    A bulb filament viewed through two blades held so that the edges form a slit
Correct answer is option 'B'. Can you explain this answer?

Nisha Kulkarni answered
The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin film interference.

Emission and absorption is best described by,
  • a)
    Particle Model
  • b)
    Wave Model
  • c)
    Dual / Schizophrenic model
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?

Pooja Mehta answered
The Particle Model of emission and absorption is based on the idea that light is composed of discrete particles called photons, and these photons carry quantized energy. This model helps us understand how atoms and molecules interact with light by either emitting or absorbing photons during electronic transitions.

In general the term diffraction is used
  • a)
    if large number of sources is added vectorially
  • b)
    if two sources are added vectorially
  • c)
    if single source is multiplied by scalar
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?

Madhurima Mishra answered
Explanation:

Diffraction is a phenomenon that occurs when a wave encounters an obstacle or passes through a narrow aperture. This phenomenon can be observed for a wide range of waves, including sound waves, electromagnetic waves, and water waves.

Definition:

Diffraction is the bending of waves around obstacles or through narrow openings.

Types of Diffraction:

There are two types of diffraction:

1. Fraunhofer diffraction: When the source of light is at infinity, and the diffracting object is in the near field, then it is called Fraunhofer diffraction.

2. Fresnel diffraction: When the source of light is not at infinity, and the diffracting object is in the near field, then it is called Fresnel diffraction.

Diffraction by a Single Slit:

When a beam of light passes through a narrow slit, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a central bright band called the central maximum and several smaller bands called secondary maxima and minima.

Diffraction by a Circular Aperture:

When a beam of light is incident on a circular aperture, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a bright central spot surrounded by a series of bright and dark rings.

Diffraction by a Grating:

When a beam of light is incident on a grating, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a series of bright and dark bands called interference fringes.

Conclusion:

The term diffraction is used when a large number of sources are added vectorially. This phenomenon can be observed for a wide range of waves, including sound waves, electromagnetic waves, and water waves. The diffraction pattern produced by an obstacle or narrow aperture depends on the size and shape of the obstacle or aperture, as well as the wavelength and angle of incidence of the wave.

wave front is
  • a)
    locus of all adjacent points at which the Electric field is the same
  • b)
    locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same
  • c)
    series of points on the wave with same amplitude
  • d)
    series of points on the wave with same frequency
Correct answer is option 'B'. Can you explain this answer?

Mihir Yadav answered
Explanation:

A wave front is defined as the locus of all adjacent points in a medium that are in the same phase of vibration of a physical quantity associated with the wave. This physical quantity can be the displacement of the particles in a medium, the pressure in a sound wave, or the electric and magnetic fields in an electromagnetic wave.

Some examples of wave fronts include the crests and troughs of a water wave, the compressions and rarefactions of a sound wave, and the peaks and valleys of an electromagnetic wave.

Properties of wave fronts:

Wave fronts have some important properties that are worth noting:

- Wave fronts are perpendicular to the direction of wave propagation.

- The distance between two adjacent wave fronts is equal to one wavelength.

- The shape of the wave front depends on the geometry of the source of the wave and the properties of the medium through which the wave propagates.

- The speed of the wave is determined by the properties of the medium and is constant for a given medium.

Applications of wave fronts:

Wave fronts are used in many areas of science and technology. Some examples include:

- Huygens' principle: This principle states that each point on a wave front can be considered as a new source of waves, and the new wave fronts can be constructed by connecting the secondary sources.

- Optics: Wave fronts are used to describe the behavior of light waves. They are used to explain phenomena such as reflection, refraction, and diffraction.

- Acoustics: Wave fronts are used to describe the behavior of sound waves. They are used to explain phenomena such as echoes, sound absorption, and noise reduction.

- Radar: Wave fronts are used in radar systems to detect and locate objects. The radar emits electromagnetic waves, and the wave fronts that are reflected by the object are used to determine its location.

In conclusion, wave fronts are an important concept in the study of waves. They are defined as the locus of all adjacent points in a medium that are in the same phase of vibration of a physical quantity associated with the wave. Wave fronts have important properties that are used in many areas of science and technology.

Chapter doubts & questions for Wave Optics - Topic-wise MCQ Tests for NEET 2025 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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