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# NEET Important Chapter - Wave Optics

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Last updated date: 19th Sep 2023
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The chapter on Wave Optics starts with the basic revision of all the concepts we have studied in our lower grades to build a connection between the advanced parts of Wave Optics. In Wave Optics, we will discuss Huygens Principle, Reflection and Refraction of Plane Waves using Huygens Principle, Interference of Light Waves, Young's Double Slit Experiment, Coherent and Non-Coherent Sources, Diffraction, etc.

We will get introduced to a few most important concepts: Diffraction and Polarisation  along with the polarisation by scattering and reflection.

In this chapter, students will also get to learn about the validity of Ray Optics and Resolving Power of Optical Instruments alongside the Doppler Effect.

Now, let us move on to the important concepts and formulae related to NEET and NEET exams along with a few solved examples.

### Important Topics of Wave Optics

• Wave Front

• Huygens's Principle

• Young's Double Slit Experiment

• Principle of Superposition

• Interference of Light

• Doppler's Effect in Light

• Polarisation

• Diffraction of Light

• Coherent And Non- Coherent Sources

### Important Concepts of Wave Optics

 Name of the Concept Key Points of the Concept Wavefront If all the particles vibrate in the same phase at a given instant then the locus of all those particles is called a wavefront.This wavefront is divided into three categories: Spherical, Cylindrical and Plane wavefront. Huygen’s Principle Huygen’s Principle is mainly a geometrical construction based on the following pointsEach point of the wavefront is a source of secondary disturbance which spreads in all directions.Each point on primary wavelets produces secondary wavelets which travel in space with the speed of light.The forward envelope of the secondary wavelets at any instant gives the new wavefront.In a homogeneous medium, the wavefront is always perpendicular to the direction of wave propagation. Interference of Light The phenomenon of redistribution of light energy in a medium when the light waves from two coherent sources superimpose one another is called interference of light wavesInterference here is considered of two types:Constructive and Destructive Coherent and Non-Coherent Sources If two sources are having a constant phase difference between them, then those sources can be considered as coherent sources.These two sources should give light of the same frequency (or wavelength). Principle of Superposition This principle states that when two or more waves with the same frequency overlap, the resultant wave is the algebraic sum of individual waves. Diffraction of Light The phenomenon of bending of light around the corners when light waves are obstructed by an obstacle is called diffraction.Diffraction is categorised into 2 types.FraunhoferFresnel Doppler’s effect in Light The change in the frequency of source when either the observer moves towards the source or far from the source is called Doppler’s Effect.This is mainly of two types:Red Shift (decreasing apparent frequency)Blue Shift (increasing apparent frequency) Polarisation The phenomenon of confining the direction of wave vibration of an electric vector in one particular direction perpendicular to wave propagation’s direction is called polarisation. Polarisation is mainly of two typesPolarisation by ScatteringPolarisation by Reflection

### List of Important Formulae

 Sl. No Name of the Concept Formula 1. Condition of maximum intensity in interference $I = I_{1}+I_{2} + 2\sqrt{I_{1}I_{2}}$ 2. Condition of minimum intensity $I = I_{1}+I_{2} - 2\sqrt{I_{1}I_{2}}$ 3. Fringe Width in vacuum $\beta = \dfrac{D\lambda}{d}$ 4. Fringe width in any medium having refractive index $\beta = \dfrac{D\lambda}{d\mu}$ 5. Resolving power of optical instruments $r_{\circ}= \dfrac{1.22f\lambda}{2a}$ 6. Malus law $I= I_{\circ}\cos^{2}\Phi$ 7. Brewester’s law $\mu= \tan(i_{p})$ 8. Red Shift $\nu^{\acute{}}= \nu(1-\dfrac{v}{c})$ 9. Blue Shift $\nu^{\acute{}}= \nu(1+\dfrac{v}{c})$

### Solved Examples

1. Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which bright fringes are separated by 8.1mm. A second light produces an interference pattern in which the fringes are separated by 7.2mm. Calculate the wavelength of the second light.

Sol:

Given, $\lambda_{1}$ = 630nm and $\beta_{1}$ = 8.1mm

and $\beta_{2}$ = 7.2mm

$\beta = \dfrac{D\lambda}{d}$ for same value of D and d

$\beta$ is directly proportional to $\lambda$

Hence $\dfrac{\beta_{1}}{\beta_{2}}$ = $\dfrac{\lambda_{1}}{\lambda_{2}}$

So $\dfrac{8.1}{7.2}$ = $\dfrac{630}{\lambda_{2}}$ and

$\lambda_{2}$ = 560nm.

Therefore, the wavelength of the second light is 560 nm.

Key point: Here only the relationship between wavelength and fringe width should be known to students.

1. Find the distance for which ray optics is a good approximation for an aperture of 4mm and wavelength 400 nm.

Sol:

Given, a = 4mm, ${\lambda}$ = 400nm

$z_{f}$= $\dfrac{a^{2}}{\lambda}$

This $z_{f}$ is called Fresnel distance for which ray optics is an good approximation for aperture a and wavelength ${\lambda}$

$z_{f}$= $\dfrac{(4\times 10^{-3})^{2}}{4\times 10^{-7}}$

= 40m

Key point: Here only the relationship between aperture and wavelength  should be known to students.

### Previous Year Questions from NEET Papers

1. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first and dark fringes on either side of the central bright fringe is

Sol:

Given wavelength = 600 nm , d = 1 mm = 10-3m and D = 2m

Hence, the central fringe width = $\dfrac{2D\lambda}{d}$

= $\dfrac{2 \times 2 \times 600 \times 10^{-9}}{1}$

=2.4mm

The distance between the first and dark fringes on either side of the central bright fringe is 2.4mm

Trick: Here the formula of diffraction in a single slit can be used directly.

1. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1m away was found to be $0.2^{\circ}$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (Refractive Index of water = 4/3)

Sol:

Given wavelength = 400 nm, Refractive index of water = 4/3, angular width in air = $0.2^{\circ}$.

Angular width in air $\theta_{\circ}$ = $\dfrac{\beta}{D}$

Angular width in water = $\dfrac{\beta}{\mu D}$

=$\dfrac{\theta_{\circ}}{\mu}$

=$\dfrac{0.2^{\circ}}{4/3}$

=$0.15^{\circ}$

Therefore, the angular width of the first minima is $0.15^{\circ}$.

Trick: Here one can see the medium dependency on angular width and how the formula changes for that.

### Practice Questions

1. Find the ratio of intensities at two points on a screen in Young’s double slit experiment when waves from the two slits have path difference of (i) 0 and (ii) $\dfrac{\lambda}{4}$   (Ans: 0 and 2:1)

1. 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 metre away. It is observed that the first minimum is at a distance of 2.5mm from the centre of the screen, find the width of the slit.   (Ans: 0.2mm)

### Conclusion

In this article, we have provided important information regarding the chapter Wave Optics, such as important concepts, formulae, etc. Students should work on more solved examples and numericals  for securing good grades in the NEET exam 2022.

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See More ## FAQs on NEET Important Chapter - Wave Optics

FAQ

1. What is the weightage of the Wave Optics chapter in NEET?

Nearly 2-3 questions appear in the exam from this chapter covering about 10 marks, which makes about 2% of the total marks.

2. What are the key points that need to be practised for solving questions from Wave Optics?

Students should practice more numericals from Wave Optics. Know the different laws and formulas pertaining to the chapter to efficiently solve questions from Wave Optics.

3. Are previous year questions enough for NEET?

Solving previous year's questions is very beneficial for NEET aspirants during their revision time. There is more probability that concepts of some questions might repeat. That's why aspirants can gain benefits from solving previous years' questions for NEET 2022. ## Notice board

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