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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.

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

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.

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.

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.

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.

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)

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)

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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NTA is responsible for the release of the NEET 2022 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for NEET 2022 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, NEET qualifying marks for 2021 ranged from 720-138 general category, while for OBC/SC/ST categories, they ranged from 137-108 for OBC, 137-108 for SC and 137-108 for ST category.

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NEET 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in private medical and dental colleges. NEET 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the NEET 2022 state rank list on the official website or on our site.

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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.

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