NEET Important Chapter - Wave Optics

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

List of Important Formulae

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.

2. 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.

2. 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)

2. 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.