# Wave Optics Chapter 10 Class 12 Physics NCERT Textbook PDF

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### Chapter 10: Wave Optics

#### 10.1 INTRODUCTION

In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. It explained the laws of reflection and refraction of light at an interface.

The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium.

This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book, the corpuscular model is very often attributed to Newton.

In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter.

As we will see, the wave model could satisfactorily explain the phenomena of reflection and refraction; however, it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium.

This is in contradiction to the prediction made by using the corpuscular model of light.

It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model; Foucault carried out this experiment in 1850.

### 10.2 Huygens Principle

We would first define a wavefront: when we drop a small stone on a calm pool of water, waves spread out from the point of impact. Every point on the surface starts oscillating with time.

At any instant, a photograph of the surface would show circular rings on which the disturbance is maximum. Clearly, all points on such a circle are oscillating in phase because they are at the same distance from the source.

Such a locus of points, which oscillate in phase is called a wavefront; thus a wavefront is defined as a surface of constant phase.

The speed with which the wavefront moves outwards from the source is called the speed of the wave. The energy of the wave travels in a direction perpendicular to the wavefront.

If we have a point source emitting waves uniformly in all directions, then the locus of points that have the same amplitude and vibrate in the same phase are spheres and we have what is known as a spherical wave as shown in Fig. 10.1(a).

### The doppler effect

We should mention here that one should be careful in constructing the wavefronts if the source (or the observer) is moving.

For example, if there is no medium and the source moves away from the observer, then later wavefronts have to travel a greater distance to reach the observer and hence take a longer time.

The time taken between the arrival of two successive wavefronts is hence longer at the observer than it is at the source.

Thus, when the source moves away from the observer the frequency as measured by the source will be smaller. This is known as the Doppler effect.

Astronomers call the increase in wavelength due to the doppler effect a red shift since a wavelength in the middle of the visible region of the spectrum moves towards the red end of the spectrum.

When waves are received from a source moving towards the observer, there is an apparent decrease in wavelength, this is referred to as blue shift.

### NCERT Solutions Class 12 Physics Chapter 10 Wave Optics

Question 3:

(i) The refractive index of glass is 1.5. What is the speed of light in glass? The speed of light in a vacuum is ( 3.0 x 108 m s-1 )

(ii) Is the speed of light in glass Independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

(i) Refractive Index of glass, \muμ = 1.5

Speed of light, c = 3 × 108 ms-1

The relation for the speed of light in a glass is: v=\frac{c}{\mu}v=μc

= \frac{3\times10^{8}}{1.5}1.53×108​ = 2\times 10^{8}m/s2×108m/s

Hence, the speed of light in glass is 2 × 108 m s-1

(ii) The speed of light is dependent on the color of the light. For white light, the refractive index of the violet component is greater than the refractive index of the red component. So, the speed of violet light is less than the speed of the red light in the glass. This will reduce the speed of violet light in glass prism when compared with red light.

Question 4:

In Young’s double-slit experiment, 0.28mm separation between the slits and the screen is placed 1.4m away. 1.2cm is the distance between the central bright fringe and the fourth bright fringe. Determine the wavelength of light used in the experiment.

Distance between the slits and the screen, D = 1.4 m

and the distance between the slits, d = 0.28 mm = 0.28 x 10-3 m

Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2cm = 1.2 × 10-2 m

For constructive interference, the following is the relation for the distance between the two fringes:u=n\;\lambda\;\frac{D}{d}u=nλdD

Where n = order of fringes

= 4\lambdaλ = Wavelength of light used

Rearranging the formula, we get\lambda =\frac{ud}{nD}λ=nDud

= \frac{1.2\times10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}4×1.41.2×10−2×0.28×10−3​

= 6 × 10-7 m = 600nm

600nm is the wavelength of the light.

Question 5:

In Young’s double-slit experiment using the monochromatic light of wavelength \lambdaλ, the intensity of light at a point on the screen where path difference is \lambdaλ is K units. What is the intensity of light at a point where path difference is \frac{\lambda}{3}3λ​?

Let I_{1}I1​ and I_{2}I2​ be the intensity of the two light waves. Their resultant intensities can be obtained as:I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phiI’=I1​+I2​+2I1​I2​​cosϕ

Where,\phiϕ = Phase difference between the two waves

For monochromatic light waves:I_{1}I1​ = I_{2}I2​

Therefore I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phiI’=I1​+I2​+2I1​I2​​cosϕ

= 2I_{1}+2I_{1}\;cos\phi2I1​+2I1​cosϕ

Phase difference = \frac{2\pi}{\lambda}\times\;Path\;differenceλ2π​×Pathdifference

Since path difference = \lambdaλ, Phase difference, \phi=2\piϕ=2π and I’ = K [Given]

Therefore I_{1}=\frac{K}{4}I1​=4K​ . . . . . . . . . . . . . . . (i)

When path difference= \frac{\lambda}{3}3λ

Phase difference, \phi=\frac{2\pi}{3}ϕ=32π

Hence, resultant intensity:I’_{g}=I_{1}+I_{1}+2\sqrt{I_{1}\;I_{1}}\; cos\frac{2\pi}{3}Ig​=I1​+I1​+2I1​I1​​cos32π​\\=2I_{1}+2I_{1}(-\frac{1}{2})=2I1​+2I1​(−21​)

Using equation (i), we can write:I_{g}=I_{1}=\frac{K}{4}Ig​=I1​=4K

Hence, the intensity of light at a point where the path difference is \frac{\lambda}{3}3λ​ is \frac{K}{4}4K​ units.