CBSE Class 12 Physics NCERT Solutions: Chapter 18 - Wave Optics

The wavelength of the <math>H_a</math> line emitted by hydrogen is given as <math>\lambda = 6563 \text{ Å}</math>. This can be written in meters as <math>\lambda = 6563 \times 10^{-10} \text{ m}</math>.

The observed red shift, which is the difference between the observed wavelength and the emitted wavelength, is given as <math>(\lambda' - \lambda) = 15 \text{ Å}</math>. In meters, this is <math>15 \times 10^{-10} \text{ m}</math>.

The speed of light is a constant, <math>c = 3 \times 10^8 \text{ m/s}</math>.

Let the velocity of the star receding away from the Earth be <math>v</math>.

The relationship between the red shift and the velocity of the receding source is given by the Doppler effect for light:

$$\backslash lambda\text{'}\; -\; \backslash lambda\; =\; \backslash frac\{v\}\{c\}\backslash lambda$$

We can rearrange this formula to solve for the velocity <math>v</math>:

$$v\; =\; \backslash frac\{c\}\{\backslash lambda\}\; \backslash times\; (\backslash lambda\text{'}\; -\; \backslash lambda)$$

Now, we substitute the given values into the formula:

$$v\; =\; \backslash frac\{3\; \backslash times\; 10^8\; \backslash text\{\; m/s\}\}\{6563\; \backslash times\; 10^\{-10\}\; \backslash text\{\; m\}\}\; \backslash times\; (15\; \backslash times\; 10^\{-10\}\; \backslash text\{\; m\})$$

Calculating the value:

$$v\; =\; \backslash frac\{3\; \backslash times\; 10^8\; \backslash times\; 15\; \backslash times\; 10^\{-10\}\}\{6563\; \backslash times\; 10^\{-10\}\}\; \backslash text\{\; m/s\}$$

$$v\; \backslash approx\; 6.87\; \backslash times\; 10^5\; \backslash text\{\; m/s\}$$

Therefore, the speed with which the star is receding away from the Earth is approximately <math>6.87 \times 10^5</math> m/s.

Newton's corpuscular theory of light, proposed by Isaac Newton, described light as consisting of tiny particles called corpuscles. According to this theory, when light corpuscles travel from a rarer medium (like air) to a denser medium (like water), they experience attractive forces perpendicular to the surface separating the two media. These forces cause the component of the corpuscles' velocity perpendicular to the surface to increase, while the component parallel to the surface remains unchanged.

Let <math>c</math> be the speed of light in vacuum (or rarer medium) and <math>v</math> be the speed of light in the denser medium (water). Let <math>i</math> be the angle of incidence and <math>r</math> be the angle of refraction. According to the corpuscular theory, the tangential component of velocity remains constant, and the normal component increases. This leads to the relation:

$$c\; \backslash sin\; i\; =\; v\; \backslash sin\; r$$

The relative refractive index (<math>\mu</math>) of water with respect to air is defined as the ratio of the speed of light in air to the speed of light in water:

$$\backslash mu\; =\; \backslash frac\{c\}\{v\}$$

From Snell's law, we know that <math>\sin i = \mu \sin r</math>. Comparing this with the relation derived from corpuscular theory, <math>c \sin i = v \sin r</math>, we can write:

$$\backslash frac\{\backslash sin\; i\}\{\backslash sin\; r\}\; =\; \backslash frac\{v\}\{c\}$$

Since <math>\mu = \frac{\sin i}{\sin r}</math> for refraction from air to water, and for water <math>\mu > 1</math>, it implies that <math>\frac{v}{c} > 1</math>, or <math>v > c</math>. Thus, Corpuscular theory predicts that the speed of light in a denser medium is greater than in vacuum.

However, experimental determinations of the speed of light in water, notably by Foucault and later experiments, have conclusively shown that the speed of light in water is actually less than the speed of light in vacuum (<math>v < c</math>). Therefore, the prediction of the Corpuscular theory is not confirmed by experimental results.

The alternative picture of light that is consistent with these experimental findings is the **Wave theory of light**. The wave theory correctly predicts that light travels slower in a denser medium than in vacuum. This is because the wavelength of light decreases in a denser medium, while its frequency remains constant, leading to a reduced speed (<math>v = f\lambda</math>).

Let us consider a point object 'O' placed at a distance 'x' from a plane mirror MO'. According to Huygens' principle, every point on a wavefront acts as a source of secondary wavelets. We can consider a spherical wavefront originating from the point object O.

Let's consider two rays OA and OB originating from O and striking the mirror at points A and B, respectively. According to the law of reflection, the angle of incidence equals the angle of reflection. Using Huygens' principle, we can show that the reflected wavefront originating from A and B will appear to come from a point 'I' located behind the mirror.

Consider the wavefront originating from O. Let the wavefront reach point A on the mirror at time <math>t</math>. During this time <math>t</math>, the object O has moved to a position such that the distance travelled by light is <math>OA</math>. Simultaneously, the secondary wavelets originating from points on the mirror will propagate outwards.

Let's consider the wavefront that starts from O and reaches point A on the mirror. Let the time taken be <math>t</math>. The distance travelled by light from O to A is <math>OA</math>. During this time <math>t</math>, the secondary wavelet originating from A will travel a distance <math>AD</math> in the direction perpendicular to the mirror, where D is a point on the reflected wavefront. The distance travelled by the wavelet is <math>AD = c \times t</math>, where <math>c</math> is the speed of light.

We can consider the triangle OAM, where M is the point on the mirror directly opposite to O. The distance OM is the object distance, let's call it <math>u</math>. Let the image distance be <math>v</math>.

Using Huygens' principle, we can construct the reflected wavefront. Consider a point A on the mirror. The wavelet starting from O reaches A. Let the time taken be <math>t</math>. The distance OA is the path length. The secondary wavelet from A travels a distance <math>AD</math> in time <math>t</math>. The distance travelled by light from O to A is <math>OA</math>. The distance travelled by the wavelet from A is <math>AD</math>.

Consider the object at O. Let the wavefront be represented by a sphere. When this wavefront strikes the plane mirror, each point on the mirror acts as a source of secondary wavelets. The envelope of these secondary wavelets forms the reflected wavefront.

Let the object be at O. Consider a point A on the mirror. The time taken for light to travel from O to A is <math>t = OA/c</math>. During this time, a secondary wavelet originating from A travels a distance <math>AD = c \times t = OA</math> in the direction perpendicular to the mirror.

Let the object distance be <math>u</math> and the image distance be <math>v</math>. For a plane mirror, it can be shown using Huygens' principle that the image distance is equal to the object distance (<math>u = v</math>). The image formed is virtual because the reflected rays only appear to diverge from this point behind the mirror; they do not actually converge there.

Specifically, consider a point object O at a distance 'x' from the mirror. Let A be a point on the mirror. The time taken for light to travel from O to A is <math>t = OA/c</math>. During this time, a secondary wavelet from A travels a distance <math>AD = c \times t = OA</math>. By constructing the envelope of these wavelets, it can be shown that the virtual image 'I' is formed such that OI = IA, and the image is virtual and located at the same distance behind the mirror as the object is in front.