CBSE · Class 12 · Physics · Wave OpticsState Huygens' Principle of wave propagation. Using this principle, prove the laws of reflection of light (Angle of incidence equals angle of reflection). Also, solve the following numerical problem: In a Young's double slit experiment using monochromatic light of wavelength $600\text{ nm}$, the fringe width is found to be $0.4\text{ mm}$. When the screen is moved forward by $0.1\text{ m}$, the new fringe width becomes $0.3\text{ mm}$. Calculate the initial distance between the screen and the slits, and the distance between the two slits.

Huygens' Principle\nHuygens' Principle is a geometrical construction used to determine the position of a wavefront at any later time, given its position at any instant. It is based on the following fundamental assumptions:

• Primary Wavefront: Every point on a given wavefront acts as a source of secondary disturbance, emitting secondary wavelets in all directions with the speed of light in that medium.
• Envelope of Wavelets: The new position of the wavefront at any subsequent time $t$ is given by the forward envelope (tangent surface) of these secondary wavelets.

Proof of Laws of Reflection Using Huygens' Principle

1. Consider a plane wavefront $AB$ incident obliquely on a reflecting surface $XY$ at an angle of incidence $i$.
2. As the wavefront progresses, point $A$ hits the surface first at time $t = 0$, and point $B$ reaches point $C$ after a time interval $t$ with velocity $v$, such that $BC = vt$.
3. According to Huygens' principle, during this time $t$, secondary wavelets originating from point $A$ travel a distance $v \times t = BC$ in the same medium, forming a hemisphere of radius $AD = BC$.
4. Drawing a tangent $CD$ from point $C$ to this hemisphere gives the reflected wavefront $CD$.
5. In triangles $\triangle ABC$ and $\triangle ADC$:
   • $\angle BAC = \angle ADC = 90^\circ$
   • Side $AC$ is common
   • $BC = AD = vt$
6. Therefore, the triangles are congruent ($\triangle ABC \cong \triangle ADC$). This gives $\angle BAC = \angle ACD$, which implies that the angle of incidence $i$ equals the angle of reflection $r$ ($i = r$). Also, the incident wavefront, the normal, and the reflected wavefront all lie in the same plane.

Numerical Problem Solution

Given data:

• Initial wavelength, $\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m}$
• Initial fringe width, $\beta_1 = 0.4\text{ mm} = 0.4 \times 10^{-3}\text{ m}$
• Change in screen distance, $\Delta D = 0.1\text{ m}$
• New fringe width, $\beta_2 = 0.3\text{ mm} = 0.3 \times 10^{-3}\text{ m}$

Formula for fringe width: $$\beta = \frac{\lambda D}{d}$$\nWhere $D$ is the distance between the slits and the screen, and $d$ is the distance between the two slits.

Step 1: Finding the initial distance $D$\nInitial condition: $$\beta_1 = \frac{\lambda D}{d} \implies \frac{D}{d} = \frac{\beta_1}{\lambda}$$ \nWhen the screen is moved forward by $0.1\text{ m}$ (assuming towards the slits, reducing the distance): $$\beta_2 = \frac{\lambda (D - \Delta D)}{d} \implies \frac{D - \Delta D}{d} = \frac{\beta_2}{\lambda}$$ \nDividing the two equations: $$\frac{D}{D - \Delta D} = \frac{\beta_1}{\beta_2}$$ $$\frac{D}{D - 0.1} = \frac{0.4}{0.3} = \frac{4}{3}$$ $$3D = 4(D - 0.1)$$ $$3D = 4D - 0.4$$ $$D = 0.4\text{ m}$|

Step 2: Finding the slit separation $d$\nUsing the fringe width formula with $D = 0.4\text{ m}$: $$\beta_1 = \frac{\lambda D}{d}$$ $$d = \frac{\lambda D}{\beta_1} = \frac{(600 \times 10^{-9}\text{ m}) \times (0.4\text{ m})}{0.4 \times 10^{-3}\text{ m}}$$ $$d = 600 \times 10^{-9} \times 10^3 = 600 \times 10^{-6}\text{ m} = 0.6\text{ mm}$$

Answer:

• Initial distance between screen and slits ($D$) = $0.4\text{ m}$
• Distance between the two slits ($d$) = $0.6\text{ mm}$ (or $6 \times 10^{-4}\text{ m}$)