Interference of Light waves– Notes for Class 12, JEE & NEET - Physicskund

By -

Interference of Light

Interference of light is the phenomenon in which two or more coherent light waves superpose and redistribute light intensity in space, producing regions of maximum and minimum intensity.

Principle of Superposition

The phenomenon of interference is based on the Principle of Superposition, which states:

When two or more light waves overlap at a point, the resultant displacement at that point is equal to the vector sum of the displacements produced by the individual waves.

Conditions for Sustained Interference

  • The sources must be coherent.
  • The waves must have the same frequency or wavelength.
  • The phase difference between the waves should remain constant.
  • The amplitudes should be nearly equal for clear interference patterns.

Types of Interference

1. Constructive Interference

Constructive interference occurs when two waves meet in the same phase. The resultant amplitude becomes maximum, producing maximum intensity.

Condition:

$\Delta = n\lambda$

where:

  • $\Delta$ = Path Difference
  • $\lambda$ = Wavelength of Light
  • $n = 0,1,2,3,\ldots$

Result: Maximum Intensity (Bright Region)

2. Destructive Interference

Destructive interference occurs when two waves meet in opposite phases. The resultant amplitude becomes minimum, producing minimum intensity.

Condition:

$\Delta = \left(n+\frac{1}{2}\right)\lambda$

where:

  • $\Delta$ = Path Difference
  • $\lambda$ = Wavelength of Light
  • $n = 0,1,2,3,\ldots$

Result: Minimum Intensity (Dark Region)

Characteristics of Interference

  • Interference is a wave phenomenon.
  • It demonstrates the wave nature of light.
  • Energy is redistributed from one region to another.
  • Bright and dark regions are formed due to variations in intensity.
  • The total energy remains conserved.

Applications of Interference

  • Measurement of wavelength of light.
  • Determination of refractive index.
  • Testing optical surfaces.
  • Thin-film and anti-reflection coatings.
  • Interferometers used in precision measurements.

Conclusion

Interference is one of the most important phenomena exhibited by light. It arises from the superposition of coherent light waves and provides strong evidence for the wave nature of light.

Frequently Asked Questions (FAQ)

1. What is interference of light?

Interference of light is the phenomenon in which two or more coherent light waves superpose and produce a redistribution of intensity.

2. What is the principle behind interference?

Interference is based on the Principle of Superposition.

3. What are coherent sources?

Coherent sources are sources that emit light waves of the same frequency and maintain a constant phase difference.

4. What is constructive interference?

Constructive interference occurs when two waves meet in the same phase, producing maximum intensity.

5. What is the condition for constructive interference?

$\Delta = n\lambda$

6. What is destructive interference?

Destructive interference occurs when two waves meet in opposite phases, producing minimum intensity.

7. What is the condition for destructive interference?

$\Delta = \left(n+\frac{1}{2}\right)\lambda$

8. Why is interference evidence of the wave nature of light?

Because interference is a result of wave superposition, a property exhibited by waves.

9. Can interference occur with ordinary light sources?

No. Sustained interference requires coherent sources.

10. What are some applications of interference?

Wavelength measurement, refractive index determination, optical testing, thin-film coatings, and interferometers.

Quiz on Interference of Light

Multiple Choice Questions

1. Interference of light is a result of:

  • A) Reflection
  • B) Refraction
  • C) Superposition of waves
  • D) Polarization

Answer: C) Superposition of waves

2. Which of the following is necessary for sustained interference?

  • A) Different frequencies
  • B) Coherent sources
  • C) High intensity
  • D) Polarized light

Answer: B) Coherent sources

3. Constructive interference occurs when the path difference is:

  • A) $\lambda/2$
  • B) $\lambda/4$
  • C) $n\lambda$
  • D) $\frac{(2n+1)\lambda}{2}$

Answer: C) $n\lambda$

4. Destructive interference occurs when the path difference is:

  • A) $n\lambda$
  • B) $\left(n+\frac{1}{2}\right)\lambda$
  • C) $2n\lambda$
  • D) Zero

Answer: B) $\left(n+\frac{1}{2}\right)\lambda$

5. In constructive interference, the resultant intensity is:

  • A) Minimum
  • B) Zero
  • C) Maximum
  • D) Constant

Answer: C) Maximum

True or False

6. Interference demonstrates the wave nature of light. Answer: True

7. Coherent sources have a constant phase difference. Answer: True

8. Destructive interference produces maximum intensity. Answer: False

9. Interference can occur between incoherent sources for a long duration. Answer: False

10. Total energy is conserved during interference. Answer: True

Short Answer Questions

11. Define interference of light.

Interference is the phenomenon in which two or more coherent light waves superpose to produce a redistribution of intensity.

12. State one condition necessary for interference.

The sources must be coherent.

13. What is meant by coherent sources?

Sources that emit waves of the same frequency and maintain a constant phase difference.

14. Write the condition for constructive interference.

$\Delta = n\lambda$

15. Write the condition for destructive interference.

$\Delta = \left(n+\frac{1}{2}\right)\lambda$

Assertion–Reason Question

Assertion (A): Interference is a consequence of the superposition principle.

Reason (R): Light waves can combine and produce regions of different intensities.

Answer: Both A and R are true, and R is the correct explanation of A.

Comments

Post a Comment

Popular posts from this blog

NCERT Solution Class 10 Science Chapter 11 Electricity -

By -
Image
NCERT Exercise Solution Class 10 Science Chapter 11 Electricity  Example 11.1 A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit. Solution: Given  I = 0.5 A ,   t = 10 min = 600 s. We have Q = It = 0.5 A × 600 s = 300 C Page Number 172 – Question with solution 1 to 3 : 1. What does an electric circuit mean? Solution : It is a continuous and closed path through which electric current flows. It contains different components like • Cell or Battery Plug Key • Wires • Electric Components (like Ammeter,Voltmeter) • Bulb etc  2. Define the unit of current. Solution :  The unit of electric current is ampere (A). I  = $\frac{Q}{t}$ When Q = 1 C and t = 1 sec then I = $\frac{1C}{1S}$ I = 1A 1 A is defined as the flow of 1 C of charge through a wire in 1 s. 3. Calculate the number of electrons constituting one coulomb of charge. Solution :  We know  Q...
Read more

Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER

By -
Image
Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER - 10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. Solution :  On Celsius scale  t(°C) = T(K) - 273.15 Triple point of neon = 24.57 - 273.15 = - 248.58°C Triple point of carbon dioxide = 216.55 - 273.15 = -56.6°C On Fahrenheit scale  $t(^{o}F) = \frac{9} {5} (t^{o}C)+32$ Triple point of neon = $\frac{9} {5} (-248.58)+32$  Triple point of neon= -447.444+32=-415.44°F Triple point of carbon dioxide = $\frac{9} {5} (-56.6)+32$  Triple point of carbon dioxide = - 101.88 +32 = -69.88°F 10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350B. What is the relation between $T_{A}$ and $T_{B}$? Solution :  Triple point of  water in  absolute scale A = 200A absolute scale B = 350 B Temperature on Kelvin scale = 273.16K 200A = 273.16K $1A = \frac{273.16}{...
Read more

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

By -
Image
NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 × 10^{4} J/g$ ?  Solution:   Given  Mass of flowing water , m = 3.0 litre/ min = 3000 g/min The geyser heats the water , raising the temperature from 27 °C to 77 °C. Initial temperature, $T_{1}$ = 27°C Final temperature $T_{2}$ = 77°C Rise in temperature ,  $T = T_{2} -  T_{1}$ = 77-27= 50°C Heat of combustion =$ 4.0 × 10^{4} J/g$ Specific heat of water , $c= 4.2 J/g^{o}C$ Total heat used ,  $$Q = mcT$$ $$Q = 3000 \frac{g}{min} \times 4.2 \frac{J}{g^{o}C}\times 50^{o}C$$ $$Q = 6.3 × 10^{5}J/ min$$ $$Rate \ of \ consumption = \frac{6.3 \times 10^{5}}{(4 \times 10^{4})}$$ $$= 15.75 g/min$$ Therefore, rate of consumption is 15.75 g/min 11.2 What amount of heat must be supplied to $2.0 × 10^{–...
Read more