Interference — Constructive, destructive, path difference
Wave Optics
8
JEE Qs
8%
Hard
75
min
Always identify the total effective path difference or phase difference, including initial phase and any reflection-induced phase shifts, before applying interference conditions.
🧮 Key Formulas
✅ Key Points for JEE
- 1Interference is the redistribution of light energy due to the superposition of two or more coherent waves; total energy is conserved, not created or destroyed.
- 2Coherent sources, which maintain a constant phase difference (usually zero) over time, are essential for observing sustained interference patterns.
- 3The total phase difference (phi) at any point depends on the geometric path difference (delta_x), any initial phase difference between the sources, and any phase changes upon reflection (e.g., a pi phase change for reflection from a denser medium).
- 4Constructive interference occurs when waves arrive in phase, resulting in maximum intensity; destructive interference occurs when waves arrive out of phase, resulting in minimum intensity.
- 5For waves reflecting from an interface, a phase change of pi (or an equivalent path change of lambda/2) occurs if the reflection is from a denser medium.
⚠️ Common Mistakes
- ✕Incorrectly converting between path difference and phase difference, often forgetting the `2n+1` for destructive interference conditions.
- ✕Failing to account for the additional phase change of pi (or path change of lambda/2) that occurs when light reflects from an optically denser medium.
- ✕Applying intensity formulas incorrectly, especially when sources are not identical, or confusing the total phase difference (phi) with just the path difference contribution.
📝 Practice Questions
See allQ34.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In Young's double slit experiment, the fringes produced by red light are closer as compared to those produced by blue light. Reason (R): The fringe width is directly proportional to the wavelength of light. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are true but (R) is NOT the (2) (A) is true but (R) is false correct explanation of (A) (3) Both (A) and (R) are true and (R) is the correct (4) (A) is false but (R) is true explanation of (A)
Q41.The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is (1) 5 (2) 4 (3) 6 (4) 8
Q43.Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion-(A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason-(R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below : (1) Both (A) and (R) are true but (R) is not the (2) Both (A) and (R) are true and (R) is the correct correct explanation of (A) explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true
Q31.A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550 nm . [Assume that the light is incident nearly perpendicular to the glass surface.] (1) 137.5 nm (2) 275 nm (3) 94.8 nm (4) 68.7 nm
Q37. Using the given P −V diagram, the work done by an ideal gas along the path ABCD is : (1) 3P0 V0 (2) −4P0 V0 (3) −3P0 V0 (4) 4P0 V0
Q29.Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be : (1) 0.23 mm (2) 0.33 mm (3) 0.63 mm (4) 0.46 mm
NCERT Chapters
- Class 12 Physics Ch 10: Wave Optics