In Young's double-slit experiment using monochromatic light of wavelength | PGMN Solutions

In Young's double-slit experiment using monochromatic light of wavelength ' λ ', the intensity of light at a point on the screen is ' k ' units, where path difference is ' λ ', The intensity of light at a point where path difference is (λ/3) is (cos⁡60^∘=0.5)
(A) zero units
(B) k units
(C) k/2 units
(D) k/4 units







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🚀 Step-by-Step Solution to Solve This Question 🚀
📌 Chapter: Wave Optics
📌 Topic: Young’s Double Slit Experiment – Path Difference and Interference Bands

🔹 Step 1: Introduce the core concept related to the question
In Young’s Double Slit Experiment, constructive interference occurs at points where the path difference between the two coherent sources is a whole number of wavelengths, forming bright bands. Destructive interference happens at positions where the path difference is a half-integer multiple of the wavelength, resulting in dark bands.

🔹 Step 2: Explain how the concept applies to the given situation
The question states that the point lies between the 4th and 15th bright bands and is a dark band. This indicates that the point corresponds to a destructive interference position located somewhere between two constructive (bright) points.

🔹 Step 3: Break down the logical steps required to solve the problem

Since bright bands occur at whole multiples, dark bands occur exactly halfway between them.

Between the 4th and 15th bright bands, the first dark band appears just after the 4th bright band.

Each subsequent dark band is evenly spaced until just before the 15th bright band.

The first dark band after the 4th bright band corresponds to a path difference slightly more than four wavelengths — precisely halfway to the next bright band.

This means the path difference at that point is between four and five full cycles, specifically half a cycle more than four, which is a standard condition for a dark fringe.

🔹 Step 4: Explain the final theoretical conclusion
Since the path difference at a dark fringe just after a given bright fringe lies exactly halfway to the next, and the question specifies that the dark fringe is between the 4th and 15th bright, the most straightforward interpretation is that it is the first dark fringe after the 4th bright one. This allows us to determine the relative number of wavelengths difference — slightly more than four, but not quite five — indicating the path difference.

✅ Final Answer: (B) 2.7 × 10⁻⁴ cm 🚀

🔥 Pro Tip:
Always remember:

Bright fringes = full cycle (constructive)

Dark fringes = half cycle between two bright fringes (destructive)
This makes it easy to track the position of dark bands between known bright ones!



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