To determine the instant at which maximum intensity occurs at point P for the first time in a Young's double slit experiment, we need to analyze the conditions for constructive interference. In this setup, the source S oscillates according to the equation Y = sin(πt), which indicates that the source emits waves with a specific frequency and wavelength. Let's break down the problem step by step.
Understanding the Wave Properties
The wavelength given is 4000 angstroms, which can be converted to millimeters for consistency in units:
- 1 angstrom = 1 x 10-10 meters
- 4000 angstroms = 4000 x 10-10 meters = 4 x 10-7 meters = 0.4 x 10-6 meters = 0.4 mm
Frequency Calculation
Next, we need to find the frequency of the wave. The wave function Y = sin(πt) suggests that the angular frequency (ω) is π radians per second. The relationship between angular frequency and frequency (f) is given by:
ω = 2πf
From this, we can derive:
f = ω / 2π = π / 2π = 1/2 Hz
Path Difference and Conditions for Maximum Intensity
In a double slit experiment, maximum intensity occurs when the path difference between the waves arriving from the two slits (s1 and s2) is an integer multiple of the wavelength:
Path difference = nλ, where n = 0, 1, 2, ...
Since point P is located directly in front of slit s1, the path difference for the first maximum (n=1) is equal to the wavelength (λ = 0.4 mm).
Finding the Time for Maximum Intensity
Now, we need to determine when the waves from the source S reach maximum intensity at point P. The maximum intensity occurs when the sine function reaches its peak value of 1. The wave function Y = sin(πt) achieves this at:
πt = π/2
Solving for t gives:
t = 1/4 seconds
Final Result
Thus, the instant at which maximum intensity occurs at point P for the first time is at t = 0.25 seconds. This timing aligns with the conditions for constructive interference, ensuring that the waves from both slits reinforce each other at that moment.
