To determine the value of \( x \) in the provided Young's double-slit experiment, the relationship between slit widths and interference pattern intensities is examined. It is assumed that the electric field amplitude at the screen is directly proportional to the slit width.
The amplitude of the wave from the first slit is defined as \(A_1 = a \cdot d\), and from the second slit as \(A_2 = a \cdot xd\), where \( a \) represents a constant proportionality factor.
The resultant amplitude \(A_{\text{resultant}}\) at any point on the screen is calculated as:
\(A_{\text{resultant}} = A_1 + A_2 = ad + axd = a(d + xd)\)
Intensity \( I \) is proportional to the square of the amplitude:
\(I \propto (a(d + xd))^2\)
Maximum intensity \( I_{\text{max}} \) occurs when the waves are in phase:
\(I_{\text{max}} = (ad + axd)^2 = (ad(1 + x))^2\)
Minimum intensity \( I_{\text{min}} \) occurs when the waves are out of phase, resulting in amplitude subtraction:
\(I_{\text{min}} = (ad - axd)^2 = (ad(1 - x))^2\)
The ratio of maximum to minimum intensity is given as 9:4:
\(\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9}{4}\)
Substituting the expressions for \( I_{\text{max}} \) and \( I_{\text{min}} \):
\(\frac{(ad(1 + x))^2}{(ad(1 - x))^2} = \frac{9}{4}\)
This simplifies to:
\(\frac{(1 + x)^2}{(1 - x)^2} = \frac{9}{4}\)
Taking the square root of both sides yields:
\(\frac{1 + x}{1 - x} = \frac{3}{2}\)
Cross-multiplication is performed to solve for \( x \):
\(2(1 + x) = 3(1 - x)\)
Expansion and simplification results in:
\(2 + 2x = 3 - 3x\)
Combining like terms gives:
\(5x = 1\)
Solving for \( x \) yields the final value:
\(x = 5\)
The value of \( x \) is determined to be 5.
In single slit diffraction pattern, the wavelength of light used is \(628\) nm and slit width is \(0.2\) mm. The angular width of central maximum is \(\alpha \times 10^{-2}\) degrees. The value of \(\alpha\) is ____.