The width of one of the two slits in Young's double-slit experiment is \( d \) while that of the other slit is \( x d \). If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4, then what is the value of \( x \)?
(Assume that the field strength varies according to the slit width.)
Show Hint
- 2
- 3
- 5
- 4
The Correct Option is C
Approach Solution - 1
To find the value of \( x \) in the given Young's double-slit experiment, we start by analyzing the relationship between slit widths and intensities in the interference pattern. Assume the electric field amplitude at the screen is proportional to the slit width.
Let's denote the amplitude of the wave from the first slit as \(A_1 = a \cdot d\) and from the second slit as \(A_2 = a \cdot xd\), where \( a \) is a constant proportionality factor.
The resulting amplitude \(A_{\text{resultant}}\) at any point on the screen is given by:
\(A_{\text{resultant}} = A_1 + A_2 = ad + axd = a(d + xd)\)
The intensity \( I \) is proportional to the square of the amplitude:
\(I \propto (a(d + xd))^2\)
The maximum intensity \( I_{\text{max}} \) occurs when the two waves are in phase:
\(I_{\text{max}} = (ad + axd)^2 = (ad(1 + x))^2\)
The minimum intensity \( I_{\text{min}} \) occurs when the two waves are out of phase, i.e., their amplitudes subtract:
\(I_{\text{min}} = (ad - axd)^2 = (ad(1 - x))^2\)
We are given the ratio of the maximum to minimum intensity as 9:4:
\(\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9}{4}\)
Substitute the expressions for \( I_{\text{max}} \) and \( I_{\text{min}} \):
\(\frac{(ad(1 + x))^2}{(ad(1 - x))^2} = \frac{9}{4}\)
This reduces to:
\(\frac{(1 + x)^2}{(1 - x)^2} = \frac{9}{4}\)
Taking the square root on both sides:
\(\frac{1 + x}{1 - x} = \frac{3}{2}\)
Cross-multiply to solve for \( x \):
\(2(1 + x) = 3(1 - x)\)
Expand and simplify:
\(2 + 2x = 3 - 3x\)
Combine like terms:
\(5x = 1\)
Finally, solve for \( x \):
\(x = 5\)
Therefore, the value of \( x \) is 5.
Approach Solution -2
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