In Young's double slit experiment, the intensity at any point on the screen is given by: \[ I = I_0 \left( 1 + \cos \delta \right) \] where \( I_0 \) is the maximum intensity and \( \delta \) is the phase difference between the two waves.
For bright and dark fringes, the intensity can be written in terms of the amplitudes \( A_1 \) and \( A_2 \) of the two waves as follows:
- For bright fringes: \( I_{\text{bright}} = I_0 \left(1 + \frac{A_1}{A_2}\right)^2 \)
- For dark fringes: \( I_{\text{dark}} = I_0 \left(1 - \frac{A_1}{A_2}\right)^2 \)
Given that the intensity ratio between bright and dark fringes is 3, we can equate the intensities: \[ \frac{I_{\text{bright}}}{I_{\text{dark}}} = 3 \] Substituting the expressions for \( I_{\text{bright}} \) and \( I_{\text{dark}} \): \[ \frac{\left(1 + \frac{A_1}{A_2}\right)^2}{\left(1 - \frac{A_1}{A_2}\right)^2} = 3 \] Taking square roots on both sides: \[ \frac{1 + \frac{A_1}{A_2}}{1 - \frac{A_1}{A_2}} = \sqrt{3} \] Solving for \( \frac{A_1}{A_2} \): \[ 1 + \frac{A_1}{A_2} = \sqrt{3} \left( 1 - \frac{A_1}{A_2} \right) \] Expanding: \[ 1 + \frac{A_1}{A_2} = \sqrt{3} - \sqrt{3} \cdot \frac{A_1}{A_2} \] Rearranging terms: \[ \frac{A_1}{A_2} \left( 1 + \sqrt{3} \right) = \sqrt{3} - 1 \] Thus: \[ \frac{A_1}{A_2} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]
Hence, the ratio of the amplitudes of the two waves is \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \).
Thus, the correct answer is: \[ \text{(A) } \frac{\sqrt{3}+1}{\sqrt{3}-1} \]
Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: