Young's double slit experiment is performed with monochromatic light of wavelength \( 6000 \text{Ã…} \). If the intensity of light at a point on the screen where the path difference is \( 2000 \text{Ã…} \) is \( I_1 \) and the intensity of light at a point where path difference is \( 1000 \text{Ã…} \) is \( I_2 \), then the ratio \( I_1: I_2 \) is:
To determine the ratio \( I_1 : I_2 \) in Young's double slit experiment, we need to analyze the interference pattern and how the path difference affects the intensity of light at different points on the screen. 1. Given Data: - Wavelength of light, \( \lambda = 6000 \text{Ã…} \) - Path difference at point 1, \( \Delta_1 = 2000 \text{Ã…} \) - Path difference at point 2, \( \Delta_2 = 1000 \text{Ã…} \) - Intensity at point 1, \( I_1 \) - Intensity at point 2, \( I_2 \) 2. Calculate the Phase Difference: - The phase difference \( \phi \) is related to the path difference \( \Delta \) by: \[ \phi = \frac{2\pi \Delta}{\lambda} \] - For point 1: \[ \phi_1 = \frac{2\pi \times 2000 \text{Ã…}}{6000 \text{Ã…}} = \frac{2\pi}{3} \] - For point 2: \[ \phi_2 = \frac{2\pi \times 1000 \text{Ã…}}{6000 \text{Ã…}} = \frac{\pi}{3} \] 3. Calculate the Intensity: - The intensity \( I \) at a point on the screen is given by: \[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \] where \( I_0 \) is the intensity of light from a single slit. - For point 1: \[ I_1 = 4I_0 \cos^2\left(\frac{\phi_1}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \left(\frac{1}{2}\right)^2 = I_0 \] - For point 2: \[ I_2 = 4I_0 \cos^2\left(\frac{\phi_2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{6}\right) = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 3I_0 \] 4. Determine the Ratio \( I_1 : I_2 \): - The ratio of the intensities is: \[ I_1 : I_2 = I_0 : 3I_0 = 1 : 3 \] 5. Final Answer: - The ratio \( I_1 : I_2 \) is: \[ \boxed{1:3} \] This corresponds to option (1).
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per meter in the solenoid is …………
In a hydraulic lift, the surface area of the input piston is 6 cm² and that of the output piston is 1500 cm². If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _________ kJ.
If \( \sqrt{5} - i\sqrt{15} = r(\cos\theta + i\sin\theta), -\pi < \theta < \pi, \) then
\[ r^2(\sec\theta + 3\csc^2\theta) = \]
The system of simultaneous linear equations :
\[ \begin{array}{rcl} x - 2y + 3z &=& 4 \\ 2x + 3y + z &=& 6 \\ 3x + y - 2z &=& 7 \end{array} \]
Calculate the determinant of the matrix: