Given: - Wavelength of the microwave: \( \lambda = 2.0 \, \text{cm} = 0.02 \, \text{m} \) - Width of the slit: \( a = 4.0 \, \text{cm} = 0.04 \, \text{m} \) - Distance from the slit to the screen is irrelevant for angular spread calculation.
The angular position \( \theta \) of the first minimum in a single-slit diffraction pattern is given by: \[ a \sin \theta = m \lambda \] where: - \( m = 1 \) for the first minimum. Substituting the given values: \[ 0.04 \sin \theta = 1 \times 0.02 \] \[ \sin \theta = \frac{0.02}{0.04} \] \[ \sin \theta = 0.5 \] Thus: \[ \theta = \sin^{-1}(0.5) = 30^\circ \]
The angular spread of the central maximum is given by \( 2\theta \): \[ \text{Angular spread} = 2 \times 30^\circ = 60^\circ \]
The angular spread of the central maxima of the diffraction pattern is \( 60^\circ \).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32