The frequency of the wave cannot change as that depends on the source producing it. Therefore, in \( v = f\lambda \), the speed and the wavelength will change.
CloseSnell's law is given by the expression \( \frac{\sin{i}}{\sin{r}} = \frac{v_1}{v_2} \), when the angle is measured between the incoming wavefronts and the normal to the boundary. Here, the question gives us the complementary angle, which is \( 90^\circ - \theta \). Using the identity \( \sin{(90^\circ - \theta)} = \cos{\theta} \), the correct expression becomes \( \frac{\cos{i}}{\cos{r}} = \frac{v_1}{v_2} \).
CloseWhen 2 waves superpose, the individual displacements will add vectorially (meaning the sign of the displacement also matters).
CloseThis is Snell's law \(\frac{\sin{i}}{\sin{r}}=\frac{n_2}{n_1}\). Critical angle means the reflected angle is \(90^\circ\), meaning \(\sin{i}=\frac{n_2}{n_1}\) and since sin can only give values less than or equal to 1, \(n_1\) must be larger than \(n_2\) for the equation to be possible, hence B.
CloseSince sound has a much longer wavelength, it experiences a much larger diffraction around a corner than light, hence C.
CloseThe frequency cannot change, as that only depends on the source. Since air has a smaller refractive index than water, the speed will increase, and since speed and wavelength are directly proportional, the wavelength will increase too.
CloseThe fringe spacing is calculated using the formula:
\[ s = \frac{\lambda D}{d} \]
Substitute the values:
\[ s = \frac{600 \times 10^{-9} \times 2.0}{0.1 \times 10^{-3}} = 1.2 \times 10^{-3} \, \mathrm{m} = 1.2 \, \mathrm{mm}.\]
CloseThe angle for the first minimum in single-slit diffraction is given by:
\[ a \sin \theta = m \lambda \quad (m = 1) \]
Substitute the values:
\[ \sin \theta = \frac{\lambda}{a} = \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} = 0.01 \]
\[ \theta = \arcsin(0.01) \approx 0.57^\circ. \]
CloseFor constructive interference, the path difference must satisfy:
\[ \Delta x = n\lambda, \quad n = 0, 1, 2, \dots \]
Here, the path difference is:
\[ \Delta x = 1500 \, \mathrm{nm}, \quad \lambda = 500 \, \mathrm{nm}. \]
\[ \Delta x = 3\lambda, \quad n = 3. \]
Since the path difference is an integer multiple of the wavelength, the interference is constructive.
CloseFor destructive interference, the path difference must satisfy:
\[ \Delta x = \left(n + \frac{1}{2}\right) \lambda, \quad n = 0, 1, 2, \dots \]
Here, the path difference is:
\[ \Delta x = 0.6 \, \mathrm{m}, \quad \lambda = 1.2 \, \mathrm{m}. \]
\[ \Delta x = \frac{1}{2} \lambda. \]
Since the path difference corresponds to \(\frac{1}{2} \lambda\), the interference is destructive.
CloseSnell’s law is given by:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2. \]
Here:
\[ n_1 = 1.00, \quad \theta_1 = 30^\circ, \quad n_2 = 1.50. \]
\[ 1.00 \times \sin(30^\circ) = 1.50 \times \sin \theta_2. \]
\[ 0.5 = 1.50 \times \sin \theta_2. \]
\[ \sin \theta_2 = \frac{0.5}{1.50} = 0.333. \]
\[ \theta_2 = \arcsin(0.333) \approx 19.5^\circ. \]
CloseThe critical angle is given by:
\[ \sin \theta_c = \frac{n_2}{n_1}, \]
where \(n_2 < n_1\). Here:
\[ \sin \theta_c = \frac{1.00}{1.50} = 0.666. \]
\[ \theta_c = \arcsin(0.666) \approx 41.8^\circ. \]
CloseThe order \(n\) of the bright fringe is determined by the condition for constructive interference:
\[ \Delta x = n \lambda. \]
Here:
\[ \Delta x = 1.8 \, \mu\mathrm{m} = 1.8 \times 10^{-6} \, \mathrm{m}, \quad \lambda = 600 \, \mathrm{nm} = 6.0 \times 10^{-7} \, \mathrm{m}. \]
\[ n = \frac{\Delta x}{\lambda} = \frac{1.8 \times 10^{-6}}{6.0 \times 10^{-7}} = 3. \]
Thus, the third-order bright fringe is observed.
Close