Wave Optics

Given:
- Wavelength of incident light: $\lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
- Refractive index of water: $\mu = 1.33$
- Speed of light in vacuum/air: $c = 3.0 \times 10^8 \text{ m s}^{-1}$

Frequency of incident light:
$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{589 \times 10^{-9}} \approx 5.09 \times 10^{14} \text{ Hz}$$

(a) Reflected Light:

Reflection occurs in the same medium (air). The frequency, wavelength, and speed do not change upon reflection.

- Wavelength: $\lambda_{\text{reflected}} = 589 \text{ nm}$
- Frequency: $\nu_{\text{reflected}} = 5.09 \times 10^{14} \text{ Hz}$
- Speed: $v_{\text{reflected}} = 3.0 \times 10^8 \text{ m s}^{-1}$

(b) Refracted Light:

When light enters a denser medium, its frequency remains unchanged but its speed and wavelength change.

- Frequency: Frequency does not change on refraction.
$$\nu_{\text{refracted}} = 5.09 \times 10^{14} \text{ Hz}$$

- Speed: Using $\mu = \dfrac{c}{v}$:
$$v = \frac{c}{\mu} = \frac{3.0 \times 10^8}{1.33} \approx 2.26 \times 10^8 \text{ m s}^{-1}$$

- Wavelength: Using $\lambda_{\text{water}} = \dfrac{v}{\nu}$:
$$\lambda_{\text{water}} = \frac{2.26 \times 10^8}{5.09 \times 10^{14}} \approx 444 \text{ nm}$$

Alternatively: $\lambda_{\text{water}} = \dfrac{\lambda}{\mu} = \dfrac{589}{1.33} \approx 443 \text{ nm}$

Summary:
| Quantity | Reflected | Refracted |
|---|---|---|
| Wavelength | 589 nm | ~443 nm |
| Frequency | $5.09 \times 10^{14}$ Hz | $5.09 \times 10^{14}$ Hz |
| Speed | $3.0 \times 10^8$ m/s | $2.26 \times 10^8$ m/s |

(a) Light diverging from a point source:

When light diverges from a point source, it spreads out equally in all directions. All points equidistant from the source are in the same phase. Therefore, the wavefront is spherical (a series of concentric spheres centred at the point source).

(b) Light emerging out of a convex lens when a point source is placed at its focus:

When a point source is placed at the focus of a convex lens, the diverging spherical waves are converted into parallel rays after refraction through the lens. All these parallel rays are in the same phase, so the wavefront is plane (flat).

(c) The portion of the wavefront of light from a distant star intercepted by the Earth:

A distant star is effectively at infinity. The spherical wavefronts originating from it have an extremely large radius by the time they reach the Earth. The small portion intercepted by the Earth is essentially a flat surface. Therefore, the wavefront is plane (flat).

(a) Speed of light in glass:

Given: $\mu_{\text{glass}} = 1.5$, $c = 3.0 \times 10^8 \text{ m s}^{-1}$

Formula: $\mu = \dfrac{c}{v}$

$$v = \frac{c}{\mu} = \frac{3.0 \times 10^8}{1.5}$$

$$\boxed{v = 2.0 \times 10^8 \text{ m s}^{-1}}$$

(b) Dependence of speed on colour:

No, the speed of light in glass is not independent of the colour (wavelength) of light. This is because the refractive index of glass varies with wavelength — a phenomenon called dispersion.

The refractive index of glass is higher for violet light than for red light (\mu_{\text{violet}} > \mu_{\text{red}}).

Since $v = \dfrac{c}{\mu}$, a higher refractive index means a lower speed.

Therefore, violet light travels slower than red light in a glass prism.

Given:
- Slit separation: $d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}$
- Distance to screen: $D = 1.4 \text{ m}$
- Distance between central fringe and 4th bright fringe: $y_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$

Concept: The position of the $n$-th bright fringe is given by:
$$y_n = \frac{n\lambda D}{d}$$

For the 4th bright fringe ($n = 4$):
$$y_4 = \frac{4\lambda D}{d}$$

Solving for $\lambda$:
$$\lambda = \frac{y_4 \cdot d}{4D}$$

$$\lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$$

$$\lambda = \frac{1.2 \times 0.28 \times 10^{-5}}{5.6}$$

$$\lambda = \frac{0.336 \times 10^{-5}}{5.6} = 0.06 \times 10^{-5} \text{ m}$$

$$\boxed{\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}}$$

Given:
- At path difference $\Delta = \lambda$, intensity $I = K$
- Find intensity at path difference $\Delta' = \lambda/3$

Concept: The phase difference $\phi$ corresponding to a path difference $\Delta$ is:
$$\phi = \frac{2\pi}{\lambda} \cdot \Delta$$

The resultant intensity in Young's double slit experiment is:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
where $I_0$ is the intensity due to each slit.

Step 1: Find $I_0$ using the condition at $\Delta = \lambda$:

Phase difference: $\phi_1 = \dfrac{2\pi}{\lambda} \times \lambda = 2\pi$

$$K = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0 \times (-1)^2 = 4I_0$$

So $I_0 = \dfrac{K}{4}$.

Step 2: Find intensity at $\Delta' = \lambda/3$:

Phase difference: $\phi_2 = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}$

$$I' = 4I_0 \cos^2\left(\frac{\phi_2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$

$$I' = 4 \times \frac{K}{4} \times \left(\frac{1}{2}\right)^2 = K \times \frac{1}{4}$$

$$\boxed{I' = \frac{K}{4}}$$

Given:
- $\lambda_1 = 650 \text{ nm}$, $\lambda_2 = 520 \text{ nm}$
- Slit separation: $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ *(standard assumed value for this problem)*
- Distance to screen: $D = 1.2 \text{ m}$ *(standard assumed value for this problem)*

*(Note: The problem does not specify $d$ and $D$; the answer is expressed in terms of $D/d$.)*

(a) Distance of 3rd bright fringe for $\lambda_1 = 650$ nm:

Formula: $y_n = \dfrac{n\lambda D}{d}$

For $n = 3$, $\lambda_1 = 650 \text{ nm} = 650 \times 10^{-9} \text{ m}$:

$$y_3 = \frac{3 \times 650 \times 10^{-9} \times D}{d}$$

Using $D = 1.2$ m and $d = 2 \times 10^{-3}$ m:

$$y_3 = \frac{3 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$

$$y_3 = \frac{2340 \times 10^{-9}}{2 \times 10^{-3}} = 1170 \times 10^{-6} \text{ m}$$

$$\boxed{y_3 = 1.17 \times 10^{-3} \text{ m} \approx 1.17 \text{ mm}}$$

(b) Least distance where bright fringes of both wavelengths coincide:

Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$:

$$n_1 \lambda_1 = n_2 \lambda_2$$

$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5}$$

So the smallest integers satisfying this are $n_1 = 4$ and $n_2 = 5$.

Least distance (using $\lambda_1$, $n_1 = 4$):

$$y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$

$$y = \frac{3120 \times 10^{-9}}{2 \times 10^{-3}} = 1560 \times 10^{-6} \text{ m}$$

$$\boxed{y = 1.56 \times 10^{-3} \text{ m} = 1.56 \text{ mm}}$$

Verification: $y = \dfrac{5 \times 520 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.56 \text{ mm}$ ✓