Given: Original wavelength $\lambda = 589.0 \text{ nm}$. Observed wavelength $\lambda' = 589.6 \text{ nm}$. The observed wavelength is longer than the original wavelength, so this is a **redshift**, meaning the galaxy is moving away from us ($v_{radial} > 0$).

Change in wavelength $\Delta\lambda = \lambda' - \lambda = 589.6 \text{ nm} - 589.0 \text{ nm} = 0.6 \text{ nm}$.

We use the Doppler shift formula for wavelength for $v_{radial} \ll c$: $\frac{\Delta\lambda}{\lambda} = \frac{v_{radial}}{c}$.

$v_{radial} = c \frac{\Delta\lambda}{\lambda}$. Speed of light $c = 3 \times 10^8 \text{ m/s}$.

$v_{radial} = (3 \times 10^8 \text{ m/s}) \times \frac{0.6 \text{ nm}}{589.0 \text{ nm}}$. Note that the units of wavelength cancel out.

$v_{radial} = 3 \times 10^8 \times \frac{0.6}{589.0} \text{ m/s} \approx 3 \times 10^8 \times 0.0010186 \text{ m/s} \approx 0.0030558 \times 10^8 \text{ m/s} = 3.0558 \times 10^5 \text{ m/s}$.

Convert to km/s: $3.0558 \times 10^5 \text{ m/s} = 305.58 \times 10^3 \text{ m/s} = 305.58 \text{ km/s}$.

The text gives 306 km/s, which is close.

The galaxy should move away from us at a speed of approximately 306 km/s.

(a) Same frequency in reflection and refraction: Reflection and refraction are phenomena that occur due to the interaction of light waves with the atoms and molecules of the medium at the interface. When the oscillating electric field of the incident light wave interacts with the electrons in the atoms, it causes these electrons to oscillate at the same frequency as the incident light. These oscillating electrons then re-emit light (scatter light) at the same frequency. The reflected and refracted waves are composed of this scattered light that interferes constructively in specific directions (governed by the laws of reflection and refraction). Since the driving frequency is the frequency of the incident light, the frequency of the induced oscillations and thus the frequency of the re-emitted (reflected and refracted) light is the same as the incident frequency.

(b) Reduction in speed and energy: No, the reduction in speed of light when it travels from a rarer to a denser medium does **not** imply a reduction in the energy carried by the light wave. The energy of a light wave in the wave picture is related to its amplitude, specifically, the intensity is proportional to the square of the amplitude ($I \propto a^2$). When light enters a denser medium, its amplitude is generally reduced due to reflection at the interface and absorption within the medium, but the speed change itself is a result of the wave interacting with the medium's properties (like its permittivity and permeability or the polarisability of atoms) which affects how the wave propagates, not the energy carried by each photon or unit area (unless intensity decreases). Energy of a photon is $E=h\nu$. Since frequency $\nu$ does not change during refraction, the energy of each photon remains unchanged.

(c) Intensity in photon picture: In the photon picture of light, intensity is related to the **number of photons** passing through a unit area per unit time. For a given frequency (and thus given energy per photon, $E=h\nu$), higher intensity means a larger number of photons crossing that unit area per unit time.

Given: Slit separation $d = 1 \text{ millimetre} = 1 \times 10^{-3} \text{ m}$. Screen distance $D = 1 \text{ metre} = 1 \text{ m}$. Wavelength of light $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$.

The fringe separation (or fringe width) $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.

$\beta = \frac{(5 \times 10^{-7} \text{ m}) \times (1 \text{ m})}{1 \times 10^{-3} \text{ m}} = \frac{5 \times 10^{-7}}{10^{-3}} \text{ m} = 5 \times 10^{-7 - (-3)} \text{ m} = 5 \times 10^{-4} \text{ m}$.

Convert to millimetres: $5 \times 10^{-4} \text{ m} = 0.5 \times 10^{-3} \text{ m} = 0.5 \text{ mm}$.

The fringe separation is 0.5 mm.

The fringe width is given by $\beta = \frac{\lambda D}{d}$. The angular fringe width is $\Delta\theta = \beta/D = \lambda/d$.

(a) Screen moved away from the slits: $D$ increases. $\beta = \frac{\lambda D}{d}$. As $D$ increases, the fringe separation $\beta$ increases. The angular separation $\Delta\theta = \lambda/d$ remains unchanged.

(b) Monochromatic source replaced by shorter wavelength: $\lambda$ decreases. $\beta = \frac{\lambda D}{d}$. As $\lambda$ decreases, the fringe separation $\beta$ decreases. Fringes become closer.

(c) Separation between two slits increased: $d$ increases. $\beta = \frac{\lambda D}{d}$. As $d$ increases, the fringe separation $\beta$ decreases. Fringes become closer.

(d) Source slit moved closer to the double-slit plane: The distance $S$ from the source slit to the double slit decreases. The width of the source slit is $s$. For the two slits $S_1, S_2$ to be coherently illuminated, the condition $s/S \le \lambda/d$ should be satisfied. As $S$ decreases, the angle $s/S$ subtended by the source at the double slit increases. If $S$ becomes too small such that $s/S > \lambda/d$, the coherence is reduced, and the interference pattern will become less sharp and eventually disappear. The fringe separation $\beta = \lambda D/d$ itself depends on $\lambda, D, d$ and is independent of $S$. So, the fringe separation remains the same, but the visibility of the fringes decreases and they may disappear if $S$ is too small.

(e) Width of the source slit increased: $s$ increases. Similar to (d), increasing the source slit width $s$ increases the angle $s/S$ subtended by the source at the double slit. If this angle becomes too large ($s/S > \lambda/d$), the interference pattern visibility decreases and it may disappear. The fringe separation remains the same.

(f) Monochromatic source replaced by white light: White light is a mixture of different wavelengths. Each wavelength produces its own interference pattern with its own fringe width ($\beta = \lambda D/d$). The central bright fringe ($n=0$) for all wavelengths is at the same position ($x=0$), so the central fringe is white. For other maxima/minima, the positions $x_n = n\lambda D/d$ depend on $\lambda$. Maxima and minima of different colours will be at different positions and will overlap. Near the center, the fringes will be coloured (spectrum-like). As we move away from the center, the overlap increases, and the distinct fringes disappear after a few bands.

From Example 10.3, slit separation $d = 1 \text{ mm}$, screen distance $D = 1 \text{ m}$, wavelength $\lambda = 500 \text{ nm}$. Fringe width for double-slit interference is $\beta = \lambda D/d$. The angular separation between double-slit maxima is $\Delta\theta_{int} = \lambda/d$.

For single-slit diffraction (slit width $a$), the central maximum is between the first minima on either side, which occur at angles $\theta = \pm \lambda/a$. The angular width of the central maximum is $2(\lambda/a) = 2\lambda/a$.

We want 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern. This means we want the 10th bright fringe from the center in the double-slit pattern to be located at or before the first minimum of the single-slit pattern.

Position of the $n$-th bright fringe in double-slit: $x_n = n\lambda D/d$. Position of the 10th bright fringe is $x_{10} = 10\lambda D/d$. The angular position is $\theta_{10\_int} = 10\lambda/d$.

Position of the first minimum in single-slit: $x_{min1} = \lambda D/a$. The angular position is $\theta_{min1} = \lambda/a$.

For the 10 maxima to be within the central maximum, the angular width covered by the central maximum must be large enough to contain the angular positions of these maxima. The central maximum extends from $-\lambda/a$ to $+\lambda/a$. We need the angular position of the 10th bright fringe from the center to be less than or equal to the angular position of the first minimum of the single slit pattern.

Let's consider one side. The bright fringes are at $0, \pm \lambda D/d, \pm 2\lambda D/d, ...$. The angular positions are $0, \pm \lambda/d, \pm 2\lambda/d, ...$. We want the 10th maximum to be within the central diffraction peak. The central diffraction peak spans from $-\lambda/a$ to $+\lambda/a$. We want the position of the 10th bright fringe ($n=10$) to be less than the first minimum ($n=1$) of the single slit pattern. Is it 10 maxima *between* the central one and the first minimum, or 10 maxima *including* the central one? "10 maxima of the double slit pattern within the central maximum". The central maximum includes the $n=0$ fringe. The question likely means there are 10 bright fringes on *one side* of the central maximum, plus the central one, for a total of 21 fringes within the central peak. Or it means the region from the center out to the 10th bright fringe on *each side* fits within the central maximum. Or it means the 10th bright fringe on each side is within the central peak. Let's assume it means the region from $x=0$ to the 10th bright fringe ($n=10$) is within the region $0$ to the first minimum ($\lambda D/a$). Or the region from the -10th to +10th bright fringe is within the central peak. This means the range of positions from $x_{-10}$ to $x_{10}$ is contained within $[-\lambda D/a, +\lambda D/a]$.

Let's assume it means that the 10th bright fringe on *each side* is just at the edge of the central maximum (or slightly inside). This means $x_{10} \le \lambda D/a$. $10\lambda D/d \le \lambda D/a$. Canceling $\lambda D$, we get $10/d \le 1/a$, or $a \le d/10$. This would give 21 bright fringes (0 to 10 on each side). This is probably not intended as "10 maxima".

Let's assume it means there are 10 bright fringes total within the central maximum, including the central one. This would mean 5 on each side. So the 5th bright fringe on each side is at the edge or inside. $x_5 \le \lambda D/a$. $5\lambda D/d \le \lambda D/a$. $a \le d/5$. This would mean 11 bright fringes (0 to 5 on each side). Again, probably not intended as "10 maxima".

Let's assume it means there are 10 bright fringes *between* the two first minima of the single slit pattern. The central fringe ($n=0$) is one of them. So there are 9 other maxima to be distributed on either side, 4.5 on each side? No. Let's assume it means the entire width covered by 10 fringes (from center to 9th max on one side, plus center to 9th max on the other side, plus the widths between them) is contained. No. Let's assume it means there are 10 bright fringes *on one side* of the central maximum, plus the central one, making 11 fringes total. Then the 10th fringe is at the edge. $x_{10} = \lambda D/a$. $10 \lambda D/d = \lambda D/a$. $a = d/10$. Let's assume it means the central maximum contains a total of 10 fringes. This would mean 5 on each side of the center. So the 5th fringe on each side is at the edge. $x_5 = \lambda D/a$. $5 \lambda D/d = \lambda D/a$. $a=d/5$. Let's re-read the text's solution: "We want a , a / a", "10 2", "d a". The first part refers to angles $\lambda/a$ and $\lambda/d$. "10 2" likely refers to a ratio. The condition is that the width of the central maximum of the single slit (2$\lambda D/a$) is large enough to contain 10 fringe spacings (which is related to $10 \times \beta$). The angular width of the central maximum is $2\lambda/a$. The angular separation of fringes is $\lambda/d$. The number of fringes in the central maximum is the width of the central maximum divided by the fringe spacing. Number of fringes $\approx (2\lambda D/a) / (\lambda D/d) = 2d/a$. If we want 10 maxima (which implies about 10 fringe spacings) within the central maximum, then $2d/a \approx 10$. So $a \approx d/5$. Let's consider the number of *bright* fringes. The central maximum spans from $-\lambda/a$ to $+\lambda/a$. Bright fringes are at $x_n = n\lambda D/d$. We want $x_n \le \lambda D/a$. $n\lambda D/d \le \lambda D/a \implies n/d \le 1/a \implies n \le d/a$. This $n$ is the maximum fringe order within the central maximum on one side. So the number of bright fringes on one side (excluding center) is $\lfloor d/a \rfloor$. Total number of bright fringes is $2\lfloor d/a \rfloor + 1$. If we want 10 maxima *within* the central maximum, perhaps this means $2\lfloor d/a \rfloor + 1 \approx 10$, which is $2\lfloor d/a \rfloor \approx 9$, $\lfloor d/a \rfloor \approx 4.5$. So the 5th bright fringe is roughly at the edge. $5 \approx d/a \implies a \approx d/5$. The text solution gives "10 maxima ... within the central maximum". It gives the relation "10 2" and "d a". This likely refers to the total number of maxima being 10. The bright fringes are at positions $x_n = n \beta = n (\lambda D/d)$, for $n = 0, \pm 1, \pm 2, ...$. The dark fringes of the single slit pattern are at $\theta = \pm \lambda/a, \pm 2\lambda/a, ...$. The central bright maximum of the single slit pattern extends from $x = -\lambda D/a$ to $x = \lambda D/a$. We want the positions of the bright fringes $|x_n| \le \lambda D/a$. So $|n| \lambda D/d \le \lambda D/a$. $|n|/d \le 1/a$. $|n| \le d/a$. The bright fringes are for $n=0, \pm 1, \pm 2, ..., \pm N_{max}$, where $N_{max} = \lfloor d/a \rfloor$. The total number of bright fringes is $2 N_{max} + 1$. If we want exactly 10 maxima, $2N_{max} + 1 = 10$, which is not possible for integer $N_{max}$. Perhaps it means 10 maxima including the central one, on one side? No. Let's assume the statement "10 maxima of the double slit pattern within the central maximum" means there are 10 bright fringes in total, including the central one. Then there are 4.5 fringes on each side, which doesn't make sense. Let's assume it means 5 bright fringes on each side of the central maximum, plus the central one. Total 11. Let's assume the text is correct in its calculation derivation steps $10 \lambda/d = 2 \lambda/a$. This is the angular position of the 10th maximum equal to the angular width of the central maximum. This means the 10th maximum is at the edge of the central peak. The bright fringes are at $0, \pm \lambda/d, \pm 2\lambda/d, ..., \pm n \lambda/d, ...$. We want the 10th bright fringe from the center ($n=10$) to be at the edge of the central maximum ($\theta = \lambda/a$). So $10 \lambda/d = \lambda/a$. $10/d = 1/a$. $a = d/10$. However, the text solution gives $a = d/5$. Let's check if $a = d/5$ gives 10 maxima in the central peak. If $a = d/5$, then the central maximum spans from $-\lambda D/(d/5)$ to $+\lambda D/(d/5)$, which is $-5\lambda D/d$ to $+5\lambda D/d$. The bright fringes are at $x_n = n \lambda D/d$. We want $|x_n| \le 5\lambda D/d$. So $|n| \lambda D/d \le 5\lambda D/d$. $|n| \le 5$. The possible integer values for $n$ are $0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$. This gives a total of $2 \times 5 + 1 = 11$ bright fringes within the central maximum (including the one at $n=5$ and $n=-5$). This doesn't give exactly 10. Perhaps "10 maxima" refers to a specific number of maxima between the first minima. If the central maximum is between $-\lambda/a$ and $+\lambda/a$, the bright fringes are at $0, \pm \lambda D/d, \pm 2\lambda D/d, ...$. Let's assume the first minimum occurs exactly at the position of a double-slit bright fringe. The first minimum is at $\theta = \lambda/a$. A bright fringe is at $\theta = n\lambda/d$. So $\lambda/a = n\lambda/d$. $d/a = n$. This means the first minimum of the single slit coincides with the $n$-th bright fringe of the double slit. The central maximum includes fringes from $n=-n$ to $n=+n$. Total $2n+1$ fringes. If this total is 10, $2n+1=10$, not possible. If the number of bright fringes *on one side* within the central peak is 10, then the 10th fringe is at the edge or inside. $|x_{10}| \le \lambda D/a$. $10\lambda D/d \le \lambda D/a$. $a \le d/10$. Let's revisit the interpretation: "10 maxima of the double slit pattern within the central maximum". The central peak contains the $n=0$ fringe. The other maxima are on either side. Perhaps it means there are 5 maxima on the positive side and 5 on the negative side (excluding the center). This is unlikely. Let's consider the possibility that the 10th bright fringe *is the first minimum*. This means $x_{10} = x_{min1}$. $10\lambda D/d = \lambda D/a$. $10/d = 1/a$. $a = d/10$. This gives 21 fringes in total in the central peak ($n=0$ to $n=\pm 10$). Let's assume the text's condition means that there are 10 interference maxima *between* the first two minima of the single-slit pattern. The first two minima are at $x = -\lambda D/a$ and $x = +\lambda D/a$. The number of interference maxima in this range is the number of bright fringes whose positions $x_n$ satisfy $-\lambda D/a < x_n < \lambda D/a$. $x_n = n \lambda D/d$. So $-\lambda D/a < n \lambda D/d < \lambda D/a$. Divide by $\lambda D$: $-1/a < n/d < 1/a$. $-d/a < n < d/a$. The number of integers $n$ in this range is related to $2d/a$. If $2d/a$ is an integer $N$, the points $\pm N/2$ would be the boundaries. The number of maxima between $-d/a$ and $d/a$ (exclusive of endpoints) is $2 \times \lfloor d/a - \epsilon \rfloor + 1$ if $d/a$ is not an integer. If $d/a$ is an integer $n$, then the minima occur at the positions of the $n$-th bright fringes. If $d/a = 5$, then the first minimum is at $\theta = \lambda/a = 5\lambda/d$. Bright fringes are at $0, \pm \lambda/d, \pm 2\lambda/d, \pm 3\lambda/d, \pm 4\lambda/d, \pm 5\lambda/d$. The minima are at $\pm 5\lambda/d$. So the bright fringes at $\pm 5\lambda/d$ are at the edge of the central peak. The integers $n$ such that $|n|\lambda D/d \le \lambda D/a$ are $|n| \le d/a$. If $d/a = 5$, then $|n| \le 5$. The integers are $0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$. Total $2 \times 5 + 1 = 11$ bright fringes. Perhaps the text meant that the 10th bright fringe *on one side* is just inside the central peak. $x_{10} < \lambda D/a$. $10 \lambda D/d < \lambda D/a$. $a < d/10$. Smallest a would be just less than $d/10$. Let's re-examine the text's calculation: "We want a , a / a", "10 2", "d a". This looks like setting an angle equal to a multiple of another angle. The angular position of the 10th bright fringe is $10 \lambda/d$. The angular half-width of the central maximum is $\lambda/a$. If the 10th bright fringe is at the edge of the central maximum, $10 \lambda/d = \lambda/a$, so $a = d/10$. If the text meant the 10th maximum *including* the central one, and on one side, it would be the 9th maximum on one side. $9 \lambda/d = \lambda/a$. $a = d/9$. If it means the 10th maximum *including* the central one, and on both sides, that's 5 on each side. $x_5 \le \lambda D/a$. $5\lambda D/d \le \lambda D/a$. $a \le d/5$. Given the text's calculation result $a = d/5$. Let's assume this is the intended meaning. The width of each slit should be $d/5$. $d = 1 \text{ mm} = 10^{-3}$ m. $a = (10^{-3} \text{ m}) / 5 = 0.2 \times 10^{-3} \text{ m} = 0.2 \text{ mm}$. The width of each slit should be 0.2 mm. Let's check if $a=d/5$ gives 10 maxima *between* the first minima (excluding the central one). No, it gives 11 total including center. Let's consider the possibility that the angle of the 10th maximum is 10 times the angular separation between fringes, so $10 (\lambda/d)$. And the angle of the first minimum is $\lambda/a$. If $10 (\lambda/d)$ is within $\lambda/a$, then $10/d \le 1/a$. Let's assume the text's solution $a=d/5$ is correct and it means the 5th bright fringe on each side coincides with the first minimum. That would give 11 fringes total in the central peak. Maybe "10 maxima" is an approximation or includes something else. Let's trust the calculation $a=d/5$. Final answer based on text calculation: Slit width $a = d/5 = 1 \text{ mm} / 5 = 0.2 \text{ mm}$.

From Example 10.3, $d = 1$ mm. We need to find the width of each slit $a$. We want 10 maxima of the double-slit pattern to be within the central maximum of the single-slit pattern. The angular positions of the bright fringes in the double-slit pattern are $\theta_n = n \lambda/d$, where $n = 0, \pm 1, \pm 2, ...$. The central maximum of the single-slit pattern is defined by the angular region between the first minima, which occur at $\theta = \pm \lambda/a$. We want the bright fringes to be within this region, so $|\theta_n| \le \lambda/a$. $|n| \lambda/d \le \lambda/a$. $|n| \le d/a$. The bright fringes are for integer values of $n$. The total number of bright fringes inside the central maximum (excluding the exactly edge points) is $2 \times \lfloor d/a - \epsilon \rfloor + 1$. If $d/a$ is an integer $N$, the first minima of single slit coincide with the $N$-th bright fringes of double slit. The central peak contains bright fringes for $|n| \le N = d/a$. The integers are $n = -N, -N+1, ..., -1, 0, 1, ..., N-1, N$. The number of such integers is $2N + 1 = 2(d/a) + 1$. If we want 10 maxima, maybe it means 10 maxima *on one side*, including the central one? No. Let's go with the text's derivation conclusion: $a = d/5$. This implies that the 5th bright fringe ($n=5$) coincides with the first minimum ($\theta = \lambda/a = \lambda/(d/5) = 5\lambda/d$). Bright fringes are at $0, \pm \lambda/d, \pm 2\lambda/d, \pm 3\lambda/d, \pm 4\lambda/d, \pm 5\lambda/d, ...$. The central peak is from $-\lambda/a$ to $+\lambda/a$, which is $-5\lambda/d$ to $+5\lambda/d$. The bright fringes within this range are for $n=0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$. There are $2 \times 5 + 1 = 11$ bright fringes in total. Maybe the question intends "10 maxima" to mean from $n=-4$ to $n=5$, total $5-(-4)+1=10$? No. Let's assume the text meant "the tenth bright fringe is just at the edge of the central maximum". This means $x_{10} = \lambda D/a$. $10 \lambda D/d = \lambda D/a$. $a = d/10$. This gives 21 fringes total. Let's assume the text meant "there are 10 bright fringes on one side of the central maximum (not including the central one) within the central maximum". This is 10 fringes on one side, so $n=1$ to $n=10$. The 10th fringe is at the edge. $x_{10} = \lambda D/a$. $10 \lambda D/d = \lambda D/a$. $a = d/10$. Let's assume the text meant "there are 10 bright fringes *between* the central maximum and the first minimum on one side". This is 10 bright fringes between $x=0$ and $x=\lambda D/a$. The bright fringes are at $n\lambda D/d$. So $0 < n\lambda D/d < \lambda D/a$. $0 < n/d < 1/a$. $0 < n < d/a$. We need 10 integers $n$ in this range. So $n=1, 2, ..., 10$. This means $10 < d/a \le 11$. $a > d/11$ and $a \le d/10$. E.g., $a = d/10.5$. Let's assume the text meant that the first minimum of the single slit pattern occurs at the location of the 10th bright fringe on *each side* of the central maximum. So $x_{\pm 10} = x_{min1}$. $|10| \lambda D/d = \lambda D/a$. $10/d = 1/a$. $a = d/10$. Given the high probability of misinterpretation and the directness of the text's calculation $a=d/5$, let's rely on that result and interpret it as intended by the question's phrasing in the source. The calculation $10 (\lambda/d) = 2 (\lambda/a)$ seems to imply setting the angular width of the region containing the first 10 fringes (angular spacing $10 \lambda/d$) equal to the angular width of the central maximum (2$\lambda/a$). No, angular spacing is $\lambda/d$. Angular position of 10th fringe is $10\lambda/d$. The calculation in the text implies $10 \times (\lambda/d) = 2 \times (\lambda/a)$. This equates the angular position of the 10th bright fringe of the double slit pattern to the angular width of the central maximum of the single slit pattern. This means the 10th bright fringe is at the edge of the central maximum. However, the text also has a step "10 2" and "d a". This looks like $10/d = 2/a$. $10a = 2d$. $a = 2d/10 = d/5$. This simple relation gives $a=d/5$. Let's trust this simple relationship. If $a=d/5$, the first minimum of the single slit is at $\theta = \lambda/a = \lambda/(d/5) = 5\lambda/d$. The bright fringes of the double slit are at $\theta_n = n\lambda/d$. So when $a=d/5$, the first minimum is at $\theta_n$ where $n=5$. This means the 5th bright fringe is at the first minimum. The central peak contains fringes $n=0, \pm 1, \pm 2, \pm 3, \pm 4$. No, it should contain $n=0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$. Total 11 fringes. Let's assume the text intended "10 fringes" meaning the central fringe plus 9 on one side (or 4 on each side plus 1 central, for total 9?). Let's assume the relationship $a=d/5$ is correct from the text's calculation steps. The width of each slit should be $d/5$. Given $d=1$ mm, $a = (1 \text{ mm})/5 = 0.2$ mm. Final Answer based on text calculation: The width of each slit should be 0.2 mm.

From Example 10.3, $d = 1$ mm. We need to find the width of each slit $a$ such that there are 10 maxima of the double slit pattern within the central maximum of the single slit pattern. The central maximum of the single slit diffraction pattern extends from the first minimum at $\theta = -\lambda/a$ to the first minimum at $\theta = +\lambda/a$. The angular positions of the bright fringes in the double slit interference pattern are given by $\theta_n = n\lambda/d$, for $n = 0, \pm 1, \pm 2, ...$. We want these bright fringes to fall within the central maximum, i.e., $|\theta_n| \le \lambda/a$. This means $|n\lambda/d| \le \lambda/a$, which simplifies to $|n|/d \le 1/a$, or $|n| \le d/a$. The maximum integer value of $n$ on either side of the central fringe is $N_{max} = \lfloor d/a \rfloor$. The total number of bright fringes within the central maximum is $2N_{max} + 1$. We want this total number to be 10. However, $2N_{max}+1=10$ gives $2N_{max}=9$, $N_{max}=4.5$, which is not an integer. There might be an issue with the number '10' in the question, or its interpretation.

Let's consider the possibility that the 10 maxima are meant to be on one side of the central maximum. Including the central one would mean 11 maxima. Not including the central one would mean 10 maxima on the positive side and 10 on the negative side, plus the central one, total 21. This also seems unlikely.

Let's assume the question implies that the first minimum of the single slit diffraction pattern coincides with the $n$-th bright fringe of the double slit pattern, and that there are 10 bright fringes in total up to this point (including the central one, and the one at the minimum). If the central maximum contains $2N_{max}+1$ bright fringes, and we want this to be approximately 10, then $N_{max} \approx 4.5$. This suggests the 5th bright fringe on each side is just at the edge of the central maximum. So, the angular position of the 5th bright fringe ($n=5$) is approximately equal to the angular position of the first minimum ($\theta = \lambda/a$). $\theta_5 = 5\lambda/d$. So, $5\lambda/d \approx \lambda/a$, which means $a \approx d/5$.

Using $d = 1$ mm and $a \approx d/5$, we get $a \approx (1 \text{ mm})/5 = 0.2$ mm.

This is consistent with the provided answer calculation in the text. Let's proceed with $a = d/5$.

The width of each slit should be approximately $d/5$.

$a = (1 \text{ mm}) / 5 = 0.2 \text{ mm}$.

The width of each slit should be 0.2 mm.

Given: Aperture width $a = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$. Wavelength $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$.

The distance for which ray optics is a good approximation is up to the Fresnel distance $z_F = a^2/\lambda$.

$z_F = \frac{(3 \times 10^{-3} \text{ m})^2}{5 \times 10^{-7} \text{ m}} = \frac{9 \times 10^{-6} \text{ m}^2}{5 \times 10^{-7} \text{ m}} = \frac{9}{5} \times 10^{-6 - (-7)} \text{ m} = 1.8 \times 10^1 \text{ m} = 18 \text{ m}$.

For distances up to about 18 meters, ray optics is a good approximation for an aperture of 3 mm and light of 500 nm wavelength. Beyond this distance, diffraction effects become significant.

Let the two crossed polaroids be $P_1$ and $P_3$. Their pass axes are perpendicular to each other (angle $90^\circ$). Let unpolarised light be incident on $P_1$. The intensity of light transmitted by $P_1$ will be $I_0 = I_{incident}/2$. This light is linearly polarised along the pass axis of $P_1$.

Now, $P_3$ is placed after $P_1$, with its pass axis perpendicular to $P_1$'s pass axis. If only $P_1$ and $P_3$ were present, the light from $P_1$ (polarised along $P_1$'s axis) would be incident on $P_3$ with the angle between polarisation direction and $P_3$'s pass axis being 90°. By Malus's law, the intensity transmitted by $P_3$ would be $I_0 \cos^2(90^\circ) = 0$. No light would pass through the combination.

Now, a third polaroid $P_2$ is inserted between $P_1$ and $P_3$. Let the pass axis of $P_2$ make an angle $\theta$ with the pass axis of $P_1$.

• Light from $P_1$ is linearly polarised with intensity $I_0$.
• This light is incident on $P_2$. The component of the electric field along the pass axis of $P_2$ (at angle $\theta$ to $P_1$'s axis) passes through $P_2$. The intensity transmitted by $P_2$ is $I_1 = I_0 \cos^2\theta$. This light is linearly polarised along the pass axis of $P_2$.
• This light of intensity $I_1$ and polarised along $P_2$'s axis is incident on $P_3$. The pass axis of $P_3$ is perpendicular to $P_1$'s axis, so the angle between $P_2$'s axis and $P_3$'s axis is $(90^\circ - \theta)$.
• The intensity of light transmitted by $P_3$ is $I_{transmitted} = I_1 \cos^2(90^\circ - \theta) = I_1 \sin^2\theta$.

Substitute $I_1 = I_0 \cos^2\theta$ into the expression for $I_{transmitted}$:

$I_{transmitted} = (I_0 \cos^2\theta) \sin^2\theta = I_0 (\cos\theta \sin\theta)^2 = I_0 \left(\frac{1}{2}\sin(2\theta)\right)^2 = \frac{I_0}{4} \sin^2(2\theta)$.

As the intermediate polaroid $P_2$ is rotated (changing $\theta$), the transmitted intensity varies. The intensity is minimum (zero) when $\sin(2\theta) = 0$, which happens when $2\theta = 0^\circ, 180^\circ, 360^\circ, ...$, so $\theta = 0^\circ, 90^\circ, 180^\circ, ...$. This occurs when $P_2$'s axis is parallel or perpendicular to $P_1$'s axis (and thus parallel or perpendicular to $P_3$'s axis). The intensity is maximum ($I_0/4$) when $\sin(2\theta) = \pm 1$, which happens when $2\theta = 90^\circ, 270^\circ, ...$, so $\theta = 45^\circ, 135^\circ, ...$. This occurs when $P_2$'s axis is at 45° to the axes of both $P_1$ and $P_3$.

Intensity of transmitted light: $\mathbf{I_{transmitted} = \frac{I_0}{4} \sin^2(2\theta)}$. As the intermediate polaroid is rotated, the intensity goes from zero to a maximum of $I_0/4$ and back to zero every 90° of rotation of the intermediate polaroid.

Given: Unpolarised light incident on a plane glass surface. Let the interface be between air (medium 1, $n_1 \approx 1$) and glass (medium 2, $n_2 = 1.5$, from Example 9.3(ii)). We want to find the angle of incidence ($i$) such that the reflected ray and the refracted ray are perpendicular to each other. This condition defines Brewster's angle ($i_B$).

According to Brewster's law, when the reflected and refracted rays are perpendicular, the angle of incidence is Brewster's angle $i_B$, and $\tan i_B = n_{21} = n_2/n_1$.

Here, $n_1 = 1$ (air) and $n_2 = 1.5$ (glass).

$\tan i_B = 1.5 / 1 = 1.5$.

$i_B = \tan^{-1}(1.5)$.

Using a calculator, $\tan^{-1}(1.5) \approx 56.3^\circ$. The text gives 57°. Let's use 56.3° or 57° as approximation.

The angle of incidence should be approximately 56.3° (or 57°) so that the reflected and refracted rays are perpendicular to each other. At this angle, the reflected light will be completely polarised perpendicular to the plane of incidence.