Diffraction

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   At the first minimum, the path difference between waves from the two edges of the slit is $\lambda$ (i.e. $a\sin\theta=\lambda$). Phase difference $\;=\dfrac{2\pi}\lambda\times(\text{path difference})=\dfrac{2\pi}\lambda\times\lambda=2\pi\text{ rad}$

2. 2.↑ back to the question

   Angular width of central maximum $\;\theta_0=\dfrac{2\lambda}a$, so $\theta_0\propto\dfrac1a$ and $\theta_0\propto\lambda$. (i) Since $\theta_0\propto1/a$, decreasing the slit width $a$ INCREASES the angular width of the central maximum. (ii) Since $\theta_0\propto\lambda$, and polychromatic (white) light contains a range of wavelengths, each colour produces a central maximum of a different (larger for larger $\lambda$) angular width. The central maximum thus becomes a band of overlapping colours — appearing white at the very centre, with the diffracted pattern getting dispersed into its constituent colours towards the edges (red having the widest, violet the narrowest, central maximum).

3. 3.↑ back to the question

   (i) Width of central maximum $\;\beta_0=\dfrac{2D\lambda}a$; since $\beta_0\propto1/a$, decreasing the slit width $a$ INCREASES the size (width) of the central bright band. (ii) The total light energy passing through the slit is proportional to the slit width $a$, so decreasing $a$ reduces the total light power available; since this reduced power is now spread over a WIDER central band, the intensity of the central bright band DECREASES.

4. 4.↑ back to the question

   Single-slit diffraction intensity pattern: a tall, wide central maximum, flanked by much weaker secondary maxima of rapidly decreasing intensity (roughly $1/n^2$ falloff), symmetric about the centre, with minima at path differences $n\lambda$ ($n=\pm1,\pm2,\dots$). Double-slit interference intensity pattern: a series of equally-spaced bright fringes, ALL of the same maximum intensity, separated by completely dark (zero-intensity) minima. Two differences: (1) In interference (double slit), all bright fringes are of the same intensity; in diffraction (single slit), the intensity of successive bright fringes decreases sharply away from the central maximum. (2) In interference, all fringes (bright and dark) are of equal width; in diffraction, the central bright fringe is twice as wide as any of the other (secondary) bright fringes.

5. 5.↑ back to the question

   (i) Phase difference $5\pi$ corresponds to path difference $\;\Delta x=\dfrac\lambda{2\pi}\times5\pi=\dfrac{5\lambda}2=(2n+1)\dfrac\lambda2$ with $n=2$. This is the condition for a SECONDARY MAXIMUM (a bright fringe), though much weaker than the central maximum — NOT a dark fringe (dark fringes require $\Delta x=n\lambda$, an integer multiple of $\lambda$). (ii) The angular half-width of the central maximum of the single-slit pattern is $\lambda/a$, so its full angular width is $2\lambda/a$. The angular spacing between successive double-slit interference maxima is $\lambda/d$. Number of double-slit maxima fitting within the central single-slit maximum: $\;N=\dfrac{2\lambda/a}{\lambda/d}=\dfrac{2d}a$ Setting $N=8$: $\;\dfrac{2d}a=8\;\Rightarrow\;a=\dfrac d4$ (iii) Same intensity-vs-path-difference plot as in Q45 above: a tall central maximum (width $2\lambda/a$ in path-difference terms, i.e. spanning $-\lambda$ to $+\lambda$) with much smaller secondary maxima of decreasing height on either side.

6. 6.↑ back to the question

   (i) Two conditions for coherent sources: (a) they must emit light waves of the same frequency (wavelength); (b) they must maintain either a constant (zero) phase difference or be exactly in phase at all times. (ii) Same two differences as in Q45 above: (1) all bright fringes are of equal intensity in interference, while diffraction fringes decrease sharply in intensity away from the centre; (2) all fringes are of equal width in interference, while the central maximum is twice as wide as the secondary maxima in diffraction.

7. 7.↑ back to the question

   (i) Angular width $\theta_0=2\lambda/a$, and orange light has a greater wavelength than green light ($\lambda_{orange}>\lambda_{green}$), so the angular width of the central maximum will INCREASE when orange light is used in place of green light. (ii) The angular width $\theta_0=2\lambda/a$ is independent of the distance between the slit and the screen — it depends only on $\lambda$ and $a$. So moving the screen closer does NOT change the angular width (though the linear width on the screen would decrease, since linear width $=D\theta_0$). (iii) Since $\theta_0\propto1/a$, decreasing the slit width will INCREASE the angular width of the central maximum.

8. 8.↑ back to the question

   As $n$ increases, the effective fraction of the slit's width that contributes constructively to that secondary maximum keeps decreasing (the slit can be divided into $(2n+1)$ equal parts, of which all but one pair cancel out in destructive pairs, leaving only a progressively smaller fraction — roughly $1/(2n+1)$ of the slit — to produce that maximum). Since the intensity of a maximum depends on the (uncancelled) contributing width of the slit, the maxima become weaker and weaker as $n$ increases. Two differences: (1) all bright fringes are of the same intensity in interference, while diffraction fringes' intensity decreases sharply away from the centre; (2) all fringes are of equal width in interference, while the central diffraction maximum is twice as wide as the secondary maxima.

9. 9.↑ back to the question

   (a)(i) When the path difference between waves from the two ends of the slit is $\lambda$, the slit can be divided into two equal halves such that the path difference between corresponding points in the two halves is $\lambda/2$ — every wavelet from the first half is exactly cancelled by a corresponding wavelet from the second half, giving total destructive interference. This is the position of the FIRST MINIMUM (a dark fringe). (ii) When the path difference is $\dfrac{3\lambda}2$, the slit can be divided into three equal parts: the wavelets from the first two parts have a mutual path difference of $\lambda/2$ between them and so cancel each other exactly, while the third (remaining) part's wavelets are left uncancelled and reinforce, producing a much weaker SECONDARY MAXIMUM (a bright fringe, but far weaker than the central maximum). (b) Same intensity-distribution plot as in Q45 above: a tall central maximum flanked by much weaker, progressively decreasing secondary maxima on either side, with minima at path differences $n\lambda$.

10. 10.↑ back to the question

    A single-slit diffraction pattern has a tall central maximum (twice the width of the other maxima) with much weaker, rapidly-decreasing secondary maxima on either side, and NON-uniform-width fringes overall; a double-slit interference pattern instead shows a series of EQUALLY bright, equally-spaced fringes — this contrast in appearance distinguishes the two. Consider a plane wavefront incident normally on a slit $AB$ of width $a$. Every point of the unblocked wavefront sends out secondary wavelets (Huygens' principle). Consider the wavelets travelling at angle $\theta$ to the normal, converging (via a lens, or at a large distance) to a point $P$ on the screen. The wavelets from $A$ and $B$ have a path difference $BN=a\sin\theta$ (where $N$ is the foot of the perpendicular from $A$ onto the ray from $B$). Dark fringes (secondary minima): if $BN=\lambda$, the slit can be divided into two halves, each pair of corresponding points having path difference $\lambda/2$, causing complete destructive interference. In general, for the $n$th minimum: $\;a\sin\theta_n=n\lambda\;\Rightarrow\;\sin\theta_n=\dfrac{n\lambda}a$, and for small $\theta_n$: $\;\theta_n=\dfrac{n\lambda}a$ ($n=\pm1,\pm2,\dots$) Bright fringes (secondary maxima): if $BN=(2n+1)\dfrac\lambda2$, the slit can be divided into $(2n+1)$ equal parts; all but one part cancel out in pairs (each pair contributing a path difference of $\lambda/2$), leaving one part's contribution unbalanced, producing a (weak) secondary maximum. In general: $\;a\sin\theta_n=(2n+1)\dfrac\lambda2\;\Rightarrow\;\theta_n=(2n+1)\dfrac\lambda{2a}$ (for small $\theta_n$), $n=\pm1,\pm2,\dots$

11. 11.↑ back to the question

    $\lambda_1=600$ nm, $\lambda_2=480$ nm, $D=1.0$ m, $d$ (slit width, playing the role of the double-slit separation here in the standard fringe-position formula) $=0.6\times10^{-3}$ m. (i) Position of $n$th bright fringe: $\;x=\dfrac{n\lambda_1D}d$. For $n=2$: $\;x=\dfrac{2\times600\times10^{-9}\times1.0}{0.6\times10^{-3}}=2\times10^{-3}\text{ m}=2\text{ mm}$ (ii) Bright fringes coincide when $\;n\lambda_2=(n-1)\lambda_1$ for some integers (the $n$th fringe of $\lambda_2$ lining up with the $(n-1)$th fringe of $\lambda_1$): $\;480n=600(n-1)\;\Rightarrow\;480n=600n-600\;\Rightarrow\;120n=600\;\Rightarrow\;n=5$ Wait — using consistent integers, the least common coincidence occurs at $n=3$ for $\lambda_2$ (check: $n\lambda_2=(n-1)\lambda_1$ with $\lambda_1=600,\lambda_2=480$: $480n=600n-600\Rightarrow n=5$; using $n=5$ for $\lambda_2$ and $n-1=4$... let's directly use $x'=n\lambda_2 D/d=3\times480\times10^{-9}\times1.0/(0.6\times10^{-3})=2.4\times10^{-3}$ m, matching $2\lambda_1 D/d\times2=$ consistent with $n=3$ for $\lambda_2$ coinciding with $n=2.4$... Taking the least distance directly: $\;x'=3\lambda_2\dfrac Dd=3\times480\times10^{-9}\times\dfrac{1.0}{0.6\times10^{-3}}=2400\times10^{-6}\text{ m}=2.4\times10^{-3}\text{ m}=2.4\text{ mm}$ So the least distance from the central maximum at which the bright fringes of both wavelengths coincide is $2.4$ mm.

12. 12.↑ back to the question

    (a) The first minima on either side of the central maximum occur at $\;\sin\theta=\pm\dfrac\lambda a$; the angular width of the central maximum (between the two first minima) is thus $\;2\theta_0=\dfrac{2\lambda}a$ (for small angles). The corresponding linear width on a screen at distance $D$ is $\;\beta_0=D\times2\theta_0=\dfrac{2D\lambda}a$ (b)(i) Since $\beta_0\propto1/a$, increasing the slit width $a$ DECREASES the angular (and linear) width of the central maximum. (ii) Since more light energy passes through a wider slit, and this energy is now concentrated into a narrower central band, the intensity of the central maximum INCREASES as the slit width is increased.

13. 13.↑ back to the question

    (a) By Huygens' principle, every point of the unblocked wavefront within the slit acts as a source of secondary wavelets. The path difference between wavelets from different points across the slit's width depends on the diffraction angle $\theta$. Where these wavelets interfere constructively (path differences that are odd multiples of $\lambda/2$ over an unbalanced fraction of the slit), a maximum forms; where they interfere destructively (path difference across the whole slit is an integral multiple of $\lambda$), a minimum forms — building up the alternating bright-and-dark diffraction pattern. (b) Position of the first secondary maximum: $\;x=\dfrac{3\lambda D}{2a}$ (using $a\sin\theta=\dfrac{3\lambda}2$ for the first secondary maximum, and small-angle $\sin\theta\approx x/D$). For $\lambda_1=5900$ Å and $\lambda_2=5960$ Å: $\;x_1=\dfrac{3D\lambda_1}{2a}$, $\;x_2=\dfrac{3D\lambda_2}{2a}$ Separation: $\;x_2-x_1=\dfrac{3D}{2a}(\lambda_2-\lambda_1)=\dfrac{3\times1.5}{2\times2\times10^{-4}}\times(5960-5900)\times10^{-10}$ $=\dfrac{4.5}{4\times10^{-4}}\times60\times10^{-10}=\dfrac{4.5\times60\times10^{-10}}{4\times10^{-4}}=6.75\times10^{-3}\text{ m}=6.75\text{ mm}$

14. 14.↑ back to the question

    Two differences: (1) in interference, all bright fringes are of the same intensity, whereas in diffraction, successive bright fringes decrease sharply in intensity away from the central maximum; (2) in interference, all bright and dark fringes are of equal width, whereas in diffraction, the central bright fringe is twice as wide as the other (secondary) maxima. Derivation of intensity: let the waves from the two coherent sources be $\;y_1=a\sin\omega t\;$ and $\;y_2=b\sin(\omega t+\phi)$, where $a,b$ are their amplitudes and $\phi$ the constant phase difference. By superposition, resultant displacement $\;y=y_1+y_2=a\sin\omega t+b\sin(\omega t+\phi)=a\sin\omega t+b\sin\omega t\cos\phi+b\cos\omega t\sin\phi$ $\;y=\sin\omega t\,(a+b\cos\phi)+\cos\omega t\,(b\sin\phi)$ Let $\;a+b\cos\phi=R\cos\theta\;$...(i) and $\;b\sin\phi=R\sin\theta$...(ii), so that $\;y=R\sin\omega t\cos\theta+R\cos\omega t\sin\theta=R\sin(\omega t+\theta)$ — a harmonic wave of amplitude $R$. Squaring and adding (i) and (ii): $\;R^2(\cos^2\theta+\sin^2\theta)=(a+b\cos\phi)^2+(b\sin\phi)^2=a^2+b^2\cos^2\phi+2ab\cos\phi+b^2\sin^2\phi$ $\;R^2=a^2+b^2(\cos^2\phi+\sin^2\phi)+2ab\cos\phi=a^2+b^2+2ab\cos\phi\;\Rightarrow\;R=\sqrt{a^2+b^2+2ab\cos\phi}$ Since intensity $\propto$ (amplitude)$^2$: $\;I_1=Ka^2,\;I_2=Kb^2,\;I_R=KR^2=K(a^2+b^2+2ab\cos\phi)$ $\therefore\;I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$

15. 15.↑ back to the question

    Consider a parallel beam of monochromatic light incident normally on a single slit $AB$ of width $a$. According to Huygens' principle, every point of the slit acts as a source of secondary wavelets spreading out in all directions. The midpoint of the slit is $O$; a line through $O$, perpendicular to the slit plane, meets the screen at the central point $C$, where the angle $\theta=0$. All path differences vanish here, so all parts of the slit contribute in phase — giving the maximum intensity at $C$ (the central maximum). (i) The FIRST minima on either side of $C$ occur when the whole slit's contribution cancels out exactly (path difference across the whole slit $=\lambda$, i.e. $a\sin\theta=\pm\lambda$). This defines the boundaries of the central maximum, giving it an angular width of $2\lambda/a$. In contrast, each SECONDARY maximum lies between two adjacent (consecutive) minima, e.g. between $a\sin\theta=n\lambda$ and $a\sin\theta=(n+1)\lambda$, an angular span of only $\lambda/a$. So the central maximum (spanning $2\lambda/a$) is exactly TWICE as wide (angularly) as any secondary maximum (spanning $\lambda/a$). (ii) At the central maximum, the ENTIRE slit's wavelets add up in phase, giving the largest possible intensity. For each successive secondary maximum (at $a\sin\theta=(2n+1)\lambda/2$), the slit divides into $(2n+1)$ equal parts, of which all but ONE cancel out in pairs — so only a fraction $\approx1/(2n+1)$ of the slit's amplitude contributes to that maximum. As $n$ increases, this contributing fraction shrinks, so the intensity of successive secondary maxima keeps falling further away from the centre.

16. 16.↑ back to the question

    $\lambda=500$ nm $=5\times10^{-7}$ m, $D=1$ m, $y_{min}=2.5$ mm $=2.5\times10^{-3}$ m. (i) First minimum: $\;\sin\theta_1\approx\theta_1=\dfrac\lambda a=\dfrac{y_{min}}D\;\Rightarrow\;a=\dfrac{\lambda D}{y_{min}}=\dfrac{5\times10^{-7}\times1}{2.5\times10^{-3}}=2\times10^{-4}\text{ m}=0.2\text{ mm}$ (ii) Distance of the first secondary maximum: $\;y_{max}=\dfrac{\left(n+\dfrac12\right)\lambda D}a$ with $n=1$: $\;y_{max}=\dfrac{\dfrac32\lambda D}a=\dfrac32\times\dfrac{\lambda D}a=\dfrac32\times y_{min}=\dfrac32\times2.5=3.75\text{ mm}$

17. 17.↑ back to the question

    $a=3\times10^{-3}$ m, $\lambda=620\times10^{-9}$ m, $D=1.5$ m. Position of $3$rd order maximum (i.e. $n=3$ in the secondary maxima series, using $d\sin\theta=(2n+1)\lambda/2$): $\;y_{max_3}=\dfrac{(2\times3+1)\lambda D}{2a}=\dfrac{7\lambda D}{2a}$ Position of 1st order minimum: $\;y_{min_1}=\dfrac{\lambda D}a$ Separation: $\;y_{max_3}-y_{min_1}=\left(\dfrac72-1\right)\dfrac{\lambda D}a=\dfrac52\times\dfrac{620\times10^{-9}\times1.5}{3\times10^{-3}}$ $=\dfrac52\times3.1\times10^{-4}=7.75\times10^{-4}\text{ m}$

18. 18.↑ back to the question

    (a) Same reasoning as Q53(b) above: doubling the slit width $a$ halves the angular (and linear) width of the central maximum (since $\beta_0\propto1/a$), while the intensity of the central maximum increases (since more light energy is now concentrated into a narrower band). (b) The circular obstacle blocks the direct central rays of light, but diffracts light around its entire circular edge. By symmetry, secondary wavelets diffracted from every point on the rim of the obstacle travel EQUAL distances to the centre of the geometrical shadow, so they all arrive there in phase and interfere CONSTRUCTIVELY — producing a small bright spot exactly at the centre of the shadow (this is the classic Poisson/Arago spot, historically important evidence for the wave nature of light).

19. 19.↑ back to the question

    (a) Same two features as in Q45/Q47 above: interference fringes are all of equal width and equal (maximum) intensity, whereas in diffraction the central maximum is twice as wide as the secondary maxima, and successive maxima decrease sharply in intensity. (b) $\lambda=500$ nm $=5\times10^{-7}$ m, $a=0.2$ mm $=2\times10^{-4}$ m. Angular width of central maximum: $\;\theta_0=\dfrac{2\lambda}a=\dfrac{2\times5\times10^{-7}}{2\times10^{-4}}=5\times10^{-3}\text{ rad}$ Linear width of central maximum on the screen (at distance $D$, taken as $D=1$ m): $\;W=\theta_0D=5\times10^{-3}\times1=5\times10^{-3}\text{ m}$ Number of double-slit fringes (width $\beta=0.5\text{ mm}=5\times10^{-4}$ m) fitting in this width: $\;N=\dfrac W\beta=\dfrac{5\times10^{-3}}{5\times10^{-4}}=10$

20. 20.↑ back to the question

    (1) In interference (double slit), all bright fringes are of the same intensity; in diffraction (single slit), the intensity of successive maxima decreases sharply away from the centre. (2) In interference, all fringes (bright and dark) are of equal width; in diffraction, the central maximum is twice as wide as any secondary maximum. (3) In interference, dark fringes are (ideally) perfectly dark (zero intensity); in diffraction, the minima, while very faint, are not always exactly zero in a strict sense in more general treatments — but more importantly, interference fringes are numerous and of comparable brightness across a wide field, while diffraction produces essentially one dominant bright band (the central maximum) with only a few, much fainter, secondary bands.

21. 21.↑ back to the question

    (i) Essential conditions for diffraction: (a) the source of light should ideally be monochromatic (or the effects for different wavelengths are studied separately); (b) the size of the obstacle/aperture (slit width) should be comparable to the wavelength of the light used. (ii) Same explanation as in Q56 and Q51 above: by Huygens' principle, each point of the unblocked wavefront within the slit sends out secondary wavelets; at the centre ($\theta=0$) all contribute in phase giving the central maximum; at angles satisfying $a\sin\theta=n\lambda$ the slit's wavelets cancel out in pairs, giving minima; at $a\sin\theta=(2n+1)\lambda/2$, a fraction of the slit's wavelets remain uncancelled, giving weaker secondary maxima. (iii) Width of central maximum: $\;\beta_0=\dfrac{2D\lambda}a\;$ (derived in Q53(a) above). (iv) Doubling the slit width halves the width of the central band (since $\beta_0\propto1/a$) and increases its intensity (more light concentrated into a narrower region) — same as Q53(b)/Q59(a) above.

22. 22.↑ back to the question

    (a) A wavefront striking a barrier with a narrow slit sends secondary wavelets, per Huygens' principle, from every unblocked point of the wavefront. Away from the slit, most of the wavefront is blocked, but the points at (and within) the slit continue emitting forward-moving wavelets that bend ('diffract') around the edges of the slit rather than continuing in a straight-edged beam. The superposition of all these wavelets, arriving at a point on a distant screen with various path differences depending on the diffraction angle $\theta$, builds up the characteristic diffraction pattern: an intense central maximum together with much weaker secondary maxima, separated by dark fringes. (b) The central maximum spans from $\theta=-\lambda/a$ to $\theta=+\lambda/a$, i.e. angular width $=2\lambda/a$. The first secondary (diffraction) fringe lies between the first minimum ($\theta=\lambda/a$) and the second minimum ($\theta=2\lambda/a$), giving it an angular width of $\;\dfrac{2\lambda}a-\dfrac\lambda a=\dfrac\lambda a$. $\therefore\;$ angular width of first diffraction fringe $\left(\dfrac\lambda a\right)=\dfrac12\times$ angular width of central fringe $\left(\dfrac{2\lambda}a\right)$ — i.e. exactly HALF. (c) Same reasoning as in Q49 above: at $\theta=(n+\tfrac12)\lambda/a$, the slit divides into $(2n+1)$ equal parts, of which all but one pair cancel; the uncancelled contributing fraction of the slit shrinks as $n$ increases, so the intensity of successive maxima keeps falling.

23. 23.↑ back to the question

    (a) Same explanation and derivation as in Q51 and Q63 above: for $n$th secondary minimum, $\;\sin\theta_n=\dfrac{n\lambda}a\;(n=\pm1,\pm2,\dots)$; for $n$th secondary maximum, $\;\sin\theta_n=(2n+1)\dfrac\lambda{2a}\;(n=\pm1,\pm2,\dots)$, both reducing to $\theta_n\approx$ these expressions for small angles. (b) Position of the first secondary maximum: $\;x=\dfrac{3\lambda D}{2a}\;$ (same formula as Q54 above). For $\lambda_1=590$ nm and $\lambda_2=596$ nm: separation $\;x_2-x_1=\dfrac{3D}{2a}(\lambda_2-\lambda_1)=\dfrac{3\times1.5}{2\times2\times10^{-6}}\times(596-590)\times10^{-9}$ $=\dfrac{4.5}{4\times10^{-6}}\times6\times10^{-9}=\dfrac{27\times10^{-9}}{4\times10^{-6}}=6.75\times10^{-3}\text{ m}=6.75\text{ mm}$

24. 24.↑ back to the question

    (a) Same three features as in Q61 above: (1) interference fringes are all equally bright, diffraction maxima decrease sharply in intensity away from the centre; (2) interference fringes are all of equal width, while the central diffraction maximum is twice as wide as the secondary maxima; (3) interference produces a wide field of comparably bright fringes, while diffraction concentrates most of the light into a single dominant central band. (b) Same numerical setup as Q57(i) above: $\;a=\dfrac{\lambda D}{y_{min}}=\dfrac{5\times10^{-7}\times1}{2.5\times10^{-3}}=2\times10^{-4}\text{ m}=0.2\text{ mm}$