201. In Young's double-slit experiment, for a point \(P\) on the screen at a small angle \(\theta\) from the central line, the path difference is first written as
ⓐ. \(\Delta x=d\sin\theta\)
ⓑ. \(\Delta x=\frac{D}{d}\sin\theta\)
ⓒ. \(\Delta x=D\sin\theta\)
ⓓ. \(\Delta x=dD\sin\theta\)
Correct Answer: \(\Delta x=d\sin\theta\)
Explanation: The path difference in Young's double-slit experiment is the difference between the distances travelled by the two waves from the two slits to the same screen point. If the slit separation is \(d\) and the point \(P\) is observed at angle \(\theta\) from the central line, the geometrical path difference is \(\Delta x=d\sin\theta\). The screen distance \(D\) enters later when \(\sin\theta\) is approximated using the screen coordinate \(y\). Writing \(D\sin\theta\) would wrongly use the slit-screen distance as if it were the slit separation. The factor that produces the path difference is the separation between the two sources, not the distance from the sources to the screen.
202. In the usual small-angle treatment of Young's double-slit experiment, the path difference at screen coordinate \(y\) is approximately
ⓐ. \(\Delta x=\frac{\lambda D}{d}\)
ⓑ. \(\Delta x=\frac{dy}{D}\)
ⓒ. \(\Delta x=\frac{Dy}{d}\)
ⓓ. \(\Delta x=\frac{dD}{y}\)
Correct Answer: \(\Delta x=\frac{dy}{D}\)
Explanation: For a point \(P\) on the screen, the path difference is \(\Delta x=d\sin\theta\). In the central region of the pattern, the angle \(\theta\) is small. Therefore, \(\sin\theta\approx\tan\theta\). From the geometry of the screen, \(\tan\theta\approx\frac{y}{D}\). Substituting this into \(\Delta x=d\sin\theta\) gives \(\Delta x\approx d\frac{y}{D}\). Thus, \(\Delta x=\frac{dy}{D}\) under the small-angle approximation.
203. A point \(P\) lies \(2.0\,\text{mm}\) above the central point in a double-slit experiment. The slit separation is \(0.50\,\text{mm}\), and the screen is \(1.0\,\text{m}\) away. The approximate path difference at \(P\) is
ⓐ. \(1.0\times10^{-6}\,\text{m}\)
ⓑ. \(4.0\times10^{-6}\,\text{m}\)
ⓒ. \(2.5\times10^{-4}\,\text{m}\)
ⓓ. \(1.0\times10^{-3}\,\text{m}\)
Correct Answer: \(1.0\times10^{-6}\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(y=2.0\,\text{mm}\), \(d=0.50\,\text{mm}\), and \(D=1.0\,\text{m}\).
\( \textbf{Required:} \) Approximate path difference \(\Delta x\).
\( \textbf{Small-angle relation:} \)
\[
\Delta x=\frac{dy}{D}
\]
\( \textbf{Unit conversion:} \)
\[
d=0.50\,\text{mm}=5.0\times10^{-4}\,\text{m}
\]
\[
y=2.0\,\text{mm}=2.0\times10^{-3}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
\Delta x=\frac{(5.0\times10^{-4})(2.0\times10^{-3})}{1.0}\,\text{m}
\]
\( \textbf{Multiplication:} \)
\[
(5.0\times10^{-4})(2.0\times10^{-3})=1.0\times10^{-6}
\]
\( \textbf{Final answer:} \) The path difference is \(1.0\times10^{-6}\,\text{m}\). Keeping both \(d\) and \(y\) in metres prevents the common millimetre-conversion error.
204. A derivation note for Young's double-slit experiment contains the following steps.
| Row | Step |
|---|
| P | \(\Delta x=d\sin\theta\) |
| Q | For small \(\theta\), \(\sin\theta\approx\tan\theta\) |
| R | \(\tan\theta\approx\frac{y}{D}\) |
| S | \(\Delta x\approx\frac{Dy}{d}\) |
The row that should be corrected is
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P gives the standard geometrical path difference for two slits separated by \(d\). Row Q is the small-angle approximation used near the central region of the pattern. Row R correctly connects the screen coordinate \(y\) with the screen distance \(D\). Combining these steps gives \(\Delta x=d\frac{y}{D}=\frac{dy}{D}\). Row S reverses the roles of \(d\) and \(D\), which would give the wrong dependence on slit separation and screen distance.
205. A graph is plotted in a Young's double-slit experiment with path difference \(\Delta x\) on the vertical axis and screen coordinate \(y\) on the horizontal axis. With fixed \(d\) and \(D\), the slope of the graph is
ⓐ. \(\lambda D\)
ⓑ. \(\frac{D}{d}\)
ⓒ. \(\frac{\lambda}{dD}\)
ⓓ. \(\frac{d}{D}\)
Correct Answer: \(\frac{d}{D}\)
Explanation: \( \textbf{Graph relation:} \) In the small-angle region,
\[
\Delta x=\frac{dy}{D}
\]
\( \textbf{Writing in straight-line form:} \)
\[
\Delta x=\left(\frac{d}{D}\right)y
\]
\( \textbf{Vertical variable:} \) The vertical axis is \(\Delta x\).
\( \textbf{Horizontal variable:} \) The horizontal axis is \(y\).
\( \textbf{Slope identification:} \)
\[
m=\frac{d}{D}
\]
\( \textbf{Dimension check:} \) Both \(d\) and \(D\) are lengths, so \(\frac{d}{D}\) is dimensionless, matching the slope of length versus length.
\( \textbf{Physical meaning:} \) Increasing slit separation makes path difference change faster with screen position, while increasing screen distance makes it change more slowly.
\( \textbf{Final answer:} \) The slope is \(\frac{d}{D}\).
206. For a double-slit setup with \(d=0.25\,\text{mm}\), \(D=1.25\,\text{m}\), and \(y=3.0\,\text{mm}\), the approximate path difference is
ⓐ. \(6.0\times10^{-5}\,\text{m}\)
ⓑ. \(6.0\times10^{-7}\,\text{m}\)
ⓒ. \(9.6\times10^{-7}\,\text{m}\)
ⓓ. \(1.5\times10^{-6}\,\text{m}\)
Correct Answer: \(6.0\times10^{-7}\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(d=0.25\,\text{mm}\), \(D=1.25\,\text{m}\), and \(y=3.0\,\text{mm}\).
\( \textbf{Required:} \) Path difference \(\Delta x\).
\( \textbf{Useful relation:} \)
\[
\Delta x=\frac{dy}{D}
\]
\( \textbf{Convert \(d\):} \)
\[
0.25\,\text{mm}=2.5\times10^{-4}\,\text{m}
\]
\( \textbf{Convert \(y\):} \)
\[
3.0\,\text{mm}=3.0\times10^{-3}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
\Delta x=\frac{(2.5\times10^{-4})(3.0\times10^{-3})}{1.25}
\]
\( \textbf{Numerator:} \)
\[
(2.5\times10^{-4})(3.0\times10^{-3})=7.5\times10^{-7}
\]
\( \textbf{Division:} \)
\[
\Delta x=\frac{7.5\times10^{-7}}{1.25}=6.0\times10^{-7}\,\text{m}
\]
\( \textbf{Final answer:} \) The path difference is \(6.0\times10^{-7}\,\text{m}\).
207. In Young's double-slit experiment, the approximation \(\Delta x=\frac{dy}{D}\) becomes unreliable far from the central region mainly because
ⓐ. the screen distance \(D\) becomes zero
ⓑ. the slits stop being separated by \(d\)
ⓒ. wavelength loses its physical meaning
ⓓ. small-angle approximation breaks down
Correct Answer: small-angle approximation breaks down
Explanation: The exact geometrical step is \(\Delta x=d\sin\theta\). To obtain \(\Delta x=\frac{dy}{D}\), we use \(\sin\theta\approx\tan\theta\approx\frac{y}{D}\). This approximation is good only when \(\theta\) is small, which usually means points near the central region and \(D\) much larger than \(y\). Far from the centre, \(\theta\) may no longer be small enough. The formula then becomes less accurate because of the approximation, not because \(d\), \(D\), or \(\lambda\) lose their meanings.
208. Use the arrangement described below.
Two slits \(S_1\) and \(S_2\) are separated vertically, with \(S_1\) above \(S_2\). The screen is placed to the right. A point \(P\) is chosen above the central point on the screen. The path difference is defined as \(\Delta x=S_2P-S_1P\).
With this convention, the sign of \(\Delta x\) at \(P\) is expected to be
ⓐ. positive, because \(P\) is closer to the upper slit \(S_1\)
ⓑ. zero for every point above the central point
ⓒ. positive, because \(S_2P\) is longer than \(S_1P\)
ⓓ. negative, because \(S_2P\) is shorter than \(S_1P\)
Correct Answer: positive, because \(S_2P\) is longer than \(S_1P\)
Explanation: The point \(P\) is above the central point, so it is closer to the upper slit \(S_1\) than to the lower slit \(S_2\). Therefore, \(S_1P\) is shorter and \(S_2P\) is longer. Since the convention is \(\Delta x=S_2P-S_1P\), a longer \(S_2P\) gives a positive path difference. The sign depends on both the geometry and the chosen definition of \(\Delta x\). If the order of subtraction were reversed, the sign would reverse but the bright-dark condition would remain physically consistent.
209. If the screen coordinate \(y\) is changed from \(+y\) to \(-y\) in a symmetric Young's double-slit setup, the path difference \(\Delta x=\frac{dy}{D}\)
ⓐ. same magnitude with opposite sign
ⓑ. becomes equal to the screen distance \(D\)
ⓒ. keeps the same sign but doubles
ⓓ. becomes independent of slit separation \(d\)
Correct Answer: same magnitude with opposite sign
Explanation: In the small-angle treatment, the path difference is proportional to the screen coordinate \(y\). If \(y\) is replaced by \(-y\), then \(\Delta x\) is replaced by \(-\Delta x\). The magnitude remains the same because \(|-y|=|y|\). This represents the symmetry of the interference pattern about the central point. The sign tells which path is longer under the chosen convention, while the magnitude controls the order of the bright or dark fringe.
210. A double-slit experiment uses \(d=0.40\,\text{mm}\) and \(D=2.0\,\text{m}\). For light of wavelength \(500\,\text{nm}\), the screen coordinate where the path difference first becomes \(1\lambda\) is
ⓐ. \(4.00\,\text{mm}\)
ⓑ. \(2.50\,\text{mm}\)
ⓒ. \(0.25\,\text{mm}\)
ⓓ. \(1.25\,\text{mm}\)
Correct Answer: \(2.50\,\text{mm}\)
Explanation: \( \textbf{Given:} \) \(d=0.40\,\text{mm}\), \(D=2.0\,\text{m}\), and \(\lambda=500\,\text{nm}\).
\( \textbf{Required:} \) Screen coordinate \(y\) for \(\Delta x=\lambda\).
\( \textbf{Path-difference relation:} \)
\[
\Delta x=\frac{dy}{D}
\]
\( \textbf{Set condition:} \)
\[
\frac{dy}{D}=\lambda
\]
\( \textbf{Solve for \(y\):} \)
\[
y=\frac{\lambda D}{d}
\]
\( \textbf{Convert units:} \)
\[
\lambda=500\,\text{nm}=5.0\times10^{-7}\,\text{m}
\]
\[
d=0.40\,\text{mm}=4.0\times10^{-4}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
y=\frac{(5.0\times10^{-7})(2.0)}{4.0\times10^{-4}}\,\text{m}
\]
\( \textbf{Calculation:} \)
\[
y=2.5\times10^{-3}\,\text{m}=2.50\,\text{mm}
\]
\( \textbf{Final answer:} \) The first \(1\lambda\) path-difference point is at \(2.50\,\text{mm}\) from the centre.
211. Bright fringes in Young's double-slit experiment occur at screen positions satisfying
ⓐ. \(y_n=\frac{n\lambda D}{d}\)
ⓑ. \(y_n=\frac{n\lambda d}{D}\)
ⓒ. \(y_n=\frac{ndD}{\lambda}\)
ⓓ. \(y_n=\frac{\left(n+\frac{1}{2}\right)\lambda D}{d}\)
Correct Answer: \(y_n=\frac{n\lambda D}{d}\)
Explanation: Bright fringes occur when the path difference is an integral multiple of wavelength. The constructive condition is \(\Delta x=n\lambda\). In Young's double-slit experiment, \(\Delta x=\frac{dy}{D}\) near the central region. Equating these gives \(\frac{dy_n}{D}=n\lambda\). Solving for \(y_n\) gives \(y_n=\frac{n\lambda D}{d}\). The dark-fringe formula contains the half-integer factor, so it should not be used for bright positions.
212. The central bright fringe in Young's double-slit experiment corresponds to the order
ⓐ. \(n=1\)
ⓑ. \(n=0\)
ⓒ. \(n=\frac{1}{2}\)
ⓓ. \(n=-1\)
Correct Answer: \(n=0\)
Explanation: The central point on the screen is symmetrically placed with respect to the two slits. The two paths from the slits to this point are equal, so the path difference is zero. The bright-fringe condition is \(\Delta x=n\lambda\). Zero path difference corresponds to \(n=0\). This is why the central bright fringe is called the zeroth-order bright fringe.
213. A Young's double-slit experiment has \(\lambda=600\,\text{nm}\), \(D=1.5\,\text{m}\), and \(d=0.30\,\text{mm}\). The position of the second bright fringe from the centre is
ⓐ. \(4.5\,\text{mm}\)
ⓑ. \(6.0\,\text{mm}\)
ⓒ. \(9.0\,\text{mm}\)
ⓓ. \(3.0\,\text{mm}\)
Correct Answer: \(6.0\,\text{mm}\)
Explanation: \( \textbf{Given:} \) \(\lambda=600\,\text{nm}\), \(D=1.5\,\text{m}\), \(d=0.30\,\text{mm}\).
\( \textbf{Required:} \) Position of the second bright fringe, so \(n=2\).
\( \textbf{Bright-fringe formula:} \)
\[
y_n=\frac{n\lambda D}{d}
\]
\( \textbf{Convert units:} \)
\[
\lambda=600\,\text{nm}=6.0\times10^{-7}\,\text{m}
\]
\[
d=0.30\,\text{mm}=3.0\times10^{-4}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
y_2=\frac{(2)(6.0\times10^{-7})(1.5)}{3.0\times10^{-4}}\,\text{m}
\]
\( \textbf{Numerator:} \)
\[
(2)(6.0\times10^{-7})(1.5)=18.0\times10^{-7}=1.8\times10^{-6}
\]
\( \textbf{Division:} \)
\[
y_2=\frac{1.8\times10^{-6}}{3.0\times10^{-4}}=6.0\times10^{-3}\,\text{m}
\]
\( \textbf{Final answer:} \) The second bright fringe is at \(6.0\,\text{mm}\) from the centre.
214. A data table lists bright-fringe positions in a double-slit experiment.
| Order \(n\) | Expected position \(y_n\) |
|---|
| \(0\) | \(0\) |
| \(1\) | \(\frac{\lambda D}{d}\) |
| \(2\) | \(\frac{2\lambda D}{d}\) |
| \(3\) | \(\frac{3\lambda d}{D}\) |
The entry that should be corrected is for order
ⓐ. \(0\)
ⓑ. \(1\)
ⓒ. \(3\)
ⓓ. \(2\)
Correct Answer: \(3\)
Explanation: Bright-fringe positions are given by \(y_n=\frac{n\lambda D}{d}\). For \(n=0\), the position is \(0\), which is the central bright fringe. For \(n=1\), the position is \(\frac{\lambda D}{d}\). For \(n=2\), the position is \(\frac{2\lambda D}{d}\). For \(n=3\), the correct position should be \(\frac{3\lambda D}{d}\), not \(\frac{3\lambda d}{D}\). The listed entry has interchanged \(D\) and \(d\).
215. If the wavelength used in a Young's double-slit experiment is doubled while \(D\) and \(d\) remain unchanged, the position of the \(n\)th bright fringe from the centre
ⓐ. becomes double
ⓑ. remains unchanged
ⓒ. becomes half
ⓓ. becomes four times
Correct Answer: becomes double
Explanation: The bright-fringe position is \(y_n=\frac{n\lambda D}{d}\). For a fixed order \(n\), fixed screen distance \(D\), and fixed slit separation \(d\), \(y_n\) is directly proportional to \(\lambda\). If \(\lambda\) is doubled, the numerator of the expression is doubled. Therefore, the fringe position from the centre becomes double. This is why longer wavelengths form bright fringes farther from the central fringe.
216. Two bright fringes of orders \(n=2\) and \(n=5\) are observed in a Young's double-slit pattern. The separation between them is
ⓐ. \(\frac{5\lambda D}{d}\)
ⓑ. \(\frac{2\lambda D}{d}\)
ⓒ. \(\frac{3\lambda D}{d}\)
ⓓ. \(\frac{7\lambda D}{d}\)
Correct Answer: \(\frac{3\lambda D}{d}\)
Explanation: \( \textbf{Bright-fringe position:} \)
\[
y_n=\frac{n\lambda D}{d}
\]
\( \textbf{Position of \(n=5\):} \)
\[
y_5=\frac{5\lambda D}{d}
\]
\( \textbf{Position of \(n=2\):} \)
\[
y_2=\frac{2\lambda D}{d}
\]
\( \textbf{Separation:} \)
\[
y_5-y_2=\frac{5\lambda D}{d}-\frac{2\lambda D}{d}
\]
\( \textbf{Simplification:} \)
\[
y_5-y_2=\frac{3\lambda D}{d}
\]
\( \textbf{Meaning:} \) The separation depends on the difference in order numbers, not on their sum.
\( \textbf{Final answer:} \) The separation is \(\frac{3\lambda D}{d}\).
217. Bright fringes appear symmetrically on both sides of the central bright fringe because
ⓐ. only one side of the screen receives light
ⓑ. the physical slit separation changes sign
ⓒ. path difference is zero at every screen point
ⓓ. equal \(|y|\) values can satisfy the same order
Correct Answer: equal \(|y|\) values can satisfy the same order
Explanation: In the small-angle treatment, \(\Delta x=\frac{dy}{D}\). Points at \(+y\) and \(-y\) have path differences of equal magnitude and opposite sign under the same convention. Constructive interference depends on the phase relation modulo \(2\pi\), so \(\Delta x=+n\lambda\) and \(\Delta x=-n\lambda\) both give bright fringes. This produces symmetric bright fringes about the central point. The physical slit separation does not change sign; only the coordinate and path-difference convention do.
218. A graph of bright-fringe position \(y_n\) versus order number \(n\) is drawn for a fixed Young's double-slit setup. The graph should be
ⓐ. a straight line through the origin with slope \(\frac{d}{\lambda D}\)
ⓑ. a parabola opening upward
ⓒ. a horizontal line
ⓓ. a straight line through the origin with slope \(\frac{\lambda D}{d}\)
Correct Answer: a straight line through the origin with slope \(\frac{\lambda D}{d}\)
Explanation: Bright-fringe positions are given by \(y_n=\frac{n\lambda D}{d}\). This is of the form \(y=mn\), where \(n\) is the order number. Therefore, the graph of \(y_n\) against \(n\) is a straight line through the origin. The slope is \(\frac{\lambda D}{d}\), which is also the separation between consecutive bright fringes. A parabolic graph would imply that fringe spacing changes with order, which is not predicted by the small-angle double-slit formula.
219. The first bright fringe in a Young's double-slit experiment is at \(1.2\,\text{mm}\) from the centre. The fourth bright fringe on the same side is at
ⓐ. \(6.0\,\text{mm}\)
ⓑ. \(3.6\,\text{mm}\)
ⓒ. \(4.8\,\text{mm}\)
ⓓ. \(2.4\,\text{mm}\)
Correct Answer: \(4.8\,\text{mm}\)
Explanation: \( \textbf{Given:} \) The first bright fringe has \(y_1=1.2\,\text{mm}\).
\( \textbf{Bright-fringe formula:} \)
\[
y_n=n\frac{\lambda D}{d}
\]
\( \textbf{First bright position:} \)
\[
y_1=\frac{\lambda D}{d}=1.2\,\text{mm}
\]
\( \textbf{Fourth bright position:} \)
\[
y_4=4\frac{\lambda D}{d}
\]
\( \textbf{Using \(y_1\):} \)
\[
y_4=4y_1
\]
\( \textbf{Substitution:} \)
\[
y_4=4(1.2\,\text{mm})
\]
\( \textbf{Calculation:} \)
\[
y_4=4.8\,\text{mm}
\]
\( \textbf{Final answer:} \) The fourth bright fringe is at \(4.8\,\text{mm}\) from the centre.
220. A point \(P\) in a double-slit experiment is located where \(y=\frac{3\lambda D}{d}\). The point \(P\) is
ⓐ. the zero-order bright fringe
ⓑ. the first-order dark fringe
ⓒ. the third-order bright fringe
ⓓ. the second-order dark fringe
Correct Answer: the third-order bright fringe
Explanation: The bright-fringe formula in Young's double-slit experiment is \(y_n=\frac{n\lambda D}{d}\). The given position is \(y=\frac{3\lambda D}{d}\). Comparing this with the formula gives \(n=3\). Therefore, the point is the third-order bright fringe. It is not a dark fringe because dark-fringe positions contain a half-integer factor such as \(\left(n+\frac{1}{2}\right)\).
