Optics MCQs with Answers – Part 8 (NEET Physics)

Explanation: \(\textbf{Identify the counting interval and fringe conditions:}\) The observation window is \[ -6.0\,\text{mm}\leq y\leq+6.0\,\text{mm} \] Both endpoints are included. For a symmetric Young's interference pattern with zero initial phase difference, \[ y_{\mathrm{bright}}=m\beta \] where \(m\) is any integer. Dark fringes occur at \[ y_{\mathrm{dark}}=\left(m+\frac12\right)\beta \] \(\textbf{Count the original bright fringes:}\) Initially, \[ \beta_1=2.0\,\text{mm} \] The bright condition is \[ y=2m\,\text{mm} \] Within the specified window, \[ -6\leq2m\leq6 \] Therefore, \[ -3\leq m\leq3 \] The allowed integer orders are \[ m=-3,-2,-1,0,1,2,3 \] Thus, \[ N_{B1}=7 \] \(\textbf{Count the original dark fringes:}\) The dark-fringe positions satisfy \[ y=2\left(m+\frac12\right)\,\text{mm} \] \[ y=(2m+1)\,\text{mm} \] The allowed positions are \[ -5,-3,-1,+1,+3,+5\,\text{mm} \] Therefore, \[ N_{D1}=6 \] \(\textbf{Determine the new fringe width after immersion:}\) With unchanged frequency, the wavelength in the liquid is reduced by its refractive index. Hence, \[ \beta_2=\frac{\beta_1}{n} \] Substituting, \[ \beta_2=\frac{2.0}{1.50} \] \[ \beta_2=\frac43\,\text{mm} \] \(\textbf{Count the new bright fringes:}\) The bright-fringe positions are \[ y=\frac{4m}{3}\,\text{mm} \] The window condition gives \[ -6\leq\frac{4m}{3}\leq6 \] Multiplying by \(\frac34\), \[ -\frac92\leq m\leq\frac92 \] The allowed integer orders are \[ m=-4,-3,-2,-1,0,1,2,3,4 \] Thus, \[ N_{B2}=9 \] \(\textbf{Count the new dark fringes:}\) For a dark fringe, \[ y=\left(m+\frac12\right)\frac43\,\text{mm} \] The window condition is \[ -6\leq\frac43\left(m+\frac12\right)\leq6 \] Therefore, \[ -\frac92\leq m+\frac12\leq\frac92 \] Subtracting \(\frac12\), \[ -5\leq m\leq4 \] There are ten allowed integers: \[ N_{D2}=10 \] \(\textbf{Check the endpoint behaviour:}\) For the immersed pattern, the extrema at the window endpoints satisfy \[ \frac{|y|}{\beta_2}=\frac{6}{4/3} \] \[ \frac{|y|}{\beta_2}=\frac92 \] They are dark fringes, not bright fringes. Both are counted because the question explicitly includes the endpoints. \(\textbf{Physical interpretation:}\) Reducing the wavelength through uniform immersion reduces the spatial fringe width. More bright and dark extrema can consequently fit inside a fixed observation window. However, the numbers of each type need not increase in the same way because the precise locations of the extrema relative to the window boundaries matter. At the initial setting, the endpoints coincide with bright fringes. After immersion, they coincide with dark fringes. Correct counting therefore requires both the changed fringe width and an explicit treatment of the boundary convention. \(\textbf{Final answer:}\) \[ \boxed{(N_{B1},N_{D1})=(7,6),\quad(N_{B2},N_{D2})=(9,10)} \]

Explanation: \(\textbf{Identify the original slit geometry:}\) The initial coordinates are \[ y_{L1}=-0.20\,\text{mm} \] \[ y_{U1}=+0.20\,\text{mm} \] The original slit separation is \[ d_1=y_{U1}-y_{L1} \] \[ d_1=0.40\,\text{mm} \] Their midpoint is \[ y_{c1}=\frac{y_{U1}+y_{L1}}2 \] \[ y_{c1}=0 \] \(\textbf{Calculate the original fringe width:}\) The wavelength and screen distance are \[ \lambda=6.0\times10^{-7}\,\text{m} \] \[ D=2.0\,\text{m} \] The fringe width is \[ \beta_1=\frac{\lambda D}{d_1} \] \[ \beta_1=\frac{(6.0\times10^{-7})(2.0)}{4.0\times10^{-4}} \] \[ \beta_1=3.0\,\text{mm} \] \(\textbf{Apply the physical movement of one slit:}\) The lower slit remains at \[ y_{L2}=-0.20\,\text{mm} \] The upper slit moves to \[ y_{U2}=+0.40\,\text{mm} \] The new separation is \[ d_2=y_{U2}-y_{L2} \] \[ d_2=0.40-(-0.20) \] \[ d_2=0.60\,\text{mm} \] \(\textbf{Determine the changed fringe width:}\) \[ \beta_2=\frac{\lambda D}{d_2} \] \[ \beta_2=\frac{(6.0\times10^{-7})(2.0)}{6.0\times10^{-4}} \] \[ \beta_2=2.0\,\text{mm} \] \(\textbf{Locate the central-order bright fringe:}\) Because the illuminating plane wave is normally incident, the two slit disturbances begin with equal phase. The central-order bright fringe is located where their geometrical distances to the screen are equal. For two slit coordinates in a common plane and a parallel screen, that point lies opposite the midpoint of the slits. \(\textbf{Calculate the new slit midpoint:}\) \[ y_{c2}=\frac{y_{U2}+y_{L2}}2 \] \[ y_{c2}=\frac{0.40-0.20}{2} \] \[ y_{c2}=+0.10\,\text{mm} \] Therefore, \[ y_{0,2}=+0.10\,\text{mm} \] \(\textbf{Write the new bright-fringe positions:}\) The approximate bright-fringe positions are \[ y_m=y_{0,2}+m\beta_2 \] Therefore, \[ y_m=0.10+2.0m \] with \(y_m\) in millimetres. The central-order fringe occurs at \(m=0\). \(\textbf{Distinguish the two effects of moving one slit:}\) Moving only the upper slit increases the separation of the pair. That decreases the fringe width. The same movement also shifts the midpoint of the pair upwards. That shifts the central-order bright fringe relative to the fixed laboratory coordinate. These changes are not the same: the width changes by \(1.0\,\text{mm}\), whereas the central-order position changes by \(0.10\,\text{mm}\). \(\textbf{Physical interpretation:}\) In Young's experiment, fringe spacing is controlled primarily by the separation of the coherent slits. The central-order bright fringe is determined by equality of the total optical paths from the two sources. A normally incident plane wave supplies no initial phase difference, so the equal-distance position is directly opposite the slit midpoint. Moving one slit changes both the midpoint and the separation. The resulting interference pattern therefore undergoes a positional shift and a change in scale. Treating the displacement of one slit as though it were a rigid translation of the complete slit pair would miss the change in fringe width. \(\textbf{Final answer:}\) \[ \boxed{d_2=0.60\,\text{mm},\quad\beta_2=2.0\,\text{mm},\quad y_{0,2}=+0.10\,\text{mm}} \]

Explanation: Trial P increases the slit-screen distance while keeping the wavelength and slit separation fixed. The standard fringe-width relation is \[ \beta=\frac{\lambda D}{d} \] Consequently, increasing \(D\) increases the distance between adjacent bright fringes. This changes the spatial scale of the pattern without independently requiring a loss of coherence. Trial Q changes the intensity transmitted through one slit. The two coherent amplitudes become unequal, so complete destructive cancellation is no longer possible at the minima. The fringe visibility is reduced from its initial value of unity, provided both slit beams remain present. Because the filter does not change the wavelength or geometrical arrangement, it does not change the fringe width. Trial R introduces a fixed extra optical-path difference in one beam. A constant phase offset shifts the positions at which constructive and destructive interference occur. It does not change the rate at which the geometrical path difference varies across the screen. Therefore, the entire pattern can move while retaining the original fringe spacing. Trial S reduces the wavelength inside the uniformly filled interference region because the wave frequency remains unchanged while its speed decreases. The fringe width consequently decreases. This effect differs from inserting a plate into only one slit path. Uniform immersion changes the wavelength relevant to the whole interference geometry; a single-path plate primarily introduces a relative phase offset. The four experiments distinguish fringe-width changes, fringe-position shifts and visibility changes as separate observables. One must identify the physical quantity altered by each modification rather than assuming that any disturbance to the apparatus changes all properties of the pattern. The correct matching is P: wider fringes; Q: reduced visibility; R: shifted fringes; S: narrower fringes.

Explanation: The Assertion is true because interference visibility depends on whether the intensity modulation produced by the coherent slit pair remains distinguishable after all contributions are observed together. An extended ordinary source consists of many independently emitting regions. Light originating from one small source region can be divided by the two slits into beams capable of producing their own interference pattern. However, other source regions illuminate the double slit from different directions. Those directions introduce different initial phase differences between the slit disturbances. The resulting interference patterns are displaced relative to one another. Because the source regions are mutually incoherent, the detector adds their intensities on time averaging. Bright fringes of one contribution may overlap dark fringes of another, reducing the observed contrast. The narrow entrance slit restricts the angular range of illuminating source points. The resulting family of interference patterns has less variation in its transverse displacement. Their bright and dark regions therefore overlap more consistently, allowing greater fringe visibility. At the same time, the entrance slit transmits less light, so increased contrast does not imply increased total illumination. The Reason is true because it identifies the spread of initial slit-to-slit phase differences as the relevant mechanism. It correctly explains why restricting the effective source size can improve the observed pattern. The double-slit separation, screen distance and mean wavelength have not changed, so an improvement in visibility does not require a corresponding change in fringe width. This demonstrates that source brightness and spatial coherence are distinct experimental considerations. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Determine the initial phase difference:}\) The original optical-path difference is \[ \Delta_1=\frac{\lambda}{3} \] The corresponding phase difference is \[ \phi_1=\frac{2\pi\Delta_1}{\lambda} \] Therefore, \[ \phi_1=\frac{2\pi}{3} \] \(\textbf{Use the coherent intensity relation:}\) The separate beam intensities are \[ I_1=16I_0,\qquad I_2=9I_0 \] The resultant intensity is \[ I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi \] The interference-term coefficient is \[ 2\sqrt{I_1I_2}=2\sqrt{144I_0^2} \] \[ 2\sqrt{I_1I_2}=24I_0 \] \(\textbf{Calculate the original resultant intensity:}\) \[ I_{\mathrm{before}}=16I_0+9I_0+24I_0\cos\frac{2\pi}{3} \] Using \[ \cos\frac{2\pi}{3}=-\frac12 \] gives \[ I_{\mathrm{before}}=25I_0-12I_0 \] \[ I_{\mathrm{before}}=13I_0 \] \(\textbf{Determine the extra optical path required:}\) The intended additional path difference is \[ \Delta_{\mathrm{plate}}=\frac{\lambda}{6} \] Using the wavelength, \[ \Delta_{\mathrm{plate}}=\frac{600}{6} \] \[ \Delta_{\mathrm{plate}}=100\,\text{nm} \] \(\textbf{Recover the plate thickness:}\) For a normally inserted plate replacing air, \[ \Delta_{\mathrm{plate}}=(n-1)t \] The plate index is \[ n=1.50 \] Therefore, \[ t=\frac{\Delta_{\mathrm{plate}}}{n-1} \] \[ t=\frac{100}{0.50} \] \[ t=200\,\text{nm} \] \[ t=0.20\,\mu\text{m} \] \(\textbf{Calculate the modified relative phase:}\) The plate is placed in the beam whose additional delay increases the magnitude of the originally specified path difference. The new difference is \[ \Delta_2=\frac{\lambda}{3}+\frac{\lambda}{6} \] \[ \Delta_2=\frac{\lambda}{2} \] Therefore, \[ \phi_2=\frac{2\pi\Delta_2}{\lambda} \] \[ \phi_2=\pi \] \(\textbf{Calculate the new resultant intensity:}\) \[ I_{\mathrm{after}}=16I_0+9I_0+24I_0\cos\pi \] \[ I_{\mathrm{after}}=25I_0-24I_0 \] \[ I_{\mathrm{after}}=I_0 \] \(\textbf{Check the destructive-interference limit:}\) At phase difference \(\pi\), the coherent amplitudes oppose each other. Their individual amplitude magnitudes are proportional to \[ 4\sqrt{I_0}\quad\text{and}\quad3\sqrt{I_0} \] They are unequal. The minimum intensity is consequently \[ I_{\min}=(4\sqrt{I_0}-3\sqrt{I_0})^2 \] \[ I_{\min}=I_0 \] \(\textbf{Physical interpretation:}\) A transparent plate changes the phase relationship of coherent waves through its additional optical path. The phase change depends on refractive-index contrast and plate thickness, not on thickness alone. Here the plate moves the relative phase from \(120^\circ\) to \(180^\circ\). This decreases the resultant intensity from \(13I_0\) to the destructive-interference minimum. The minimum remains nonzero because the interfering amplitudes are unequal. The problem combines optical-path calculation, phase conversion and coherent intensity addition under a controlled change of experimental conditions. \(\textbf{Final answer:}\) \[ \boxed{t=0.20\,\mu\text{m},\quad I_{\mathrm{before}}=13I_0,\quad I_{\mathrm{after}}=I_0} \]

Explanation: \(\textbf{Separate the background from the interference pattern:}\) The measured maximum and minimum are \[ I_{\max,\mathrm{obs}}=19I_0 \] \[ I_{\min,\mathrm{obs}}=7I_0 \] The incoherent background is independently measured as \[ I_b=3I_0 \] A uniform incoherent background adds the same intensity at every phase. \(\textbf{Recover the coherent maximum and minimum:}\) Subtracting the background, \[ I_{\max,\mathrm{coh}}=19I_0-3I_0 \] \[ I_{\max,\mathrm{coh}}=16I_0 \] Similarly, \[ I_{\min,\mathrm{coh}}=7I_0-3I_0 \] \[ I_{\min,\mathrm{coh}}=4I_0 \] \(\textbf{Use amplitude addition to recover the two beams:}\) For coherent beams of the same polarization, \[ I_{\max,\mathrm{coh}}=(\sqrt{I_1}+\sqrt{I_2})^2 \] \[ I_{\min,\mathrm{coh}}=(\sqrt{I_1}-\sqrt{I_2})^2 \] Take \[ I_1\geq I_2 \] The measured extremes give \[ \sqrt{I_1}+\sqrt{I_2}=4\sqrt{I_0} \] \[ \sqrt{I_1}-\sqrt{I_2}=2\sqrt{I_0} \] \(\textbf{Calculate the individual beam intensities:}\) Adding the equations, \[ 2\sqrt{I_1}=6\sqrt{I_0} \] \[ I_1=9I_0 \] Subtracting, \[ 2\sqrt{I_2}=2\sqrt{I_0} \] \[ I_2=I_0 \] \(\textbf{Determine visibility with background present:}\) The experimentally observed visibility is \[ V_{\mathrm{obs}}=\frac{I_{\max,\mathrm{obs}}-I_{\min,\mathrm{obs}}}{I_{\max,\mathrm{obs}}+I_{\min,\mathrm{obs}}} \] Substituting, \[ V_{\mathrm{obs}}=\frac{19-7}{19+7} \] \[ V_{\mathrm{obs}}=\frac{12}{26} \] \[ V_{\mathrm{obs}}=\frac6{13} \] \(\textbf{Determine visibility after removing the background:}\) With the background switched off, the coherent extremes are \(16I_0\) and \(4I_0\). Therefore, \[ V_{\mathrm{coh}}=\frac{16-4}{16+4} \] \[ V_{\mathrm{coh}}=\frac{12}{20} \] \[ V_{\mathrm{coh}}=\frac35 \] \(\textbf{Identify what has actually changed:}\) Removing the background increases visibility from \(\frac6{13}\) to \(\frac35\). Neither coherent beam's intensity changes. Their amplitude ratio remains \[ \frac{\sqrt{I_1}}{\sqrt{I_2}}=3 \] The remaining fringe minimum is nonzero because those coherent amplitudes are still unequal. \(\textbf{Physical interpretation:}\) A measured interference pattern can include light that does not participate in the coherent superposition. That background must be separated before using interference extremes to infer the individual coherent-beam intensities. The background raises both maxima and minima but leaves their absolute difference unchanged. It nevertheless reduces the measured visibility because the average detected intensity increases. Removing the background improves contrast without changing beam coherence or making the coherent amplitudes equal. The independent background measurement provides the additional information needed to reconstruct the two beam intensities correctly. \(\textbf{Final answer:}\) \[ \boxed{I_1=9I_0,\quad I_2=I_0,\quad V_{\mathrm{obs}}=\frac6{13},\quad V_{\mathrm{coh}}=\frac35} \]

Explanation: \(\textbf{Separate slit separation from individual slit width:}\) The distance between the centres of the two slits is \[ d=0.30\,\text{mm} \] Their individual width is \[ a=0.10\,\text{mm} \] The slit separation controls the fine interference-fringe spacing. The individual slit width controls the broad single-slit diffraction envelope. These two quantities must not be interchanged. \(\textbf{Calculate the Young's interference fringe width:}\) The wavelength and screen distance are \[ \lambda=6.0\times10^{-7}\,\text{m} \] \[ D=2.0\,\text{m} \] The interference fringe width is \[ \beta=\frac{\lambda D}{d} \] Substituting, \[ \beta=\frac{(6.0\times10^{-7})(2.0)}{3.0\times10^{-4}} \] \[ \beta=4.0\times10^{-3}\,\text{m} \] \[ \beta=4.0\,\text{mm} \] \(\textbf{Locate the first single-slit diffraction minima:}\) For a slit of width \(a\), the first minima satisfy \[ a\sin\theta=\lambda \] For small angles, \[ x_1\approx\frac{\lambda D}{a} \] Therefore, \[ x_1=\frac{(6.0\times10^{-7})(2.0)}{1.0\times10^{-4}} \] \[ x_1=1.2\times10^{-2}\,\text{m} \] \[ x_1=12\,\text{mm} \] The first diffraction minima occur at \[ x=-12\,\text{mm}\quad\text{and}\quad x=+12\,\text{mm} \] \(\textbf{Determine the corresponding interference orders:}\) Bright interference fringes would occur at \[ x_m=m\beta \] At the positive diffraction minimum, \[ m=\frac{x_1}{\beta} \] \[ m=\frac{12}{4} \] \[ m=3 \] Therefore, the third positive interference order coincides with the first positive diffraction minimum. The third negative order similarly coincides with the first negative diffraction minimum. \(\textbf{Explain the missing orders:}\) At a single-slit diffraction minimum, the resultant contribution from each individual slit vanishes in the ideal model. Two-slit constructive phase alignment cannot create a bright fringe there because the individual slit amplitudes are zero. Thus, the nominal interference maxima of orders \[ m=+3\quad\text{and}\quad m=-3 \] are absent at those diffraction minima. \(\textbf{Count the interference maxima inside the central envelope:}\) The central diffraction maximum lies strictly between \[ -12\,\text{mm}\lt x\lt +12\,\text{mm} \] The bright interference positions within it are \[ x=-8,-4,0,+4,+8\,\text{mm} \] These correspond to \[ m=-2,-1,0,+1,+2 \] Therefore, five interference maxima lie strictly within the central diffraction maximum. \(\textbf{Check the central diffraction width:}\) The width between the first diffraction minima is \[ W=2x_1 \] \[ W=24\,\text{mm} \] The ratio of this width to the interference spacing is \[ \frac{W}{\beta}=\frac{24}{4} \] \[ \frac{W}{\beta}=6 \] This ratio describes the span of the central envelope in fringe-width units; it does not mean that six bright fringes must occur within it. The two boundary positions are diffraction minima and cannot be counted as bright interference fringes. \(\textbf{Physical interpretation:}\) A pair of finite-width coherent slits produces interference modulated by the diffraction behaviour of each slit. The fine fringe spacing depends on the distance between the slit centres. The broad diffraction envelope depends on their individual widths. When a nominal interference maximum coincides with a single-slit diffraction minimum, that interference order is suppressed. This example integrates the conditions for constructive interference and the first minima of single-slit diffraction without requiring a separate unfamiliar diffraction formula. The appropriate conclusion is that the third interference orders are missing and five interference maxima remain inside the central diffraction envelope. \(\textbf{Final answer:}\) \[ \boxed{\beta=4.0\,\text{mm},\quad x_{\mathrm{first\ minima}}=\pm12\,\text{mm},\quad m_{\mathrm{missing}}=\pm3,\quad N_{\mathrm{central}}=5} \]

Explanation: \(\textbf{Calculate the original fringe width:}\) The wavelength, slit separation and screen distance are \[ \lambda=4.0\times10^{-7}\,\text{m} \] \[ d=2.5\times10^{-4}\,\text{m} \] \[ D=1.0\,\text{m} \] The fringe width is \[ \beta=\frac{\lambda D}{d} \] \[ \beta=\frac{(4.0\times10^{-7})(1.0)}{2.5\times10^{-4}} \] \[ \beta=1.6\times10^{-3}\,\text{m} \] \[ \beta=1.6\,\text{mm} \] \(\textbf{Determine the upper plate's original optical-path excess:}\) The upper plate has \[ n_U=1.50,\qquad t_U=2.0\,\mu\text{m} \] Its excess optical path relative to the air it replaces is \[ \Delta_U=(n_U-1)t_U \] \[ \Delta_U=(1.50-1)(2.0) \] \[ \Delta_U=1.0\,\mu\text{m} \] \(\textbf{Determine the lower plate's optical-path excess:}\) The lower plate has \[ n_L=1.25,\qquad t_L=4.0\,\mu\text{m} \] Therefore, \[ \Delta_L=(n_L-1)t_L \] \[ \Delta_L=(1.25-1)(4.0) \] \[ \Delta_L=1.0\,\mu\text{m} \] \(\textbf{Determine the original relative path difference:}\) At the geometrical screen centre, the free-space paths from the two slits are equal. The relative path difference supplied by the plates is \[ \Delta_{\mathrm{rel},1}=\Delta_U-\Delta_L \] \[ \Delta_{\mathrm{rel},1}=1.0-1.0 \] \[ \Delta_{\mathrm{rel},1}=0 \] Thus, the centre is bright. The initial unequal plate thicknesses do not shift the pattern because their excess optical paths are equal. \(\textbf{Replace the upper plate:}\) Its thickness remains \[ t'_U=2.0\,\mu\text{m} \] Its refractive index becomes \[ n'_U=1.60 \] The new upper optical-path excess is \[ \Delta'_U=(n'_U-1)t'_U \] \[ \Delta'_U=(1.60-1)(2.0) \] \[ \Delta'_U=1.2\,\mu\text{m} \] \(\textbf{Recover the changed relative path difference:}\) The lower plate remains unchanged: \[ \Delta_L=1.0\,\mu\text{m} \] Therefore, \[ \Delta_{\mathrm{rel},2}=\Delta'_U-\Delta_L \] \[ \Delta_{\mathrm{rel},2}=0.2\,\mu\text{m} \] \[ \Delta_{\mathrm{rel},2}=200\,\text{nm} \] \(\textbf{Compare with the wavelength:}\) \[ \frac{\Delta_{\mathrm{rel},2}}{\lambda}=\frac{200}{400} \] \[ \frac{\Delta_{\mathrm{rel},2}}{\lambda}=\frac12 \] A half-wavelength path difference gives destructive interference for equal slit intensities. The geometrical centre therefore becomes dark. \(\textbf{Calculate the fringe-pattern displacement:}\) A constant added relative optical path shifts the fringes by \[ s=\beta\frac{|\Delta_{\mathrm{rel},2}-\Delta_{\mathrm{rel},1}|}{\lambda} \] Substituting, \[ s=1.6\left(\frac{200}{400}\right) \] \[ s=0.8\,\text{mm} \] \(\textbf{Check the fringe-width invariance:}\) The transparent plates affect the relative optical path of the slit beams. The wavelength in the surrounding observation region, slit separation and screen distance remain unchanged. Consequently, \[ \beta'=1.6\,\text{mm} \] \(\textbf{Physical interpretation:}\) Different plate thicknesses can introduce identical optical-path delays when their refractive indices differ appropriately. The original plates therefore compensate each other even though they are not physically identical. Changing one plate's refractive index without changing its thickness breaks that compensation. The resulting half-wavelength relative delay changes the geometrical centre from bright to dark and shifts the whole pattern by half a fringe width. The fringe spacing remains unaffected because the spatial interference geometry is unchanged. The decisive optical quantity is the difference between the two plates' excess optical paths, not the difference between their physical thicknesses. \(\textbf{Final answer:}\) \[ \boxed{\beta=1.6\,\text{mm},\quad\Delta_{\mathrm{rel},1}=0,\quad\Delta_{\mathrm{rel},2}=\frac{\lambda}{2},\quad s=0.8\,\text{mm}} \]

Explanation: \(\textbf{Determine the optical-path delay introduced by the plate:}\) The plate index and thickness are \[ n=1.50 \] \[ t=0.30\,\mu\text{m} \] Its excess optical path relative to air is \[ \Delta_p=(n-1)t \] Substituting, \[ \Delta_p=(1.50-1)(0.30) \] \[ \Delta_p=0.15\,\mu\text{m} \] \[ \Delta_p=150\,\text{nm} \] \(\textbf{Convert the plate delay into a phase difference:}\) The wavelength is \[ \lambda=600\,\text{nm} \] Therefore, \[ \frac{\Delta_p}{\lambda}=\frac{150}{600} \] \[ \frac{\Delta_p}{\lambda}=\frac14 \] The associated phase difference has magnitude \[ \phi_p=\frac{2\pi\Delta_p}{\lambda} \] \[ \phi_p=\frac{\pi}{2} \] \(\textbf{Calculate the initial centre intensity:}\) With normally incident illumination, the two slit disturbances begin in phase before the plate delay is applied. Their geometrical distances to the screen centre are equal. Thus, the plate supplies the only relative phase difference at that point. For equal individual intensities, \[ I=2I_0(1+\cos\phi) \] Therefore, \[ I_{\mathrm{initial}}=2I_0\left(1+\cos\frac{\pi}{2}\right) \] \[ I_{\mathrm{initial}}=2I_0 \] The initial centre is neither a bright maximum nor a dark minimum. \(\textbf{Determine the compensating illumination tilt:}\) A positive tilt makes the incident wavefront reach the upper slit earlier. That phase advance can compensate for the additional delay of the upper slit plate. For the central phase difference to vanish, \[ d\sin\theta=\Delta_p \] The slit separation is \[ d=0.30\,\text{mm}=3.0\times10^{-4}\,\text{m} \] Therefore, \[ \sin\theta=\frac{\Delta_p}{d} \] \[ \sin\theta=\frac{1.5\times10^{-7}}{3.0\times10^{-4}} \] \[ \sin\theta=5.0\times10^{-4} \] \(\textbf{Apply the small-angle approximation:}\) Since the required angle is very small, \[ \theta\approx\sin\theta \] Thus, \[ \theta\approx5.0\times10^{-4}\,\text{rad} \] \[ \theta\approx+0.50\,\text{mrad} \] \(\textbf{Calculate the centre intensity after compensation:}\) The upper slit receives an incident phase advance equivalent to the plate's additional optical-path delay. The two contributions cancel. The net path difference at the geometrical centre becomes \[ \Delta_{\mathrm{net},+}=0 \] Therefore, \[ \phi_+=0 \] and \[ I_+=2I_0(1+\cos0) \] \[ I_+=4I_0 \] \(\textbf{Reverse the illumination tilt:}\) For an equal negative tilt, the incident wavefront reaches the upper slit later instead of earlier. The illumination delay and the plate delay now act in the same direction. Their total path-difference magnitude is \[ \Delta_{\mathrm{net},-}=\Delta_p+\Delta_p \] \[ \Delta_{\mathrm{net},-}=300\,\text{nm} \] \[ \Delta_{\mathrm{net},-}=\frac{\lambda}{2} \] \(\textbf{Calculate the resulting centre intensity:}\) The corresponding phase difference has magnitude \[ \phi_-=\pi \] Therefore, \[ I_-=2I_0(1+\cos\pi) \] \[ I_-=0 \] The geometrical centre becomes dark. \(\textbf{Physical interpretation:}\) Young's interference depends on the complete relative phase of the two slit disturbances. That phase can contain contributions acquired before the slits as well as additional optical paths introduced after their illumination. An obliquely incident wavefront produces an initial slit-to-slit phase difference, while a transparent plate introduces a separate delay. The two effects can cancel or reinforce depending on their signs. Here equal positive and negative illumination tilts produce opposite outcomes because only one compensates the plate delay. The experiment demonstrates that interference positions depend on the algebraic combination of optical-path contributions, not merely on their magnitudes. \(\textbf{Final answer:}\) \[ \boxed{\theta_{\mathrm{compensating}}\approx+0.50\,\text{mrad},\quad I_+=4I_0,\quad I_-=0} \]

Explanation: \(\textbf{Apply coherent superposition when both slits are open:}\) In experiment P, the individual intensities are \[ I_1=I_2=I_0 \] For coherent beams with the same polarization, \[ I_P=I_1+I_2+2\sqrt{I_1I_2}\cos\phi \] Therefore, \[ I_P=2I_0(1+\cos\phi) \] \(\textbf{Calculate the interference extremes:}\) At constructive interference, \[ I_{P,\max}=2I_0(1+1) \] \[ I_{P,\max}=4I_0 \] At destructive interference, \[ I_{P,\min}=2I_0(1-1) \] \[ I_{P,\min}=0 \] Experiment P produces a sustained pattern with complete contrast under the stated ideal conditions. \(\textbf{Identify the crucial change in experiment Q:}\) The shutters are operated so that the two slits are never open simultaneously. At any instant, only one slit contributes light to the screen. The interference term involving both electric fields is therefore absent. A stable monochromatic source cannot produce two-beam interference when the required fields are not present together at the observation point. \(\textbf{Determine the average contribution of the first slit:}\) The first slit is open for half the exposure time. Its time-averaged intensity is \[ \langle I_1\rangle=\frac12 I_0 \] \(\textbf{Determine the average contribution of the second slit:}\) The second slit also operates for half the exposure time. Thus, \[ \langle I_2\rangle=\frac12 I_0 \] \(\textbf{Calculate the recorded intensity:}\) Because the slit contributions occur at different times, \[ \langle I_Q\rangle=\langle I_1\rangle+\langle I_2\rangle \] \[ \langle I_Q\rangle=\frac12 I_0+\frac12 I_0 \] \[ \langle I_Q\rangle=I_0 \] The average is uniform over the stated observation region. \(\textbf{Explain why rapid switching does not restore interference:}\) Increasing the switching frequency does not make the two slit fields overlap. The detector simply averages a sequence of single-slit illumination intervals. Each interval lacks the second field required for constructive or destructive two-beam interference. The resulting average therefore contains no sustained Young's fringes. \(\textbf{Physical interpretation:}\) Coherent illumination is necessary for sustained interference, but it is not sufficient unless the relevant fields overlap in space and time. A detector may receive light from both slits during an exposure without ever receiving both slit fields simultaneously. Those successive contributions add as recorded intensities rather than coherent amplitudes. Experiment Q therefore has uniform average illumination despite using exactly the same source and slits as experiment P. The experiment distinguishes temporal overlap from source coherence and shows why rapid alternation cannot substitute for simultaneous superposition. \(\textbf{Final answer:}\) \[ \boxed{I_{P,\max}=4I_0,\quad I_{P,\min}=0,\quad \langle I_Q\rangle=I_0,\quad V_Q=0} \]

Explanation: \(\textbf{Write the pattern during the first half-exposure:}\) For equal coherent intensities, \[ I=2I_0[1+\cos\Phi] \] where \(\Phi\) is the total relative phase at the observation point. During the first half, \[ \Phi_1=\phi(x) \] Therefore, \[ I_1(x)=2I_0[1+\cos\phi(x)] \] This pattern has complete interference contrast. \(\textbf{Apply the controlled phase reversal:}\) During the second half, the device adds \[ \pi \] to the relative phase. Thus, \[ \Phi_2=\phi(x)+\pi \] Using \[ \cos(\phi+\pi)=-\cos\phi \] the second pattern is \[ I_2(x)=2I_0[1-\cos\phi(x)] \] Its maxima coincide with the minima of the first pattern. \(\textbf{Average the equal-duration observations:}\) The ordinary long-exposure detector records \[ \langle I(x)\rangle=\frac12 I_1(x)+\frac12 I_2(x) \] Substituting, \[ \langle I(x)\rangle=I_0[1+\cos\phi(x)]+I_0[1-\cos\phi(x)] \] \[ \langle I(x)\rangle=2I_0 \] The cosine contributions cancel at every screen position. \(\textbf{Determine the long-exposure visibility:}\) The averaged intensity has no spatial maxima or minima. Consequently, \[ I_{\max,\mathrm{long}}=I_{\min,\mathrm{long}}=2I_0 \] and \[ V_{\mathrm{long}}=\frac{2I_0-2I_0}{2I_0+2I_0} \] \[ V_{\mathrm{long}}=0 \] \(\textbf{Analyse synchronized detection:}\) The synchronized detector records only the intervals with zero additional phase. It therefore samples the first intensity pattern: \[ I_{\mathrm{sync}}(x)=2I_0[1+\cos\phi(x)] \] At constructive interference, \[ I_{\max,\mathrm{sync}}=4I_0 \] At destructive interference, \[ I_{\min,\mathrm{sync}}=0 \] \(\textbf{Interpret the apparent loss and recovery of fringes:}\) The fields remain coherent within each phase-controlled interval. The disappearance of the long-exposure fringes results from averaging two complementary patterns. Selecting one phase state prevents that cancellation in the recorded observation. \(\textbf{Physical interpretation:}\) Sustained interference requires a sufficiently stable relative phase over the relevant observation interval. A system may produce excellent instantaneous interference while failing to show fringes in a long exposure. Here the phase changes in a controlled manner rather than fluctuating randomly, but the equal-duration average still removes all spatial modulation. A synchronized detector can recover one interference pattern by selecting a single phase state. The experiment distinguishes optical coherence during a measurement interval from contrast after averaging measurements made under different phase conditions. \(\textbf{Final answer:}\) \[ \boxed{\langle I\rangle_{\mathrm{long}}=2I_0,\quad V_{\mathrm{long}}=0,\quad I_{\max,\mathrm{sync}}=4I_0,\quad I_{\min,\mathrm{sync}}=0} \]

Explanation: Statement I is correct because complete destructive interference requires equal coherent field amplitudes when the two beams have the same polarization. If the original beams have unequal intensities, their destructive-interference minimum remains nonzero. Attenuating the stronger beam to match the weaker can make the minimum completely dark. The resulting visibility can increase even though less total light reaches the apparatus. Statement II is incorrect because fringe width and fringe visibility describe different properties. For standard Young's interference, \[ \beta=\frac{\lambda D}{d} \] Increasing the screen distance increases fringe spacing. However, that geometrical change does not necessarily alter the relative slit intensities or restore a fluctuating phase relation. It therefore does not automatically improve the intrinsic interference visibility. Statement III is correct because stable fringes require an adequately stable relative phase during the observation. A fluctuating path difference changes the positions of constructive and destructive interference across the screen. If many different phase states are averaged in a long exposure, the observed modulation can become weak or disappear. This can occur without changing either beam's individual intensity. Statement IV is correct because different points of an extended ordinary source can introduce different relative phases at the two slits. Their individual interference patterns can be displaced relative to one another. Reducing the angular extent of the illuminating source reduces that spread of fringe positions. Under suitable conditions, the superposed pattern becomes clearer. The accompanying reduction in transmitted light does not contradict an improvement in contrast. Together the four procedures distinguish unequal amplitudes, geometrical fringe spacing, phase instability and spatial coherence as separate aspects of an interference experiment. Only Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the relevant path-difference geometry:}\) For two coherent slits separated by a transverse vector, the path difference near the screen centre depends on the projection of the screen displacement along that vector. Let the rotated slit-separation components be \[ d_x=\frac d{\sqrt2} \] \[ d_y=\frac d{\sqrt2} \] because the separation direction makes \(45^\circ\) with both screen axes. \(\textbf{Write the paraxial optical-path difference:}\) At screen coordinates \((x,y)\), the geometric path difference is approximately \[ \Delta(x,y)=\frac{d_xx+d_yy}{D} \] Therefore, \[ \Delta(x,y)=\frac{d(x+y)}{\sqrt2D} \] The sign depends on the labelling of the slits, but bright-fringe spacing does not. \(\textbf{Determine the shape and orientation of the fringes:}\) Constructive interference requires \[ \Delta(x,y)=m\lambda \] Thus, \[ x+y=m\frac{\sqrt2\lambda D}{d} \] For each fixed integer \(m\), this represents a straight line. Its slope on the screen is \[ \frac{dy}{dx}=-1 \] The bright-fringe lines are therefore perpendicular to the rotated slit-separation direction. \(\textbf{Calculate the original normal fringe spacing:}\) The wavelength, screen distance and slit separation are \[ \lambda=6.0\times10^{-7}\,\text{m} \] \[ D=1.5\,\text{m} \] \[ d=3.0\times10^{-4}\,\text{m} \] The spacing measured perpendicular to the fringe lines remains \[ \beta=\frac{\lambda D}{d} \] \[ \beta=3.0\,\text{mm} \] \(\textbf{Find the spacing along a horizontal scan:}\) Along the horizontal axis, \[ y=0 \] Successive maxima satisfy \[ x_m=m\sqrt2\beta \] Hence, \[ \Delta x=\sqrt2\beta \] \[ \Delta x=3\sqrt2\,\text{mm} \] \(\textbf{Find the spacing along a vertical scan:}\) Along the vertical axis, \[ x=0 \] The maxima satisfy \[ y_m=m\sqrt2\beta \] Therefore, \[ \Delta y=3\sqrt2\,\text{mm} \] \(\textbf{Resolve the apparent spacing difference:}\) The shortest distance between adjacent parallel bright-fringe lines remains \(3.0\,\text{mm}\). A horizontal or vertical scan cuts the rotated lines obliquely. The resulting distances between intersections are larger by a factor of \(\sqrt2\). No physical wavelength or slit separation change is required to produce this difference in measured spacing. \(\textbf{Physical interpretation:}\) Young's interference fringes are loci of constant optical-path difference. Their orientation is determined by the direction joining the two coherent slits. Rotating that direction rotates the fringe lines, while the spacing measured perpendicular to them remains controlled by \(\lambda D/d\). A detector scanning along a different direction measures a projected fringe spacing rather than the true perpendicular spacing. The example distinguishes a change in the optical pattern's orientation from a change in its intrinsic fringe width. \(\textbf{Final answer:}\) \[ \boxed{\text{fringes perpendicular to slit separation},\quad \Delta x=\Delta y=3\sqrt2\,\text{mm},\quad\beta_{\perp}=3.0\,\text{mm}} \]

Explanation: \(\textbf{Establish the original coherent intensity limits:}\) Each slit contributes \[ I_1=I_2=4I_0 \] The interference extremes are \[ I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2 \] \[ I_{\max}=(2\sqrt{I_0}+2\sqrt{I_0})^2 \] \[ I_{\max}=16I_0 \] Similarly, \[ I_{\min}=(2\sqrt{I_0}-2\sqrt{I_0})^2 \] \[ I_{\min}=0 \] \(\textbf{Interpret trial P's unchanged contrast and spacing:}\) Trial P preserves \[ I_{\max}=16I_0,\qquad I_{\min}=0 \] It also preserves the fringe width of \(2.0\,\text{mm}\). However, the geometrical centre changes from bright to dark. A phase change of \[ \Delta\phi=\pi \] interchanges the constructive and destructive conditions at fixed screen coordinates. It does not change the separation between successive maxima. Trial P is therefore the fixed-phase-shift modification. \(\textbf{Use trial Q's extreme intensities to recover its beam strengths:}\) Trial Q has \[ I_{\max}=9I_0,\qquad I_{\min}=I_0 \] These satisfy \[ \sqrt{I'_1}+\sqrt{I'_2}=3\sqrt{I_0} \] \[ \sqrt{I'_1}-\sqrt{I'_2}=\sqrt{I_0} \] Adding, \[ 2\sqrt{I'_1}=4\sqrt{I_0} \] \[ I'_1=4I_0 \] Subtracting, \[ 2\sqrt{I'_2}=2\sqrt{I_0} \] \[ I'_2=I_0 \] The stronger beam remains unchanged, and the other is attenuated. \(\textbf{Check why trial Q's fringe width stays fixed:}\) The attenuation changes the field amplitudes but not the wavelength or geometrical path difference. Therefore, \[ \beta_Q=2.0\,\text{mm} \] The unequal amplitudes prevent complete destructive cancellation, explaining the nonzero minimum. \(\textbf{Analyse trial R's changed spatial scale:}\) Trial R retains the original maximum and minimum intensities and a bright geometrical centre. Its fringe width increases from \[ \beta_1=2.0\,\text{mm} \] to \[ \beta_R=3.0\,\text{mm} \] For fixed wavelength and slit separation, \[ \beta\propto D \] Therefore, \[ \frac{D_R}{D_1}=\frac{\beta_R}{\beta_1} \] \[ \frac{D_R}{D_1}=\frac{3.0}{2.0} \] \[ \frac{D_R}{D_1}=1.5 \] This corresponds to increasing the screen distance by \(50\%\). \(\textbf{Physical interpretation:}\) An interference pattern contains several experimentally independent observables. Its maximum and minimum intensities reveal amplitude balance and background illumination. Its central bright-or-dark condition reveals the phase relation at a fixed reference point. Its fringe width reveals the spatial rate at which path difference changes across the screen. A phase shift can move fringes without changing their contrast or spacing. Attenuation can reduce contrast without changing spacing. Changing screen distance can alter spacing without destroying the coherent amplitude relation. The three measurements therefore permit the modifications to be identified without knowing which adjustment was made in advance. The correct matching is P: phase shift; Q: one-beam attenuation; R: increased screen distance.

Explanation: \(\textbf{Determine the first plate's excess optical path:}\) For a plate replacing an equal thickness of air, the additional optical path is \[ \Delta=(n-1)t \] The first plate has \[ n=1.50 \] \[ t_1=1.2\,\mu\text{m} \] Therefore, \[ \Delta_1=(1.50-1)(1.2) \] \[ \Delta_1=0.60\,\mu\text{m} \] \[ \Delta_1=600\,\text{nm} \] \(\textbf{Evaluate the first plate for red light:}\) The red wavelength is \[ \lambda_R=600\,\text{nm} \] The plate-induced path difference is \[ \frac{\Delta_1}{\lambda_R}=\frac{600}{600} \] \[ \frac{\Delta_1}{\lambda_R}=1 \] An integer number of wavelengths produces constructive interference at the geometrical centre. The centre is therefore bright for red light. \(\textbf{Evaluate the first plate for violet light:}\) The violet wavelength is \[ \lambda_V=400\,\text{nm} \] Therefore, \[ \frac{\Delta_1}{\lambda_V}=\frac{600}{400} \] \[ \frac{\Delta_1}{\lambda_V}=\frac32 \] A half-integer number of wavelengths gives destructive interference. The centre is dark for violet light. \(\textbf{Double the plate thickness:}\) The replacement plate has \[ t_2=2t_1=2.4\,\mu\text{m} \] Its excess optical path is \[ \Delta_2=(1.50-1)(2.4) \] \[ \Delta_2=1.2\,\mu\text{m} \] \[ \Delta_2=1200\,\text{nm} \] \(\textbf{Re-evaluate the red centre:}\) \[ \frac{\Delta_2}{\lambda_R}=\frac{1200}{600} \] \[ \frac{\Delta_2}{\lambda_R}=2 \] The red component remains constructive. \(\textbf{Re-evaluate the violet centre:}\) \[ \frac{\Delta_2}{\lambda_V}=\frac{1200}{400} \] \[ \frac{\Delta_2}{\lambda_V}=3 \] The violet component is now also constructive. \(\textbf{Explain why the two colours need separate tests:}\) The same physical plate introduces the same excess optical path because its index is specified to be the same for both wavelengths. However, the associated phase difference is \[ \phi=\frac{2\pi\Delta}{\lambda} \] The smaller violet wavelength therefore experiences a larger phase change in radians for the same optical path. Consequently, a plate can preserve constructive interference for one colour while producing destructive interference for another. \(\textbf{Physical interpretation:}\) An optical-path difference must be judged in wavelength units rather than by its physical length alone. The first plate adds one red wavelength but one and a half violet wavelengths. The geometrical centre is consequently bright in red and dark in violet. Doubling the thickness makes the added path an integer multiple of both wavelengths. This restores a bright centre for both colours without changing the slit separation, screen distance or source frequencies. The experiment combines wavelength-dependent phase classification with a controlled change of plate thickness. \(\textbf{Final answer:}\) \[ \boxed{\Delta_1=600\,\text{nm}:\ R\text{ bright},\ V\text{ dark};\quad\Delta_2=1200\,\text{nm}:\ R\text{ bright},\ V\text{ bright}} \]

Explanation: \(\textbf{Write the intensity distribution including the initial phase:}\) For two equal coherent beam intensities, \[ I(y)=2I_0[1+\cos\Phi(y)] \] The total relative phase is \[ \Phi(y)=\phi_0+\frac{2\pi y}{\beta} \] The unknown is \(\phi_0\). \(\textbf{Use the centre measurement to restrict the phase:}\) At \[ y=0 \] the measured intensity is \[ I(0)=2I_0 \] Therefore, \[ 2I_0[1+\cos\phi_0]=2I_0 \] \[ \cos\phi_0=0 \] Within the specified phase interval, this permits \[ \phi_0=\frac{\pi}{2} \] or \[ \phi_0=\frac{3\pi}{2} \] \(\textbf{Use the quarter-fringe measurement to resolve the ambiguity:}\) At \[ y=+\frac{\beta}{4} \] the geometrical phase contribution is \[ \frac{2\pi y}{\beta}=\frac{\pi}{2} \] The measured intensity is zero: \[ I\left(\frac{\beta}{4}\right)=0 \] Thus, \[ 1+\cos\left(\phi_0+\frac{\pi}{2}\right)=0 \] \[ \cos\left(\phi_0+\frac{\pi}{2}\right)=-1 \] \(\textbf{Identify the permitted initial phase:}\) Destructive interference requires \[ \phi_0+\frac{\pi}{2}=\pi\pmod{2\pi} \] Therefore, \[ \phi_0=\frac{\pi}{2} \] The alternative \(\frac{3\pi}{2}\) would make this detector position bright, contradicting the measurement. \(\textbf{Check the half-fringe observation:}\) At \[ y=+\frac{\beta}{2} \] the geometric phase contribution is \[ \pi \] The total phase is \[ \Phi=\frac{\pi}{2}+\pi \] \[ \Phi=\frac{3\pi}{2} \] Since \[ \cos\frac{3\pi}{2}=0 \] the intensity is \[ I=2I_0 \] This agrees with the third measurement. \(\textbf{Find the nearest bright-fringe location:}\) Constructive interference requires \[ \Phi(y)=2m\pi \] Therefore, \[ \frac{\pi}{2}+\frac{2\pi y}{\beta}=2m\pi \] Solving, \[ \frac y\beta=m-\frac14 \] The integer choice \(m=0\) gives \[ y=-\frac{\beta}{4} \] This is the closest bright fringe to the geometrical centre. \(\textbf{Physical interpretation:}\) The intensity at the geometrical centre does not always determine the sign of an initial phase difference. A half-maximum centre intensity permits two phase values within a full cycle. An additional measurement at a known fractional fringe displacement can distinguish them. Once the phase is determined, the actual bright-fringe positions follow by combining the initial phase with the geometrical phase variation. The experiment reconstructs an unobserved optical phase from measured intensity data. \(\textbf{Final answer:}\) \[ \boxed{\phi_0=\frac{\pi}{2},\quad y_{\mathrm{nearest\ bright}}=-\frac{\beta}{4}} \]

Explanation: Statement I is correct because equal optical paths give zero path difference for the two slit beams. A zero path difference is an integer multiple of every wavelength. Consequently, all wavelengths can contribute constructively at the geometrical centre, provided each wavelength has coherent contributions from both slits. The central fringe of an ideal white-light double-slit pattern can therefore appear white. Statement II is correct because the fringe width depends on wavelength: \[ \beta=\frac{\lambda D}{d} \] Red light has a longer wavelength than violet light and therefore produces wider fringes for the same slit geometry. Their noncentral bright-fringe positions generally differ. Statement III is incorrect because the position of the \(m\)-th bright fringe is \[ y_m=m\frac{\lambda D}{d} \] For equal nonzero order \(m\), this position depends on wavelength. Red and violet fringes of the same noncentral order therefore do not generally coincide. Statement IV is correct because white light contains contributions over a range of wavelengths. At a noncentral point, one wavelength may be constructive while another is destructive or intermediate. Because different colours have different optical frequencies, their measured intensities add over the observation time. Overlapping differently positioned coloured patterns can reduce the contrast of the combined illumination. This is different from the central position, where zero geometrical path difference favours constructive interference simultaneously for all component wavelengths. The experiment explains why a white-light interference pattern has a prominent central fringe and colour-dependent structure away from it. It also distinguishes coherence of the two paths for each wavelength from mutual interference between light waves of different optical frequencies. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Identify how two inclined plane waves create a spatial phase difference:}\) The beams propagate at equal angles above and below the horizontal. Let their propagation directions be inclined at \[ +\theta\quad\text{and}\quad-\theta \] respectively. Their total crossing angle is \[ \gamma=2\theta \] \(\textbf{Determine the path-difference variation along the screen:}\) At a vertical displacement \(y\), the phase difference associated with the two directions is \[ \phi(y)=\frac{2\pi}{\lambda}(2y\sin\theta) \] Adjacent constructive-interference positions differ in phase by \(2\pi\). Therefore, \[ \frac{2\pi}{\lambda}(2\beta\sin\theta)=2\pi \] \(\textbf{Recover the fringe-spacing relation:}\) Simplifying, \[ 2\beta\sin\theta=\lambda \] Thus, \[ \beta=\frac{\lambda}{2\sin\theta} \] Unlike the standard Young's double-slit formula, this spacing follows directly from the angular difference between the two interfering plane waves. \(\textbf{Calculate the individual inclination angle:}\) The wavelength is \[ \lambda=600\,\text{nm}=6.0\times10^{-7}\,\text{m} \] The measured fringe width is \[ \beta=0.30\,\text{mm}=3.0\times10^{-4}\,\text{m} \] Therefore, \[ \sin\theta=\frac{\lambda}{2\beta} \] \[ \sin\theta=\frac{6.0\times10^{-7}}{2(3.0\times10^{-4})} \] \[ \sin\theta=1.0\times10^{-3} \] For small angles, \[ \theta\approx1.0\,\text{mrad} \] \(\textbf{Determine the complete beam crossing angle:}\) \[ \gamma=2\theta \] \[ \gamma=2.0\,\text{mrad} \] \(\textbf{Calculate the original coherent intensity extremes:}\) The individual intensities are \[ I_1=I_2=I_0 \] For coherent beams of identical polarization, \[ I=2I_0(1+\cos\phi) \] At constructive interference, \[ I_{\max}=4I_0 \] At destructive interference, \[ I_{\min}=0 \] \(\textbf{Apply the single-beam condition:}\) When one beam is blocked, only one plane wave illuminates the screen. There is no cross term between two fields. The intensity becomes \[ I_{\mathrm{blocked}}=I_0 \] It is uniform across the stated observation region. \(\textbf{Physical interpretation:}\) Stable interference fringes do not require two physical slits; they require overlapping coherent waves with a suitable phase relationship. Two plane waves crossing at a small angle produce a periodic transverse phase difference. Their angular separation determines the spatial fringe width. Equal coherent intensities can give complete destructive cancellation and a maximum four times the individual intensity. Blocking either wave removes the interference modulation, leaving the intensity of the surviving wave. This arrangement distinguishes the essential mechanism of interference from the particular experimental geometry of Young's slits. \(\textbf{Final answer:}\) \[ \boxed{\theta=1.0\,\text{mrad},\quad\gamma=2.0\,\text{mrad},\quad I_{\max}=4I_0,\quad I_{\min}=0,\quad I_{\mathrm{blocked}}=I_0} \]

Explanation: Statement I is correct because two-slit interference requires contributions from both coherent slit fields. When one slit is completely blocked, the cross term between the two slit fields disappears. However, light passing through the remaining slit still has wave character. Because that slit has finite width, its transmitted light can produce a single-slit diffraction pattern. The disappearance of fine Young's fringes therefore does not imply that all wave-optical spreading must disappear. Statement II is correct because the fields must overlap in the region where interference is observed. Maintaining coherence at the sources cannot create interference at a screen point reached by only one field. The partition eliminates the necessary spatial superposition in the stated observation region. Statement III is incorrect because increasing the intensity of a single surviving beam cannot create the missing second coherent amplitude. Interference is not restored by brightness alone when the other slit contribution is exactly zero. The remaining intensity may increase, but the two-beam cross term remains absent. Statement IV is correct because a stable interference pattern also requires a sufficiently stable relative phase during observation. Even with both slit fields overlapping, random phase changes shift the instantaneous pattern. A long exposure can average those changes and reduce or remove the recorded modulation. The four modifications test different requirements rather than different numerical fringe-width formulae. The presence of two fields, their spatial overlap and their phase stability are distinct experimental conditions. Equal intensity is favourable for high contrast but is not a substitute for any of those requirements. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Calculate the original fringe width:}\) The given quantities are \[ \lambda=5.0\times10^{-7}\,\text{m} \] \[ d=4.0\times10^{-4}\,\text{m} \] \[ D=1.6\,\text{m} \] For Young's experiment, \[ \beta=\frac{\lambda D}{d} \] Substituting, \[ \beta=\frac{(5.0\times10^{-7})(1.6)}{4.0\times10^{-4}} \] \[ \beta=2.0\times10^{-3}\,\text{m} \] \[ \beta=2.0\,\text{mm} \] \(\textbf{Determine the original detector order:}\) The detector coordinate is \[ y=+5.0\,\text{mm} \] Therefore, \[ \frac{y}{\beta}=\frac{5.0}{2.0} \] \[ \frac{y}{\beta}=\frac52 \] A half-integer interference order is a dark fringe. \(\textbf{Check the requested plate condition:}\) The detector is already dark before insertion. A plate of zero thickness leaves this condition unchanged. However, the question explicitly asks for a least positive plate thickness that produces darkness after a plate is inserted. For a fixed detector initially at a dark fringe, darkness recurs whenever the added optical path is an integer number of wavelengths. \(\textbf{Determine the first strictly positive recurrence:}\) The smallest positive added optical path that preserves the dark condition is \[ \Delta=\lambda \] For a plate replacing air, \[ \Delta=(n-1)t \] Therefore, \[ t=\frac{\lambda}{n-1} \] Substituting, \[ t=\frac{500\,\text{nm}}{1.50-1} \] \[ t=1000\,\text{nm} \] \[ t=1.0\,\mu\text{m} \] \(\textbf{Determine the fringe-pattern displacement:}\) An added path difference of one wavelength translates the pattern by one fringe width. Therefore, \[ s=\beta \] \[ s=2.0\,\text{mm} \] \(\textbf{Physical interpretation:}\) The detector lies at a position corresponding to a half-integer number of fringe widths from the centre. It is therefore dark before the plate is inserted. A plate can preserve this dark condition when it adds a complete number of wavelengths to the relative optical path. The least positive recurrence requires an excess path equal to one wavelength. This represents a shift of the entire pattern by one fringe width, so a dark fringe again occupies the detector position. \(\textbf{Final answer:}\) \[ \boxed{\text{initially dark},\quad t_{\min,+}=1.0\,\mu\text{m},\quad s=2.0\,\text{mm}} \]