Optics MCQs with Answers – Part 6 (NEET Physics)

Explanation: \(\textbf{Determine the diffraction-limited angular scale:}\) The aperture width is \[ a=0.20\times10^{-3}\,\text{m} \] The wavelength is \[ \lambda=500\times10^{-9}\,\text{m} \] The first minimum of the single-slit pattern occurs approximately at \[ \theta_1=\frac{\lambda}{a} \] Therefore, \[ \theta_1=\frac{500\times10^{-9}}{0.20\times10^{-3}} \] \[ \theta_1=2.5\times10^{-3}\,\text{rad} \] \[ \theta_1=2.5\,\text{mrad} \] \(\textbf{Convert diffraction angle to focal-plane distance:}\) The lens focal length is \[ f=0.40\,\text{m} \] The first minimum is displaced from the corresponding central maximum by \[ h=f\theta_1 \] Substituting, \[ h=(0.40)(2.5\times10^{-3}) \] \[ h=1.0\,\text{mm} \] \(\textbf{Interpret the specified resolution criterion:}\) The criterion requires the separation between the two central maxima to be at least \(h\). In angular terms, this means \[ \Delta\theta\geq\theta_1 \] In the focal plane, it means \[ s\geq1.0\,\text{mm} \] \(\textbf{Evaluate experiment P:}\) The angular separation is \[ \Delta\theta_P=2.0\,\text{mrad} \] Its focal-plane peak separation is \[ s_P=f\Delta\theta_P \] \[ s_P=(0.40)(2.0\times10^{-3}) \] \[ s_P=0.80\,\text{mm} \] Since \[ 0.80\lt 1.0 \] experiment P fails the defined criterion. \(\textbf{Evaluate experiment Q:}\) The angular separation is \[ \Delta\theta_Q=3.0\,\text{mrad} \] Therefore, \[ s_Q=f\Delta\theta_Q \] \[ s_Q=(0.40)(3.0\times10^{-3}) \] \[ s_Q=1.20\,\text{mm} \] Since \[ 1.20\gt 1.0 \] experiment Q satisfies the criterion. \(\textbf{Distinguish angular and linear descriptions:}\) The aperture fixes the diffraction angular scale. The lens focal length converts that angular scale into a focal-plane displacement. Changing focal length alone would enlarge both the diffraction pattern and the separation of the two central images by the same factor. It would not change the angular comparison underlying the stated criterion. \(\textbf{Physical interpretation:}\) A finite aperture causes light from each distant direction to spread into a diffraction pattern instead of forming an infinitely sharp geometrical point. The ability to distinguish two nearby directions therefore depends on their angular separation relative to the diffraction spread. The stated resolution criterion avoids the ambiguity of judging merely whether two bright regions overlap. The comparison integrates single-slit diffraction with focal-plane image formation by a convex lens. \(\textbf{Final answer:}\) \[ \boxed{h=1.0\,\text{mm},\quad s_P=0.80\,\text{mm}\lt h,\quad s_Q=1.20\,\text{mm}\gt h} \]

Explanation: \(\textbf{Use the minimum condition at a fixed observation angle:}\) For single-slit diffraction, \[ a\sin\theta=m\lambda \] The wavelength is \[ \lambda=500\,\text{nm} \] The detector direction satisfies \[ \sin\theta=0.005 \] At fixed angle, the minimum order is determined by \[ m=\frac{a\sin\theta}{\lambda} \] A diffraction minimum occurs only when this result is a positive integer. \(\textbf{Test width P:}\) For \[ a_P=0.10\,\text{mm} \] the path difference across the complete aperture is \[ \Delta_P=a_P\sin\theta \] \[ \Delta_P=(0.10\times10^{-3})(0.005) \] \[ \Delta_P=500\,\text{nm} \] Therefore, \[ m_P=\frac{500}{500} \] \[ m_P=1 \] P is correctly identified as the first minimum. \(\textbf{Test width Q:}\) For \[ a_Q=0.15\,\text{mm} \] \[ \Delta_Q=(0.15\times10^{-3})(0.005) \] \[ \Delta_Q=750\,\text{nm} \] Thus, \[ m_Q=\frac{750}{500} \] \[ m_Q=1.5 \] This is not an integer. Q is correctly identified as a non-minimum. \(\textbf{Test width R:}\) For \[ a_R=0.20\,\text{mm} \] \[ \Delta_R=(0.20\times10^{-3})(0.005) \] \[ \Delta_R=1000\,\text{nm} \] Therefore, \[ m_R=\frac{1000}{500} \] \[ m_R=2 \] R is correctly classified as the second minimum. \(\textbf{Test width S:}\) For \[ a_S=0.25\,\text{mm} \] \[ \Delta_S=(0.25\times10^{-3})(0.005) \] \[ \Delta_S=1250\,\text{nm} \] Thus, \[ m_S=\frac{1250}{500} \] \[ m_S=2.5 \] This is not an integer, so S is not a second minimum. \(\textbf{Interpret the changing intensity:}\) As the slit width increases, more secondary-wave contributions reach the detector. However, their relative phases also change because the total path difference across the aperture increases. At particular widths, those contributions cancel completely. At intermediate widths, cancellation is incomplete and the detected intensity is nonzero. \(\textbf{Physical interpretation:}\) Increasing aperture width does not guarantee increasing intensity in every fixed off-axis direction. The larger aperture changes both the number of contributing wavelets and their phase relationships. A detector can therefore move from a diffraction zero to a nonzero region and then back to another zero as the slit is widened. The minimum order at a fixed angle is determined by the ratio of edge-to-edge path difference to wavelength. The table tests that changing-order interpretation rather than a simple width-scaling calculation. \(\textbf{Final answer:}\) P is the first minimum, Q is not a minimum, R is the second minimum, and S is not a minimum.

Explanation: Statement I is correct because the total edge-to-edge path difference is \(2\lambda\). If the aperture is divided into four equal sections, corresponding points in neighbouring sections are separated across the aperture by \(a/4\). Their optical-path difference is therefore one-quarter of the total: \[ \Delta_{\mathrm{pair}}=\frac{2\lambda}{4} \] \[ \Delta_{\mathrm{pair}}=\frac{\lambda}{2} \] Such paired contributions differ in phase by \(\pi\). For uniform illumination, neighbouring equal-width sections can therefore be paired to obtain complete cancellation. Statement II is incorrect because corresponding points in the two halves are separated by \(a/2\). Their path difference is one wavelength: \[ \Delta_{\mathrm{halves}}=\frac{2\lambda}{2} \] \[ \Delta_{\mathrm{halves}}=\lambda \] That corresponds to a phase difference of \(2\pi\), not \(\pi\). The two halves cannot be shown to cancel merely by pairing corresponding points in this way. Statement III is correct because the single-slit minimum condition is \[ a\sin\theta=m\lambda \] An edge-to-edge difference of \(2\lambda\) means \(m=2\), the second minimum. Statement IV is correct because a slit of width \(a/2\) has half the edge-to-edge path difference at the same observation angle. For the replacement aperture, \[ \frac a2\sin\theta=\lambda \] This satisfies the first-minimum condition. The question distinguishes a valid cancellation pairing from an invalid one and connects the mechanism to the diffraction order. The same angular direction can correspond to different minimum orders for different aperture widths. Statements I, III and IV are correct.

Explanation: The Assertion is true because knowing the positions of two neighbouring minima does not automatically locate the intervening maximum at their midpoint. The diffraction-minimum condition identifies angles at which the contributions from the aperture cancel completely. Between those angles, the contributions generally combine to produce nonzero resultant intensity. The intensity must rise from one minimum and fall again towards the next. However, the detailed variation is determined by the phase distribution of contributions across the entire slit. The position of greatest intensity between two minima need not divide their angular separation into equal halves. The Reason correctly distinguishes determining zeros from determining maxima. A cancellation condition alone does not specify the complete intensity function between its zeros. The weaker secondary maxima are therefore not located by simply taking arithmetic midpoints of successive minima. This differs from the ideal equally spaced bright-fringe positions of Young's two-slit interference in the small-angle approximation. The familiar single-slit pattern has a broad central maximum and progressively weaker secondary bright regions. Its dark-minimum positions and secondary-maximum positions must be treated as distinct features. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Determine how immersion changes the lens power:}\) For a thin lens surrounded by a medium of refractive index \(n_m\), \[ \frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \] The glass index is \[ n_g=1.50 \] The lens surface radii do not change between experiments. Therefore, the curvature factor is common. \(\textbf{Compare the relative refractive-index factors:}\) In air, \[ n_{m1}=1 \] The lens-power factor is \[ \frac{n_g}{n_{m1}}-1=1.50-1 \] \[ =0.50 \] In the liquid, \[ n_{m2}=1.20 \] The factor becomes \[ \frac{n_g}{n_{m2}}-1=\frac{1.50}{1.20}-1 \] \[ =0.25 \] The lens power therefore becomes one-half of its original value. \(\textbf{Determine the new focal length:}\) Initially, \[ f_1=20\,\text{cm} \] Because focal length is inversely proportional to lens power, \[ \frac{f_2}{f_1}=\frac{0.50}{0.25} \] \[ \frac{f_2}{f_1}=2 \] Thus, \[ f_2=40\,\text{cm} \] \(\textbf{Calculate the first diffraction-minimum displacement in air:}\) The first-minimum angle satisfies \[ \theta_1\approx\frac{\lambda_1}{a} \] A lens maps this angular direction to its focal plane at \[ x_1=f_1\theta_1 \] Therefore, \[ x_1=\frac{f_1\lambda_1}{a} \] Using \[ f_1=0.20\,\text{m},\qquad\lambda_1=600\,\text{nm} \] and \[ a=0.30\,\text{mm} \] gives \[ x_1=\frac{(0.20)(600\times10^{-9})}{0.30\times10^{-3}} \] \[ x_1=0.40\,\text{mm} \] \(\textbf{Determine the wavelength in the liquid:}\) Frequency remains unchanged. Therefore, \[ \lambda_2=\frac{\lambda_0}{n_{m2}} \] \[ \lambda_2=\frac{600}{1.20} \] \[ \lambda_2=500\,\text{nm} \] \(\textbf{Calculate the new focal-plane displacement:}\) \[ x_2=\frac{f_2\lambda_2}{a} \] Substituting, \[ x_2=\frac{(0.40)(500\times10^{-9})}{0.30\times10^{-3}} \] \[ x_2=\frac23\,\text{mm} \] \(\textbf{Compare angular and linear diffraction widths:}\) The diffraction angle decreases because the wavelength is smaller in the liquid. However, the lens focal length doubles. The larger focal length more than compensates for the reduced angular diffraction spread. The measured focal-plane pattern therefore becomes larger. \(\textbf{Physical interpretation:}\) Immersion changes two independent features of this apparatus. It reduces the light wavelength in the surrounding medium, which narrows the angular diffraction distribution. It also changes the relative refractive index of the lens, reducing its converging power and increasing its focal length. The focal plane converts angular spread into a linear displacement proportional to that focal length. The final pattern size must therefore be calculated using both changes. This integrates lens-maker reasoning with the angular interpretation of single-slit diffraction. \(\textbf{Final answer:}\) \[ \boxed{f_1=20\,\text{cm},\quad f_2=40\,\text{cm},\quad x_1=0.40\,\text{mm},\quad x_2=\frac23\,\text{mm}} \]

Explanation: The edge-to-edge geometrical path difference is specified as \(600\,\text{nm}\). For a single slit, diffraction minima occur when that path difference equals a positive integral multiple of the wavelength. For red light, the wavelength is \(600\,\text{nm}\). The edge-to-edge path difference is therefore exactly one red wavelength. The observation direction is the red first diffraction minimum. For violet light, the wavelength is \(400\,\text{nm}\). The same geometrical path difference corresponds to one and a half violet wavelengths. That is not a positive integer number of violet wavelengths, so the usual single-slit minimum condition is not satisfied. The violet intensity is consequently nonzero under the stated ideal uniform-illumination conditions. However, a nonzero diffraction intensity need not be a local maximum. The exact positions of secondary maxima depend on the detailed variation of resultant field amplitude with angle. They cannot be inferred merely by selecting the midpoint between two consecutive minimum-order values. The shorter wavelength also does not make every observation point darker or brighter in a simple monotonic way. The correct classification is a red first minimum and a violet non-minimum, without asserting that the violet point is a secondary maximum.

Explanation: The Assertion is true because the angular diffraction pattern of a uniformly illuminated single slit depends on the slit width and wavelength. Moving the complete aperture sideways without rotation does not change those two quantities. The relative distances between different contributing points within the slit also remain unchanged. At a fixed far-field observation angle, all those points are translated by the same amount. The translation can alter the overall phase reference of the diffracted field. However, it does not alter the phase differences responsible for the slit's angular cancellation conditions. The first and higher diffraction minima therefore retain their angular positions in the ideal far-field description. A different situation occurs when only one slit of a double-slit aperture is moved. That movement can alter the centre-to-centre slit separation and the midpoint of the pair. The two-slit interference pattern can consequently change in fringe width and position. The Reason correctly explains why rigid translation of an entire single aperture differs from changing the relative geometry of two coherent apertures. The result does not imply that every form of aperture motion leaves diffraction unchanged. Rotating the slit, changing its width, or changing the direction of illumination can affect the observed angular pattern. The stated translation leaves those properties fixed. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: In arrangement P, one normally incident coherent plane wave illuminates the slit. The field contributions from different parts of the aperture have a definite relative phase for each observation direction. They cancel completely at the ideal single-slit diffraction minima. In arrangement Q, different narrow source regions illuminate the same slit at different angles. Each narrow source independently generates a diffraction pattern centred on its own incidence direction. Those individual patterns are displaced relative to one another on the observation screen. Because the source regions are mutually incoherent, their time-averaged intensities add. A direction that is a minimum for one individual pattern need not be a minimum for another. Consequently, contributions from other source directions can fill in a minimum that was dark under the single-plane-wave illumination. The combined pattern may therefore have weaker contrast and less sharply defined minima. This does not mean that individual source components have ceased to diffract. It means that the detector receives a superposition of differently positioned diffraction intensity distributions. The situation parallels the effect of source extent on the visibility of interference patterns, while retaining the distinct single-aperture diffraction mechanism. The correct outcome is that diffraction remains present but its observed minima and contrast may become less distinct.

Explanation: \(\textbf{Infer the original aperture width from the focal-plane minimum:}\) For a single slit, \[ a\sin\theta_1=\lambda \] Under paraxial conditions, \[ x_1=f\theta_1 \] Therefore, \[ x_1=\frac{f\lambda}{a} \] The measured values are \[ f=0.50\,\text{m} \] \[ \lambda=500\times10^{-9}\,\text{m} \] \[ x_1=1.0\times10^{-3}\,\text{m} \] Rearranging, \[ a=\frac{f\lambda}{x_1} \] Substituting, \[ a=\frac{(0.50)(500\times10^{-9})}{1.0\times10^{-3}} \] \[ a=2.5\times10^{-4}\,\text{m} \] \[ a=0.25\,\text{mm} \] \(\textbf{Double the transmitting aperture width:}\) The new width is \[ a'=2a \] \[ a'=0.50\,\text{mm} \] The wavelength and lens focal length remain unchanged. \(\textbf{Determine the new first-minimum position:}\) \[ x'_1=\frac{f\lambda}{a'} \] \[ x'_1=\frac{(0.50)(500\times10^{-9})}{0.50\times10^{-3}} \] \[ x'_1=0.50\,\text{mm} \] \(\textbf{Classify the fixed detector after replacement:}\) The detector remains at \[ x_D=+1.0\,\text{mm} \] Relative to the new first-minimum displacement, \[ \frac{x_D}{x'_1}=\frac{1.0}{0.50} \] \[ \frac{x_D}{x'_1}=2 \] The detector therefore lies at the second diffraction minimum. \(\textbf{Compare the on-axis field amplitudes:}\) For uniform incident illumination, the on-axis diffracted field is proportional to slit width. Therefore, \[ E_{\mathrm{axis}}\propto a \] Doubling the width doubles the field amplitude: \[ \frac{E'_{\mathrm{axis}}}{E_{\mathrm{axis}}}=2 \] \(\textbf{Determine the peak-intensity ratio:}\) Since intensity is proportional to the square of the field amplitude, \[ \frac{I'_{\max}}{I_{\max}}=\left(\frac{E'_{\mathrm{axis}}}{E_{\mathrm{axis}}}\right)^2 \] \[ \frac{I'_{\max}}{I_{\max}}=4 \] \(\textbf{Physical interpretation:}\) A larger aperture produces a narrower angular diffraction pattern. The fixed detector can remain at zero intensity while its diffraction order changes. At the same time, the greater aperture width permits coherent addition of more field contributions in the forward direction. The central peak therefore becomes more intense even though its angular width decreases. The question separates the change in diffraction-minimum order from the change in central peak brightness. \(\textbf{Final answer:}\) \[ \boxed{a=0.25\,\text{mm},\quad a'=0.50\,\text{mm},\quad m'_{\mathrm{detector}}=2,\quad\frac{I'_{\max}}{I_{\max}}=4} \]

Explanation: \(\textbf{Recover P's angular diffraction scale:}\) For a single slit, the full central width on a distant screen is \[ W_P=\frac{2D\lambda}{a_P} \] The given values are \[ W_P=8.0\times10^{-3}\,\text{m} \] \[ D=2.0\,\text{m} \] \[ \lambda=600\times10^{-9}\,\text{m} \] Therefore, \[ a_P=\frac{2D\lambda}{W_P} \] Substituting, \[ a_P=\frac{2(2.0)(600\times10^{-9})}{8.0\times10^{-3}} \] \[ a_P=0.30\,\text{mm} \] \(\textbf{Determine P's first-minimum angle:}\) \[ \theta_{1P}\approx\frac{\lambda}{a_P} \] \[ \theta_{1P}=\frac{600\times10^{-9}}{0.30\times10^{-3}} \] \[ \theta_{1P}=2.0\,\text{mrad} \] \(\textbf{Interpret Q's focal-plane width:}\) For an ideal lens of focal length \(f\), the first-minimum displacement is \[ x_1=\frac{f\lambda}{a_Q} \] The full central width is therefore \[ W_Q=\frac{2f\lambda}{a_Q} \] The focal length is \[ f=0.40\,\text{m} \] The measured central width is \[ W_Q=1.60\,\text{mm} \] \(\textbf{Recover Q's slit width:}\) \[ a_Q=\frac{2f\lambda}{W_Q} \] Substituting, \[ a_Q=\frac{2(0.40)(600\times10^{-9})}{1.60\times10^{-3}} \] \[ a_Q=0.30\,\text{mm} \] \(\textbf{Compare the diffraction angles:}\) Both slits have the same width and use the same wavelength. Consequently, \[ \theta_{1Q}\approx\frac{\lambda}{a_Q} \] \[ \theta_{1Q}=2.0\,\text{mrad} \] Therefore, \[ \theta_{1P}=\theta_{1Q} \] \(\textbf{Explain the unequal recorded pattern widths:}\) Experiment P converts diffraction angle into screen displacement using distance \(D=2.0\,\text{m}\). Experiment Q uses the lens focal length \(f=0.40\,\text{m}\). The linear widths differ by the scale ratio \[ \frac{W_P}{W_Q}=\frac{2.0}{0.40} \] \[ \frac{W_P}{W_Q}=5 \] The intrinsic angular diffraction patterns are nevertheless identical. \(\textbf{Physical interpretation:}\) A diffraction pattern has an intrinsic angular distribution determined by aperture width and wavelength. Different observing arrangements can produce different linear pattern sizes from the same angular distribution. A distant screen uses physical propagation distance to convert angles into coordinates. A lens focal plane uses focal length for the same purpose. The unequal measured central widths in these experiments therefore do not imply different slit widths. They arise entirely from the different observation scales. \(\textbf{Final answer:}\) \[ \boxed{a_P=a_Q=0.30\,\text{mm},\quad\theta_{1P}=\theta_{1Q}\approx2.0\,\text{mrad}} \]

Explanation: Item P matches (4) because the first diffraction minimum occurs when the optical-path difference across the complete slit equals one wavelength. This gives the condition \(a\sin\theta=\lambda\). It is a cancellation condition and therefore identifies a dark direction. Item Q matches (3) because the central bright maximum contains the forward propagation direction for ordinary normally incident uniform-phase illumination. At the forward direction, contributions from all parts of the slit are in phase. The central region extends until cancellation occurs at the first minimum on either side. Item R matches (2) because a secondary maximum occurs between neighbouring noncentral minima. The cancellation among slit contributions is incomplete in this region. Its peak is weaker than the broad central maximum in the ideal uniformly illuminated single-slit pattern. The secondary maximum must not be confused with a direction satisfying the minimum condition. Item S matches (1) because the full central angular width extends between the negative and positive first-minimum directions. For small diffraction angles, those directions are approximately \(-\lambda/a\) and \(+\lambda/a\). Their angular separation is therefore \(2\lambda/a\). The matching distinguishes a single dark direction, an extended central bright region, a weaker secondary bright region and a measured angular width. These represent different features of the same intensity distribution. The correct matching is P–(4), Q–(3), R–(2), S–(1).

Explanation: \(\textbf{Determine the original first-minimum position:}\) For a single slit, \[ x_1=\frac{D\lambda}{a} \] in the small-angle approximation. The original slit width is \[ a_1=0.40\times10^{-3}\,\text{m} \] The wavelength and screen distance are \[ \lambda=600\times10^{-9}\,\text{m} \] \[ D=2.0\,\text{m} \] Therefore, \[ x_{1,\mathrm{old}}=\frac{(2.0)(600\times10^{-9})}{0.40\times10^{-3}} \] \[ x_{1,\mathrm{old}}=3.0\,\text{mm} \] \(\textbf{Classify the detector initially:}\) The detector coordinate is \[ x_D=+3.0\,\text{mm} \] This coincides with the positive first diffraction minimum. Therefore, it is initially dark in the ideal model. \(\textbf{Calculate the original central width:}\) The full width extends between the first minima on opposite sides: \[ W_1=2x_{1,\mathrm{old}} \] \[ W_1=6.0\,\text{mm} \] \(\textbf{Reduce the slit width:}\) The adjusted slit width is \[ a_2=0.20\,\text{mm} \] The wavelength and screen distance remain fixed. The new first-minimum position is \[ x_{1,\mathrm{new}}=\frac{D\lambda}{a_2} \] Substituting, \[ x_{1,\mathrm{new}}=\frac{(2.0)(600\times10^{-9})}{0.20\times10^{-3}} \] \[ x_{1,\mathrm{new}}=6.0\,\text{mm} \] \(\textbf{Reclassify the fixed detector:}\) The detector still lies at \[ x_D=+3.0\,\text{mm} \] The new central maximum extends between \[ x=-6.0\,\text{mm} \] and \[ x=+6.0\,\text{mm} \] Since the detector lies strictly between these first minima, it is now inside the central diffraction region. It is not at the new first minimum. \(\textbf{Determine the new full width:}\) \[ W_2=2x_{1,\mathrm{new}} \] \[ W_2=12.0\,\text{mm} \] Therefore, \[ \frac{W_2}{W_1}=\frac{12.0}{6.0} \] \[ \frac{W_2}{W_1}=2 \] \(\textbf{Physical interpretation:}\) Narrowing the transmitting aperture increases diffraction spreading. The first minima move farther away from the central axis, causing the central maximum to become wider. A fixed detector that was initially positioned at a diffraction zero may consequently lie inside the central bright region after the aperture is narrowed. The central maximum broadens even though less light is admitted through the slit. Aperture transmission and angular spreading must therefore be treated as different physical consequences of changing slit width. \(\textbf{Final answer:}\) \[ \boxed{x_{1,\mathrm{old}}=3.0\,\text{mm},\quad x_{1,\mathrm{new}}=6.0\,\text{mm},\quad W_1=6.0\,\text{mm},\quad W_2=12.0\,\text{mm}} \]

Explanation: \(\textbf{Extract the central width from the measured minima:}\) The first negative-side minimum is at \[ x_-=-1.0\,\text{mm} \] The first positive-side minimum is at \[ x_+=+7.0\,\text{mm} \] The full central diffraction width is their separation: \[ W=x_+-x_- \] \[ W=7.0-(-1.0) \] \[ W=8.0\,\text{mm} \] \(\textbf{Recover the central diffraction direction:}\) For a uniformly illuminated slit receiving an inclined plane wave, the central maximum lies along the incident propagation direction. Its screen coordinate is also the midpoint of the two first minima: \[ x_c=\frac{x_++x_-}{2} \] \[ x_c=\frac{7.0-1.0}{2} \] \[ x_c=+3.0\,\text{mm} \] This agrees with the stated intensity maximum. \(\textbf{Find the incident angle:}\) The screen distance is \[ D=2.0\,\text{m} \] For a small incident angle \(\alpha\), \[ x_c\approx D\alpha \] Therefore, \[ \alpha=\frac{x_c}{D} \] \[ \alpha=\frac{3.0\times10^{-3}}{2.0} \] \[ \alpha=1.5\times10^{-3}\,\text{rad} \] \[ \alpha=+1.5\,\text{mrad} \] The positive sign indicates incidence towards the right. \(\textbf{Determine the first-minimum displacement relative to the shifted centre:}\) The central half-width is \[ h=\frac W2 \] \[ h=4.0\,\text{mm} \] The diffraction angle measured relative to the incident direction is \[ \theta_1\approx\frac hD \] \[ \theta_1=\frac{4.0\times10^{-3}}{2.0} \] \[ \theta_1=2.0\times10^{-3}\,\text{rad} \] \(\textbf{Recover the slit width:}\) For the first single-slit minimum, \[ a\sin\theta_1=\lambda \] Under the small-angle approximation, \[ a=\frac{\lambda}{\theta_1} \] The wavelength is \[ \lambda=500\times10^{-9}\,\text{m} \] Therefore, \[ a=\frac{500\times10^{-9}}{2.0\times10^{-3}} \] \[ a=2.5\times10^{-4}\,\text{m} \] \[ a=0.25\,\text{mm} \] \(\textbf{Check both measured minimum positions:}\) The predicted positions are \[ x_\pm=D(\alpha\pm\theta_1) \] Thus, \[ x_-=2.0(1.5-2.0)\times10^{-3} \] \[ x_-=-1.0\,\text{mm} \] and \[ x_+=2.0(1.5+2.0)\times10^{-3} \] \[ x_+=+7.0\,\text{mm} \] Both agree with the graph. \(\textbf{Physical interpretation:}\) The centre of a single-slit diffraction pattern identifies the direction of the incident plane wave. The spacing of the first minima about that centre determines the angular diffraction spread and hence the aperture width. An asymmetric pattern relative to the geometrical screen centre does not necessarily mean the slit itself is asymmetric. Uniform oblique illumination can shift the entire diffraction pattern while preserving symmetry about its new centre. The inverse measurement therefore separates wavefront steering from aperture-width determination. \(\textbf{Final answer:}\) \[ \boxed{a=0.25\,\text{mm},\quad\alpha=+1.5\,\text{mrad},\quad W=8.0\,\text{mm}} \]

Explanation: \(\textbf{Relate the diffraction and interference spatial scales:}\) For two-slit interference, the fringe width is \[ \beta=\frac{\lambda D}{d} \] For either finite-width slit, the first diffraction minimum occurs at \[ x_{\mathrm{diff},1}=\frac{\lambda D}{a} \] Taking the ratio eliminates the wavelength and screen distance: \[ \frac{x_{\mathrm{diff},1}}{\beta}=\frac da \] Substituting the aperture dimensions, \[ \frac{x_{\mathrm{diff},1}}{\beta}=\frac{0.50}{0.20} \] \[ \frac{x_{\mathrm{diff},1}}{\beta}=2.5 \] Statement I is correct. \(\textbf{Test whether the first diffraction minimum removes an interference maximum:}\) Two-slit bright maxima occur at integral fringe orders: \[ x_m=m\beta \] The first diffraction minimum occurs at \[ x_{\mathrm{diff},1}=2.5\beta \] This is not an integral interference order. It lies between the second and third nominal interference bright-fringe positions. Therefore, Statement II is incorrect. \(\textbf{Locate the second diffraction minimum:}\) The second minimum of an individual slit occurs at \[ x_{\mathrm{diff},2}=\frac{2\lambda D}{a} \] Therefore, \[ \frac{x_{\mathrm{diff},2}}{\beta}=\frac{2d}{a} \] Substituting, \[ \frac{x_{\mathrm{diff},2}}{\beta}=\frac{2(0.50)}{0.20} \] \[ \frac{x_{\mathrm{diff},2}}{\beta}=5 \] Thus, the fifth nominal two-slit bright fringe coincides with a zero of the single-slit envelope. Statement III is correct. \(\textbf{Count interference maxima within the central diffraction envelope:}\) The first diffraction minima occur at \[ x=\pm2.5\beta \] Interference maxima strictly inside this region have integer orders satisfying \[ -2.5\lt m\lt 2.5 \] The allowed values are \[ m=-2,-1,0,+1,+2 \] Thus, five interference bright-fringe centres lie in the central diffraction region. Statement IV is correct. \(\textbf{Physical interpretation:}\) The diffraction envelope and interference fringes depend on different physical dimensions of the same aperture. A missing interference order occurs only when an interference maximum coincides with a diffraction zero. The first diffraction minimum need not itself suppress a nominal interference maximum. In this arrangement, the first such coincidence occurs at a higher diffraction minimum. The central-envelope count and the missing-order condition must therefore be evaluated separately. Statements I, III and IV are correct.

Explanation: \(\textbf{Translate the film's intensity transmission into field transmission:}\) The film reduces intensity through the upper half to one-quarter. Because intensity is proportional to the square of electric-field amplitude, \[ I\propto E^2 \] its field-amplitude transmission factor is \[ t_E=\sqrt{\frac14} \] \[ t_E=\frac12 \] Thus, the transmitted upper-half field amplitude is halved, not quartered. \(\textbf{Represent the original forward-axis field:}\) Let the original resultant forward-axis amplitude be \(E_0\). For uniform illumination, the two equal-width halves each contribute half this amplitude in phase: \[ E_{\mathrm{lower},0}=\frac{E_0}{2} \] \[ E_{\mathrm{upper},0}=\frac{E_0}{2} \] After attenuation, their forward contributions become \[ E_{\mathrm{lower}}=\frac{E_0}{2} \] \[ E_{\mathrm{upper}}=\frac{E_0}{4} \] \(\textbf{Calculate the modified forward intensity:}\) The fields still arrive in phase on the forward axis. Therefore, \[ E_{\mathrm{forward}}=\frac{E_0}{2}+\frac{E_0}{4} \] \[ E_{\mathrm{forward}}=\frac{3E_0}{4} \] Consequently, \[ \frac{I_{\mathrm{forward}}}{I_{\max,0}}=\left(\frac{3E_0/4}{E_0}\right)^2 \] \[ \frac{I_{\mathrm{forward}}}{I_{\max,0}}=\frac9{16} \] \(\textbf{Analyse the former first diffraction minimum:}\) Originally, the direction satisfied \[ a\sin\theta_1=\lambda \] Corresponding points in the upper and lower halves differ in geometrical path by \[ \frac a2\sin\theta_1=\frac{\lambda}{2} \] Their contributions therefore have opposite phases. The two original half-aperture resultant fields have equal magnitudes and cancel. \(\textbf{Introduce unequal half-aperture amplitudes:}\) Let the magnitude of either original half-aperture field resultant in that direction be \(H\). The attenuated upper-half contribution has magnitude \(H/2\), while the lower-half contribution remains \(H\). Their opposite phases give \[ E_{\mathrm{old\ minimum}}=H-\frac H2 \] \[ E_{\mathrm{old\ minimum}}=\frac H2 \] \(\textbf{Relate the half-aperture contribution to the original central field:}\) At the original first minimum, the geometrical path difference across one half-aperture is \(\lambda/2\). For a uniformly illuminated half-slit, coherent integration gives its resultant field magnitude \[ H=\frac{E_0}{\pi} \] Therefore, \[ E_{\mathrm{old\ minimum}}=\frac{E_0}{2\pi} \] Its intensity relative to the original central peak is \[ \frac{I_{\mathrm{old\ minimum}}}{I_{\max,0}}=\frac1{4\pi^2} \] \(\textbf{Physical interpretation:}\) The attenuation makes the two halves contribute unequal field amplitudes. Their forward contributions still reinforce, but the central peak decreases. At the former first minimum, the contributions retain opposite phases but no longer have equal magnitudes. Thus, the previous exact cancellation is destroyed. The residual intensity must be evaluated using the actual coherent contribution from each half in that direction. It cannot be obtained by treating the half-slit fields there as though each had its original forward-axis amplitude. \(\textbf{Final answer:}\) \[ \boxed{I_{\mathrm{forward}}=\frac9{16}I_{\max,0},\quad I_{\mathrm{old\ minimum}}=\frac1{4\pi^2}I_{\max,0}} \]

Explanation: \(\textbf{Recover the diffraction angle from the original arrangement:}\) For the first minimum of a single slit, \[ a\sin\theta_1=\lambda \] Under the small-angle approximation, \[ \theta_1\approx\frac{\lambda}{a} \] The slit width and wavelength are \[ a=0.25\times10^{-3}\,\text{m} \] \[ \lambda=500\times10^{-9}\,\text{m} \] Therefore, \[ \theta_1=\frac{500\times10^{-9}}{0.25\times10^{-3}} \] \[ \theta_1=2.0\,\text{mrad} \] \(\textbf{Check the original minimum coordinate:}\) At initial screen distance \[ D_1=2.0\,\text{m} \] the positive first minimum occurs at \[ x_{1,\mathrm{old}}=D_1\theta_1 \] \[ x_{1,\mathrm{old}}=(2.0)(2.0\times10^{-3}) \] \[ x_{1,\mathrm{old}}=+4.0\,\text{mm} \] This agrees with the detector position. \(\textbf{Move the screen while keeping normal incidence:}\) The new distance is \[ D_2=3.0\,\text{m} \] Without changing incidence direction, the positive first minimum would move to \[ x_{1,\mathrm{normal}}=D_2\theta_1 \] \[ x_{1,\mathrm{normal}}=(3.0)(2.0\times10^{-3}) \] \[ x_{1,\mathrm{normal}}=+6.0\,\text{mm} \] The detector remains at \(+4.0\,\text{mm}\). \(\textbf{Use incident-wavefront steering to restore the minimum:}\) Let the new incident angle be \(\alpha\), positive towards increasing \(x\). For small angles, the first positive-side diffraction minimum is at \[ x_1=D_2(\alpha+\theta_1) \] The required detector condition is \[ 4.0\times10^{-3}=3.0(\alpha+2.0\times10^{-3}) \] Therefore, \[ \alpha=\frac{4.0\times10^{-3}}{3.0}-2.0\times10^{-3} \] \[ \alpha=-\frac23\times10^{-3}\,\text{rad} \] \[ \alpha=-\frac23\,\text{mrad} \] The illumination must tilt towards negative \(x\). \(\textbf{Determine the new central maximum:}\) The central diffraction maximum follows the incident propagation direction. Thus, \[ x_c=D_2\alpha \] \[ x_c=3.0\left(-\frac23\times10^{-3}\right) \] \[ x_c=-2.0\,\text{mm} \] \(\textbf{Verify the minimum geometry:}\) The positive first minimum is \(D_2\theta_1=6.0\,\text{mm}\) to the right of the shifted central maximum. Therefore, \[ x_+=-2.0+6.0 \] \[ x_+=+4.0\,\text{mm} \] The detector is again at the first positive-side minimum. The negative first minimum is at \(-8.0\,\text{mm}\), so the full central width is \(12.0\,\text{mm}\). \(\textbf{Physical interpretation:}\) Moving the screen changes the linear diffraction scale but not the intrinsic angular spread. Tilting the incident plane wave translates the angular diffraction distribution relative to the original slit normal. The specified detector condition can therefore be restored by steering the incident wave in the opposite direction. The central maximum no longer coincides with the geometrical screen centre. The problem combines the changed screen scale with an inverse wavefront-steering condition rather than simply substituting into the original diffraction formula. \(\textbf{Final answer:}\) \[ \boxed{\alpha=-\frac23\,\text{mrad},\quad x_c=-2.0\,\text{mm}} \]

Explanation: Statement I is correct because the electric field of an electromagnetic light wave is transverse to its direction of propagation. The beam travels along the \(x\)-axis, while its transmitted electric field vibrates along the vertical \(z\)-direction. A linear polarizer selects the field component parallel to its transmission axis. Statement II is correct because an ideal analyzer transmits the component of an incident linearly polarized electric field parallel to its own transmission axis. When the analyzer axis is aligned with the electric field, transmission is maximum. When it is perpendicular, the projected electric-field amplitude is zero and the ideal transmitted intensity vanishes. Statement III is correct because the ability to restrict the vibration direction is characteristic of transverse waves. The existence of polarization is therefore evidence for the transverse nature of light. Statement IV is incorrect because the polarizer's transmission axis refers to a permitted electric-field vibration direction. It is not the direction along which the beam propagates. Passing through the polarizer need not redirect the beam from the horizontal \(x\)-axis to the vertical \(z\)-axis. Confusing propagation direction with vibration direction would give an incorrect geometrical description of linearly polarized light. The analyzer's rotation changes the transmitted field component while the incident beam's propagation direction remains fixed. Statements I, II and III are correct.

Explanation: \(\textbf{Identify the polarization state of the incident beam:}\) The incident light is unpolarized. Its electric-field vibrations have no single fixed transverse direction. An ideal first Polaroid transmits half its incident intensity under the standard unpolarized-light model. \(\textbf{Calculate transmission through the first Polaroid:}\) The incident intensity is \[ I_{\mathrm{in}}=16\,\text{W m}^{-2} \] Therefore, \[ I_1=\frac{I_{\mathrm{in}}}{2} \] \[ I_1=8\,\text{W m}^{-2} \] The transmitted beam is now linearly polarized along the vertical axis. \(\textbf{Apply Malus' law to the second Polaroid:}\) For already linearly polarized light, \[ I_2=I_1\cos^2\theta \] Initially, the relative angle between transmission axes is \[ \theta=60^\circ \] Therefore, \[ I_2=8\cos^260^\circ \] \[ I_2=8\left(\frac12\right)^2 \] \[ I_2=2\,\text{W m}^{-2} \] \(\textbf{Rotate the second transmission axis into alignment:}\) After rotation, the axes are parallel: \[ \theta'=0^\circ \] Thus, \[ I'_2=I_1\cos^20^\circ \] \[ I'_2=8(1)^2 \] \[ I'_2=8\,\text{W m}^{-2} \] \(\textbf{Check the role of the first Polaroid:}\) Rotating only the second Polaroid cannot restore the intensity lost when unpolarized light passed through the first ideal Polaroid. The maximum available to the second Polaroid is the intensity emerging from the first. Therefore, the largest final intensity is \(8\,\text{W m}^{-2}\), not the original \(16\,\text{W m}^{-2}\). \(\textbf{Physical interpretation:}\) The first Polaroid converts unpolarized illumination into a linearly polarized beam while reducing its intensity. The second Polaroid acts as an analyzer of that already polarized beam. Its transmission depends on the angle between its axis and the polarization direction set by the first Polaroid. The one-half rule applies to unpolarized input, whereas the cosine-squared rule applies to linearly polarized input. Identifying the light's polarization state at each stage is essential for obtaining the correct sequence of intensities. \(\textbf{Final answer:}\) \[ \boxed{I_1=8\,\text{W m}^{-2},\quad I_2=2\,\text{W m}^{-2},\quad I'_2=8\,\text{W m}^{-2}} \]

Explanation: \(\textbf{Determine the intensity after the first Polaroid:}\) The incident beam is unpolarized. Therefore, \[ I_1=\frac{I_0}{2} \] The beam emerging from the first Polaroid is linearly polarized along its vertical transmission axis. \(\textbf{Transmit through the intermediate Polaroid:}\) The second axis makes angle \(\theta\) with the first. Malus' law gives \[ I_2=I_1\cos^2\theta \] Thus, \[ I_2=\frac{I_0}{2}\cos^2\theta \] The polarization direction after the second Polaroid is along its transmission axis. \(\textbf{Determine the angle to the final Polaroid:}\) The first and third transmission axes are perpendicular. The angle between the intermediate axis and the third axis is therefore \[ 90^\circ-\theta \] Applying Malus' law again, \[ I_3=I_2\cos^2(90^\circ-\theta) \] \[ I_3=\frac{I_0}{2}\cos^2\theta\sin^2\theta \] \(\textbf{Simplify the angular dependence:}\) Using \[ \sin2\theta=2\sin\theta\cos\theta \] we obtain \[ \cos^2\theta\sin^2\theta=\frac14\sin^22\theta \] Therefore, \[ I_3=\frac{I_0}{8}\sin^22\theta \] \(\textbf{Find the maximum:}\) The maximum value of \(\sin^22\theta\) is unity. Within the specified angular interval, this occurs when \[ 2\theta=90^\circ \] \[ \theta=45^\circ \] Therefore, \[ I_{3,\max}=\frac{I_0}{8} \] \(\textbf{Evaluate the }30^\circ\textbf{ position:}\) For \[ \theta=30^\circ \] \[ I_3=\frac{I_0}{2}\cos^230^\circ\sin^230^\circ \] \[ I_3=\frac{I_0}{2}\left(\frac34\right)\left(\frac14\right) \] \[ I_3=\frac{3I_0}{32} \] \(\textbf{Interpret why an intermediate Polaroid permits transmission:}\) The first and last axes are crossed, so without the intermediate Polaroid the final transmitted intensity is zero. The intermediate Polaroid selects a new polarization direction that has nonzero projections along both neighbouring axes. It does not create energy; each stage transmits only a component of the incident field. The maximum occurs when the intermediate axis divides the right angle equally. \(\textbf{Final answer:}\) \[ \boxed{\theta_{\max}=45^\circ,\quad I_{3,\max}=\frac{I_0}{8},\quad I_3(30^\circ)=\frac{3I_0}{32}} \]

Explanation: \(\textbf{Apply Brewster's law in the original direction:}\) For light incident from P into Q, \[ \tan i_B=\frac{n_Q}{n_P} \] The indices are \[ n_P=1.6,\qquad n_Q=1.2 \] Therefore, \[ \tan i_B=\frac{1.2}{1.6} \] \[ \tan i_B=\frac34 \] This agrees with the stipulated sine and cosine: \[ \tan i_B=\frac{3/5}{4/5} \] \[ \tan i_B=\frac34 \] \(\textbf{Determine the refraction angle:}\) Snell's law gives \[ n_P\sin i_B=n_Q\sin r \] Therefore, \[ \sin r=\frac{n_P}{n_Q}\sin i_B \] Substituting, \[ \sin r=\frac{1.6}{1.2}\times\frac35 \] \[ \sin r=\frac45 \] Since both angles are acute, \[ \cos r=\frac35 \] Consequently, \[ i_B+r=90^\circ \] \(\textbf{Interpret the reflected and refracted directions:}\) The reflected ray makes angle \(i_B\) with the interface normal. The refracted ray makes angle \(r\) with the normal in the second medium. Their propagation directions form an angle \[ 180^\circ-(i_B+r) \] \[ =90^\circ \] Thus, the reflected and refracted rays are perpendicular. \(\textbf{Reverse the incident medium:}\) For incidence from Q into P, Brewster's law becomes \[ \tan i'_B=\frac{n_P}{n_Q} \] \[ \tan i'_B=\frac{1.6}{1.2} \] \[ \tan i'_B=\frac43 \] The new Brewster angle is complementary to the original one. \(\textbf{Physical interpretation:}\) Brewster incidence depends on the ordered pair of refractive indices: incident medium first and transmitted medium second. Reversing the propagation direction therefore reverses the refractive-index ratio in Brewster's law. At Brewster incidence, Snell's law simultaneously makes the reflected and refracted rays perpendicular. This geometric condition is consistent with the polarization property of the reflected beam. The possibility of Brewster incidence does not disappear merely because light initially travels from a rarer medium to a denser one. \(\textbf{Final answer:}\) \[ \boxed{\sin r=\frac45,\quad \angle(\text{reflected,refracted})=90^\circ,\quad\tan i'_B=\frac43} \]