Your Answer : (Not Answered)

Correct Answer : D

The fringe width is \(\lambda\).

Two plane waves interfere with intensity \(\propto 1 + \cos(\Delta\vec{k} \cdot \vec{r})\), where \(\Delta\vec{k} = \vec{k}_1 - \vec{k}_2\). The \(\hat{\jmath}\) parts are equal and cancel:

\[ \begin{aligned} \Delta\vec{k} &= \frac{2\pi}{\lambda}\left(\tfrac{1}{2} + \tfrac{1}{2}\right)\hat{\imath} \\ &= \frac{2\pi}{\lambda}\hat{\imath} \end{aligned} \]

On the screen (the \(xz\)-plane) the phase difference is \(\frac{2\pi}{\lambda}x\), so bright fringes repeat every time \(x\) increases by \(\lambda\). Equivalently, each wave makes 30° with the screen normal, the angle between them is 60°, and the fringe spacing \(\frac{\lambda}{2\sin 30^\circ} = \lambda\).

The \(\hat{\jmath}\) components play no part, since they are identical in both waves and only set the common direction of travel; \(\frac{\lambda}{2}\), \(2\lambda\) and \(\frac{\sqrt{3}\lambda}{2}\) do not match the spacing \(2\pi/|\Delta k_x|\) fixed by the \(\hat{\imath}\) components.

Your Answer : (Not Answered)

Correct Answer : C

The correct option is that the ray would not enter the 3rd layer.

A ray at 20° to the horizontal makes 70° with the normal. For parallel layers, Snell's law keeps \(n \sin\theta\) the same in every layer:

\[ \begin{aligned} n\sin\theta &= 1.45 \sin 70^\circ = 1.363 \\ \text{layer 2:}\ \sin\theta_2 &= 1.363/1.95 = 0.699 \\ \text{layer 3:}\ \sin\theta_3 &= 1.363/1.33 = 1.025 \end{aligned} \]

The ray enters layer 2 at about 44° from the normal, but a sine greater than 1 is impossible, so at the boundary with layer 3 it is totally internally reflected.

The trap is the 20° option: layer 4 has the same index as layer 1, so if the ray did get through it would leave at 20° again, but it never reaches layer 4. 44° is the angle from the normal inside layer 2, not an angle in layer 4.

Your Answer : (Not Answered)

Correct Answer : C

The answer is \(V_{I} \gt V_{O}\); the image moves away from the lens.

From the lens equation \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\), differentiating with respect to time gives

\[ \begin{aligned} \frac{dv}{dt} &= \frac{v^{2}}{u^{2}}\frac{du}{dt} \\ &= m^{2}\frac{du}{dt} \end{aligned} \]

For an object between \(2f\) and \(f\) the image is real, beyond \(2f\) on the other side, and magnified (\(|m| \gt 1\)). So the image speed is \(m^{2}\) times the object speed, greater than \(V_{O}\). As the object approaches \(f\) the image runs off towards infinity, so it moves away from the lens.

Equal speeds occur only at \(u = 2f\), where \(|m| = 1\), and a slower image would need \(|m| \lt 1\), which happens only for objects beyond \(2f\).

Your Answer : (Not Answered)

Correct Answer : A

The wave is elliptically polarised with its axes along \(\hat{\imath}\) and \(\hat{\jmath}\) when \(\phi = 0\) and \(A_1 \neq A_2\).

Write \(\theta = kz - \omega t\). With \(\phi = 0\) the components are \(E_x = A_1 \cos\theta\) and \(E_y = A_2 \sin\theta\), so

\[ \frac{E_x^{2}}{A_1^{2}} + \frac{E_y^{2}}{A_2^{2}} = 1 \]

The tip of the field traces an ellipse whose semi-axes \(A_1\) and \(A_2\) lie along \(\hat{\imath}\) and \(\hat{\jmath}\). With \(A_1 \neq A_2\) it is a true ellipse; with \(A_1 = A_2\) it becomes a circle, so the \(\phi = 0\), \(A_1 = A_2\) option describes circular polarisation.

With \(\phi = \pi/2\), \(\sin(\theta + \pi/2) = \cos\theta\), so both components oscillate in phase and the wave is linearly polarised, whether or not the amplitudes are equal.

Your Answer : (Not Answered)

Correct Answer : B

The difference in refractive indices is \(3.16 \times 10^{-7}\).

In an optically active medium the two circular components travel with indices \(n_L\) and \(n_R\). Over a thickness \(d\) they build up a phase difference \(\frac{2\pi}{\lambda}(n_L - n_R)d\), and the plane of polarisation turns through half of it:

\[ \begin{aligned} \theta &= \frac{\pi d}{\lambda}\,|n_L - n_R| \\ |n_L - n_R| &= \frac{\theta \lambda}{\pi d} \end{aligned} \]

With \(\theta = 18^\circ = \pi/10\) rad, \(\lambda = 632 \times 10^{-9}\) m and \(d = 0.20\) m:

\[ \begin{aligned} |n_L - n_R| &= \frac{(\pi/10)(632 \times 10^{-9})}{\pi \times 0.20} \\ &= 3.16 \times 10^{-7} \end{aligned} \]

Putting the angle in as 18 degrees instead of radians gives \(1.81 \times 10^{-5}\), and the other two options are exactly double these two values, so they do not satisfy the rotation formula either.

Your Answer : (Not Answered)

Correct Answer : B

The plate is \(\lambda/2\), with \(\alpha = 30^\circ\).

A half-wave plate reflects the plane of polarisation about its axis: light polarised at angle \(\alpha\) to the axis comes out at \(-\alpha\), a rotation of \(2\alpha\). A rotation of \(60^\circ\) therefore needs \(\alpha = 30^\circ\).

A quarter-wave plate does not rotate a plane of polarisation; in general it produces elliptically polarised light, so both \(\lambda/4\) options fail. A pass-axis at \(120^\circ\) is the same line as one at \(-60^\circ\); it turns the plane by \(-120^\circ\), which lands on the same line only because a plane of polarisation is defined to within \(180^\circ\). The key takes the direct reading, the smallest axis angle giving a \(60^\circ\) rotation, \(30^\circ\).

Your Answer : (Not Answered)

Correct Answer : C

The arrangement is light, \(\lambda/4\) plate, analyser.

A quarter-wave plate turns circularly polarised light into plane polarised light, because it adds a further \(\pi/2\) between the two components that were already \(\pi/2\) apart. Rotating the analyser then gives complete extinction at one position. Unpolarised light remains unpolarised after the plate, so its intensity through the analyser stays constant. The two can therefore be told apart.

A half-wave plate turns circular light into circular light of the opposite hand, so the analyser shows no variation for either beam. Putting a polariser first destroys the difference, since both beams become plane polarised before reaching the plate.

Your Answer : (Not Answered)

Correct Answer : C

The field is elliptically polarised with a ratio of major to minor axis of 2.

Write the components: \(E_{x} = E_{0}\cos\omega t\) and \(E_{y} = 2E_{0}\cos(\omega t + \pi/2) = -2E_{0}\sin\omega t\). Then

\[ \begin{aligned} &\frac{E_{x}^{2}}{E_{0}^{2}} \\ &+ \frac{E_{y}^{2}}{4E_{0}^{2}} = 1 \end{aligned} \]

The tip of \(\vec{E}\) traces an ellipse with semi-axes \(2E_{0}\) along y and \(E_{0}\) along x, so the axis ratio is 2. Linear polarisation needs the components in phase or opposite in phase. Circular polarisation needs a quarter-period shift and equal amplitudes, which these are not. A wave with a fixed phase difference between its components is polarised, not unpolarised.

Your Answer : (Not Answered)

Correct Answer : C

The emergent wave is linearly polarized with polarization as that of incident wave.

The incident field \((\hat{x} + \hat{y})E_{0}\) is linearly polarized at 45° to the \(x\)-axis. The optic axis is at 135°, which is exactly perpendicular to that direction.

A wave plate only changes the polarization when the field has components both parallel and perpendicular to the optic axis, because the plate delays one relative to the other. Here the field has no component along the optic axis: it lies entirely along the perpendicular direction, so the plate merely delays it as a whole and the polarization is unchanged.

Circular polarization would need equal components along and across the optic axis (field at 45° to it), and other angles give elliptical polarization. A rotation by 90° is what a half-wave plate does to light at 45° to its axis, not a quarter-wave plate with the field along an eigen-axis.