PHYS7021
Burdwan · Electricity, Magnetism & Wave Optics · 2018–2024

GE-4 Borderline — Visibility, Resolving Power, Optical Activity

Borderline GE-4 questions outside the PHYS7021 text. Solve only after confirming with your lecturer. Fringe visibility, resolving power, optical activity.

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Confirm with your lecturer — outside PHYS7021 text. This page holds GE-4 borderline questions (B8–B10). They appeared in old GE-4 papers (2019–2024) but are NOT in the PHYS7021 text. Ask your lecturer whether to study them. If short on time, finish the core Interference, Diffraction and Polarization pages first.

What is inside: B8 fringe visibility and numeric (×2: 2019, 2023); B9 resolving power of grating (2022) and of an optical instrument (2024); B10 optical activity (2023). Each solution below is full-length so you can answer even without the textbook chapter.

B8. Fringe visibility

Formula B8. Michelson fringe visibility: $V = \dfrac{I_{max}-I_{min}}{I_{max}+I_{min}} = \dfrac{2\sqrt{I_1 I_2}}{I_1+I_2}$

Here $V$ = fringe visibility ($1$ = perfect dark minima, $0$ = no fringes). $I_{max} = I_1+I_2+2\sqrt{I_1I_2}$ (in-phase intensity). $I_{min} = I_1+I_2-2\sqrt{I_1I_2}$ (out-of-phase intensity). $I_{1,2} \propto (\text{amplitude})^2$.

Meaning in words. Visibility measures contrast. Equal beams give $V = 1$ (minima fully dark). One beam much stronger gives $V \to 0$ (washout). Daily example: TV contrast setting.

B9. Resolving power

Formula B9a. Grating resolving power: $R = \dfrac{\lambda}{\Delta\lambda} = Nn$
Formula B9b. Optical instrument (circular aperture $D$): $\theta_{\min} = \dfrac{1.22\lambda}{D},\qquad R = \dfrac{1}{\theta_{\min}} = \dfrac{D}{1.22\lambda}$

Here $N$ = total number of illuminated grating lines; $n$ = order; $\Delta\lambda$ = smallest wavelength gap the instrument can separate; $\theta_{min}$ = smallest angular separation of two just-resolvable stars (radians).

B10. Optical activity

Optical activity is the rotation of the plane of polarization of light by a substance (solution, crystal or vapour). Dextro (d-, right) rotates clockwise as seen toward the source (e.g. cane sugar, d-glucose). Laevo (l-, left) rotates anticlockwise (e.g. fructose). Examples: quartz crystal (along its optic axis), sugar solution, tartaric acid, turpentine.

Solved PYQs

Q-B8

What is meant by visibility of interference fringes? Two waves of amplitudes $2$ cm and $1$ cm interfere. Find the visibility.

20192023×2

Recall

Eq. B8: $V = (I_{max}-I_{min})/(I_{max}+I_{min})$. Intensity is proportional to amplitude squared.

Steps

  1. Definition: $V = (I_{max}-I_{min})/(I_{max}+I_{min})$ — ratio of fringe depth to average brightness (Michelson visibility).
  2. Intensities: $I_1 \propto 2^2 = 4$ units, $I_2 \propto 1^2 = 1$ unit.
  3. Max and min: $I_{max} = (\sqrt{4}+\sqrt{1})^2 = (2+1)^2 = 9$; $I_{min} = (2-1)^2 = 1$.
  4. Visibility: $V = (9-1)/(9+1) = 8/10 = 0.8$. Equivalently $2\sqrt{4\cdot1}/(4+1) = 4/5 = 0.8$.
  5. Sense: $0.8$ is high contrast (minima nearly dark) — sensible since amplitudes are comparable; $0 \le V \le 1$.

Answer

$\boxed{V = 0.8}$

Exam tip

Do NOT use $(I_1-I_2)/(I_1+I_2) = 3/5$ — that forgets the cross term. Always build $I_{max}$ and $I_{min}$ from $(\sqrt{I_1} \pm \sqrt{I_2})^2$ first. The correct Michelson value is $0.8$, not $0.6$.
Q-B9a

State the resolving power of a plane diffraction grating. Show that $R = Nn$.

2022×2

Recall

Resolving power $R = \lambda/\Delta\lambda$ = smallest wavelength gap the grating can separate; $N$ = total illuminated lines; $n$ = order.

Steps

  1. Definition: $R = \lambda/\Delta\lambda$, where $\Delta\lambda$ is the smallest wavelength gap still seen as two lines (Rayleigh criterion: maximum of one falls on minimum of the other).
  2. Result: for a grating, $R = Nn$. (Proof uses the width of a principal maximum, $\propto 1/N$, vs the angular dispersion, $\propto n$.)
  3. Meaning: double the lines OR work in 2nd order → double the resolution — maxima sharpen with $N$ and spread more with $n$.
  4. Example: $N = 15000$ lines in 1st order gives $R = 15000$, so at $\lambda = 600$ nm it separates $\Delta\lambda = 600/15000 = 0.04$ nm.

Answer

$\boxed{R = \dfrac{\lambda}{\Delta\lambda} = Nn.}$

Exam tip

Write the definition line first. "Resolving" (separate close lines) and "dispersive" (spread per nm) are different — do not swap them.
Q-B9b

What do you mean by resolving power of an optical instrument?

2024×2

Recall

An instrument (telescope, microscope) makes an Airy disc, not a point; two points resolve when discs just separate (Rayleigh criterion).

Steps

  1. Rayleigh criterion: two point images just resolve when the central maximum of one falls on the first minimum of the other.
  2. Telescope (circular aperture $D$): smallest resolvable angle $\theta_{min} = 1.22\lambda/D$ (first dark ring of the Airy pattern).
  3. Resolving power is the inverse: $R = 1/\theta_{min} = D/1.22\lambda$ — larger $D$ means finer detail.
  4. Microscope form (for reference): $R \propto 2\mu\sin\theta/\lambda$ — oil and a wide cone help.

Answer

$\boxed{\theta_{\min} = \dfrac{1.22\lambda}{D},\qquad R = \dfrac{D}{1.22\lambda}.}$

Exam tip

Confirm with lecturer whether they want $\theta_{min}$ or its inverse — write both and you cover either marking scheme.
Q-B10

What do you mean by optical activity?

2023

Recall

Plane-polarised light = electric vector vibrating in one plane; some substances rotate that plane as light passes through.

Steps

  1. Definition: rotation of the plane of polarisation by a substance (solution, crystal or vapour) is optical activity — seen with crossed polariser and analyser: the dark position shifts by angle $\theta$.
  2. Dextro (d-, right): rotates clockwise as seen toward the source (e.g. cane sugar, d-glucose). Laevo (l-, left): rotates anticlockwise (e.g. fructose). Names from Latin dexter / laevus.
  3. Examples: quartz crystal (along its optic axis), sugar solution, tartaric acid, turpentine.
  4. Note: activity needs asymmetric molecules or crystals; heating or wrong solvent can reduce it.

Answer

Optical activity = rotation of the plane of polarisation. $\boxed{\text{Dextro (right, e.g. cane sugar); Laevo (left, e.g. fructose); e.g. quartz, sugar solution.}}$

Exam tip

One diagram (polariser → tube of sugar solution → analyser rotated by $\theta$) plus two examples = full marks.

Reference. Brij Lal, Subrahmanyam & Avadhanulu — Optics (fringe visibility; grating resolving power; optical activity — outside the PHYS7021 chapters; confirm with your lecturer). Fallback: Ajoy Ghatak — Optics.

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