Exam 3 — All Questions, All Variations

Q1: Benzene on Water — What Is Happening?

The exam question: "A researcher measures the thickness of benzene (n=1.50) floating on water (n=1.33) by shining monochromatic light (575 nm) onto the film. She finds 575 nm is reflected most strongly. What is the minimum thickness? Draw a diagram."

Before any calculation, understand the physical setup. Light hits a thin benzene layer floating on top of water. The film has two surfaces: a TOP surface (air meets benzene) and a BOTTOM surface (benzene meets water). At EACH surface, the incoming light partially reflects and partially transmits.

Two reflected rays travel back upward: Ray 1 reflects off the TOP surface. Ray 2 enters the film, reflects off the BOTTOM surface, then exits back through the top. These two rays travel different path lengths, so they arrive with a phase difference. That phase difference determines whether reflection is strong (constructive) or weak (destructive).

The exam says the reflection is most strong — this is constructive interference of the reflected waves.

Q1: Which Reflections Cause Phase Shifts?

The phase-shift rule: when light reflects off a surface going from a medium of lower n into a medium of higher n, the reflected ray undergoes a 180° phase shift (equivalent to a half-wavelength shift). When light reflects going from higher n to lower n, there is NO phase shift.

Apply to Benzene-on-Water

• Reflection 1 (at top surface): light travels from air (n=1.00) into benzene (n=1.50). This is LOW → HIGH. The reflected Ray 1 gets a 180° phase shift = ½λ.
• Reflection 2 (at bottom surface): light travels from benzene (n=1.50) into water (n=1.33). This is HIGH → LOW (1.50 > 1.33). The reflected Ray 2 gets NO phase shift.

Net result: Ray 1 and Ray 2 have a built-in phase difference of ½λ just from the reflections alone, before we even count the extra path. They start out of phase. For them to arrive back IN phase (constructive interference), the extra path 2t must compensate by adding another ½λ (bringing the total to a full λ, λ = in phase).

Q1: Minimum Thickness of Benzene Film

Now put it all together. With ONE net phase shift (the rays start ½λ out of phase), the condition for constructive interference (maximum reflection) is:

For minimum thickness, use the smallest value of m, which is m = 0:

Numerical Verification

Check: optical path difference = 2 × t_min × n_film = 2 × 95.8 nm × 1.50 = 287.5 nm = ½ × 575 nm ✓. The extra path equals exactly ½λ, which together with the ½λ already from the phase shift gives a full λ — constructive.

Q1: Every Thin Film Variation You Might See

What if BOTH boundaries are LOW→HIGH (e.g., oil on glass, n_oil < n_glass)?

Both reflections produce a 180° shift. Two shifts cancel each other out — the net phase difference from reflections alone is ZERO (the two rays start in phase). Now the path difference 2t must be a full multiple of λ/n for the rays to constructively interfere:

What if λ = 500 nm instead of 575 nm (same benzene/water setup)?

What if n_film = n_water (same index as water, e.g., film "disappears")?

If the film has the same refractive index as the substrate below (n_film = n_water = 1.33), there is no optical interface at the bottom. Ray 2 never reflects back. With only one ray, there is no interference at all — the film is invisible in reflection. This is the principle behind anti-reflection coatings (though those use a different mechanism).

What if t = 2 × t_min (second order, m = 1)?

So if the film were twice as thick, it would strongly reflect UV light (383 nm) instead of yellow-green (575 nm). The film appears a different color for each order m.

Minimum thickness for DESTRUCTIVE reflection (for same benzene film)?

Q2: Concave Mirror — Image Calculation

The exam question: "A concave mirror has R = 10 cm. Object: 16 mm tall, distance 10 cm. Find image location and size. Describe (real/virtual, upright/inverted). Draw 3-ray diagram."

Given → Focal Length

Key observation: xₒ = 10 cm = 2f. The object sits exactly at the center of curvature C.

Mirror Equation

Magnification and Image Height

Image Description

• Location: 10 cm in front of mirror (same as object)
• Real or Virtual: xᵢ > 0 → REAL (image forms in front of mirror, on same side as object)
• Upright or Inverted: m = −1 < 0 → INVERTED
• Size: |m| = 1 → same size as object, hᵢ = 16 mm

Q2: The Three Mirror Rays — Memorize These Rules

For any concave mirror problem, the three standard rays are always drawn from the TIP of the object. The image is where all three rays (or their extensions) converge.

Ray 1 — The Parallel Ray

Draw the ray from the object tip, traveling parallel to the principal axis. After reflecting off the mirror, it passes through the focal point F. For a concave mirror, this reflected ray aims through F toward the axis.

Ray 2 — The Focal Ray

Draw the ray from the object tip through (or toward) the focal point F. After reflecting off the mirror, it travels parallel to the principal axis. This is the reverse of Ray 1.

Ray 3 — The Center Ray

Draw the ray from the object tip through the center of curvature C (which is at 2f). This ray hits the mirror perpendicular to its surface and reflects straight back through C. Because the object is at C in this problem, Ray 3 simply goes straight out and straight back.

The key fact to memorize: when an object is placed at C (= 2f), the image also forms at C, real, inverted, and same size (m = −1). This is a classic exam checkpoint.

Q2: What-If Scenarios for Any Mirror Problem

The concave mirror has f = 5 cm. The behavior depends entirely on where the object is placed relative to F and C. Here are all five cases you must know:

Case 1: xₒ = 5 cm (object AT focal point F)

This is how flashlights and headlights work: place a bulb at F and the mirror projects a parallel beam.

Case 2: xₒ = 7 cm (between F and C)

Real, inverted, magnified (|m| = 2.5). The image is farther from the mirror than the object.

Case 3: xₒ = 3 cm (object INSIDE F — makeup mirror / magnifying mirror)

Virtual (xᵢ < 0 → image behind mirror), upright (m > 0), magnified. This is how a makeup mirror gives a right-side-up magnified image.

Case 4: xₒ = 20 cm (far beyond C)

Real, inverted, reduced. The farther the object, the more compressed the image toward F.

Q3: Reading the Lens Diagram — Focal Length and Type

The exam question: "The picture below shows an object and its image formed by a thin lens. What is the focal length and type? What is the image height? Draw a 3-ray diagram."

From the exam diagram: the object is 8 cm to the LEFT of the lens, object height = 6.5 mm. The image appears on the SAME SIDE as the object (also to the left of the lens), 3 cm from the lens. This is the critical observation: an image forming on the same side as the object for a lens means xᵢ is NEGATIVE (virtual image).

Lens Equation

Since f < 0, this is a DIVERGING (concave) lens.

Magnification and Image Height

m > 0 → upright. |m| < 1 → reduced (smaller than object). Image is virtual, upright, and reduced — this is always the case for a diverging lens with a real object.

Q3: Drawing the Three Rays for a Diverging Lens

For a diverging lens, the three rays behave differently than for a converging lens. The key is that a diverging lens bends rays AWAY from the axis, so the refracted rays never actually cross — you must extend them backward (dashed lines) to find the virtual image.

Ray 1 — Parallel Ray (Diverging Version)

Incident ray travels parallel to the principal axis from the object tip. After passing through the diverging lens, it bends AWAY from the axis — but it appears to come FROM the near focal point F₁ on the same side as the object. Draw the refracted ray diverging, then extend backward (dashed) to show it appears to originate at F₁.

Ray 2 — Aimed-at-Far-Focal-Point Ray

Incident ray from object tip is aimed at the far focal point F₂ (on the opposite side of the lens from the object). After the lens, this ray refracts to travel parallel to the principal axis. The refracted ray goes parallel; its backward extension passes through the image point.

Ray 3 — Center Ray

Incident ray through the exact center of the lens passes straight through without bending. This is true for both converging and diverging lenses.

The three backward extensions (dashed) all meet at the virtual image location: 3 cm to the left of the lens, upright, reduced. This confirms xᵢ = −3 cm, m = +0.375.

Q3: Lens Variations and Converging vs Diverging

What if the same diagram showed a CONVERGING lens (f = +4.8 cm)?

With xₒ = 8 cm and f = +4.8 cm (object outside focal point):

What if xₒ = 15 cm with same diverging lens (f = −4.8 cm)?

Diverging lenses ALWAYS give virtual, upright, reduced images for any positive (real) object distance. The image is always between F₁ and the lens.

What if xₒ = 3 cm (inside |f| = 4.8 cm) for the diverging lens?

Still virtual, still upright, still reduced. Diverging lenses are completely predictable: no matter where the object is, the image is virtual, upright, and closer to the lens than the object.

Key Diverging Lens Rule to Memorize

Q4: Single Slit — Two Triangles the Exam Requires

The exam question: "Parallel light λ=600 nm falls on a single slit. Screen is 3 m away. Distance between first dark fringes on both sides = 4.5 mm. Find slit width. Draw BOTH triangles."

Extract y₁ from the Given Measurement

The "distance between first dark fringes on both sides" means the total span from −y₁ to +y₁. So: 2y₁ = 4.5 mm → y₁ = 2.25 mm = 2.25 × 10⁻³ m.

The Two Triangles (exactly as the professor requires)

Triangle 1 — at the slit (sine equation): Inside the slit, divide the slit into two halves, each of width a/2. The condition for the FIRST dark fringe is that each wavelet from the top half of the slit cancels a wavelet from the bottom half. The path difference between the top edge and the middle of the slit must equal λ/2. Geometrically, this gives:

Triangle 2 — at the screen (tangent equation): The same angle θ appears in the triangle formed by the screen distance L and the fringe position y₁:

For small angles (θ small in radians), tan θ ≈ sin θ, so we can combine:

Q4: Single Slit Width Calculation

Full Step-by-Step Solution

Unit Verification

What-If Variations

What if λ doubles (λ = 1200 nm, everything else same)?

What if L = 6 m (screen twice as far)?

What if total dark fringe separation = 9 mm?

Where are the SECOND dark fringes?

Q5: Multiple Choice — Constructive Interference at Point P

The exam question: "Two sources A and B emit wavelength λ, in phase. Constructive at P if (may be more than one): a) x=y, b) x+y=2λ, c) x-y=λ, d) x-y=5λ."

Here x = distance from source A to point P, and y = distance from source B to point P.

The Core Principle

Analyze Each Answer Choice

(a) x = y → path difference = x − y = 0 = 0 × λ. YES — constructive (m=0). ✓
Both sources are equidistant from P. Waves arrive in phase. This is the zero-order maximum.

(b) x + y = 2λ → this is a SUM, not a difference. NO. ✗
This is the classic trap. The condition involves the PATH DIFFERENCE |x−y|, never the sum x+y. Knowing x+y=2λ tells you nothing about whether interference is constructive or destructive.

(c) x − y = λ → path difference = λ = 1 × λ. YES — constructive (m=1). ✓
The wave from A travels exactly one wavelength farther than the wave from B. It arrives at P one full cycle behind — which means in phase again.

(d) x − y = 5λ → path difference = 5λ = 5 × λ. YES — constructive (m=5). ✓
Five full wavelengths of path difference — still in phase.

Q6: Interference Conditions — Exact Answers

The exam asks you to complete these sentences with precise physics language. This is not a calculation — it's a definitions question that must be answered exactly right.

Part (a): Destructive Interference

When one wave travels half a wavelength (λ/2) farther than the other, its crest arrives at the same time as the other's trough. They cancel completely → destructive.

Part (b): Constructive Interference

When the path difference is a whole number of wavelengths, both waves arrive in phase (crest with crest, trough with trough) → constructive. The m=0 case is just the two sources equidistant from the point.

Q7: Is TIR Possible Going from Air to Water?

The exam question: "Is it possible to have total internal reflection for light incident from air to water? Explain!"

Answer: NO — it is not possible.

Full Explanation (Words + Equations)

Total internal reflection (TIR) is a phenomenon that requires light to be traveling from a medium of HIGHER refractive index into a medium of LOWER refractive index. It occurs when the angle of incidence in the denser medium exceeds the critical angle θ_c.

For air-to-water: n_air = 1.00, n_water = 1.33. We have n_incident = n_air = 1.00 and n_transmitted = n_water = 1.33. Applying the formula:

But sinθ cannot exceed 1.00 (the sine function has a maximum value of 1). Since 1.33 > 1, there is no valid critical angle. The critical angle simply does not exist for this direction of travel. TIR cannot occur.

Physically: light going from air into water is going from a less optically dense medium into a more optically dense medium. At the interface, Snell's law always has a solution (the refracted angle θ₂ is always smaller than θ₁, since n₂ > n₁). Light always partially refracts into the water regardless of the angle of incidence.

Q7: TIR — All Scenarios and Critical Angles

TIR is possible whenever n_incident > n_transmitted. The critical angle θ_c satisfies sin θ_c = n₂/n₁. For angles greater than θ_c, 100% of the light is reflected with no refraction.

Common TIR Examples

Glass → Air (n_glass=1.52, n_air=1.00):

Any ray inside the glass hitting the glass-air surface at more than 41.1° undergoes TIR. This is how optical fibers work.

Glass → Water (n_glass=1.52, n_water=1.33):

TIR is still possible but requires a larger angle because water is more optically dense than air.

Diamond → Air (n_diamond=2.42, n_air=1.00):

Diamond has a very small critical angle — almost any internal ray undergoes TIR. This is why diamonds sparkle: they are cut so that most light entering from the top bounces internally many times before escaping out the top again.

The Prism Extra Problem (from exam PDF)

Glass prism n=1.52. Find maximum angle δ for a ray entering the slanted face without undergoing TIR.

When the prism is submerged in water, the critical angle is larger, so the maximum allowed angle without TIR is smaller.

Q8: What Happens to the Diffraction Pattern in Water?

The exam question: "A monochromatic laser falls on a slit and produces bright and dark spots. If the apparatus is submerged in water, the dark spots: a) won't change, b) move away from center, c) move toward center. Explain!!!!"

Answer: (c) The dark spots MOVE TOWARD THE CENTER.

Full Explanation

The position of the first dark fringe is given by:

Here λ is the wavelength in the medium where the light is propagating. When the apparatus is submerged in water, the wavelength changes:

The wavelength in water is SHORTER than in air (by a factor of 1.33). Plugging this into the fringe position formula:

The fringe position decreases by a factor of 1.33. Every dark fringe is now closer to the center — the entire diffraction pattern COMPRESSES inward. The central maximum stays at the center but becomes narrower.

The frequency of the light does NOT change when entering water (only wavelength and speed change). The slit width a is the same. The screen distance L is the same. Only λ changes.

Q9: Frequency and Wavelength of Light in Glass

The exam question: "Light of wavelength 256 nm travels from vacuum to glass (n=1.6). What is the frequency and wavelength of the light in glass?"

Step 1: Find the Frequency (It Does NOT Change)

The frequency of light is an intrinsic property of the electromagnetic wave. When light crosses from one medium to another, the frequency stays the same because the wavefronts must match at the boundary — one wavefront enters the glass every time one exits the vacuum. If frequency changed, wavefronts would pile up or disappear at the interface, which is physically impossible.

Step 2: Find the Wavelength in Glass

The speed of light in glass is v = c/n. The wavelength in glass can be found two ways:

What-If Variations

If n = 2.0 (denser glass): λ_glass = 256/2.0 = 128 nm. f unchanged.

If λ₀ = 500 nm (visible green), n = 1.5: λ_glass = 500/1.5 = 333 nm (UV range in glass). f = 3×10⁸/500×10⁻⁹ = 6×10¹⁴ Hz.

Color of light: Color is determined by frequency (or equivalently, the vacuum wavelength), NOT by the wavelength in a medium. If you put the 256 nm light in glass where λ_glass = 160 nm, the color is still determined by the 256 nm vacuum wavelength.

Q10 & Q11: Wavelength and Speed from a Refraction Diagram

Both questions Q10 and Q11 ask you to read a diagram showing light bending at an interface and deduce a relationship. The key is to identify whether the light bends TOWARD or AWAY from the normal.

The Diagnostic Rule

• If θ₂ < θ₁ (bends TOWARD normal): sin θ₂ < sin θ₁ → n₂ > n₁ (medium 2 is denser)
• If θ₂ > θ₁ (bends AWAY from normal): sin θ₂ > sin θ₁ → n₂ < n₁ (medium 2 is less dense)

For the Exam Diagram (light bends TOWARD normal, θ₂ < θ₁, so n₂ > n₁)

Q10 — Wavelength: λ = λ₀/n. Larger n → smaller λ. Since n₂ > n₁:

Q11 — Speed: v = c/n. Larger n → smaller v. Since n₂ > n₁:

Reverse Case (light bends AWAY, θ₂ > θ₁, so n₂ < n₁)

Putting It All Together — How to Answer Every Question

The professor's rule is your template. Every single answer must have all three components. If you skip diagrams on a show-your-work problem, you lose points even if the number is right.

Template for Calculation Problems (Q1, Q2, Q3, Q4, Q9)

1. Draw the diagram. Label all given quantities on the diagram itself. For mirrors/lenses: draw the ray diagram with all 3 rays. For thin films: draw the two paths with phase shift labels. For single slit: draw both triangles.
2. Write the equation. Write the formula symbolically BEFORE substituting numbers.
3. List knowns and unknowns. State what you're solving for.
4. Solve algebraically. Rearrange for the unknown first, then substitute.
5. Check units. Dimensional analysis saves partial credit.
6. Describe the result in words. "The image is real, inverted, and located 10 cm in front of the mirror."

Template for Conceptual/Explanation Problems (Q5, Q6, Q7, Q8, Q10, Q11)

1. State the principle. "TIR requires n_incident > n_transmitted."
2. Apply to the case. "Here n_air = 1.00 < n_water = 1.33, so the condition is not met."
3. Show the math. "sin θ_c = 1.33/1.00 = 1.33 > 1, impossible."
4. Draw a diagram. Even a quick sketch showing refracted rays (not TIR) earns points.
5. State the conclusion. "Therefore TIR is NOT possible from air to water."

Complete Equation Sheet for Exam 3

Light in Media

Snell's Law

Applied at every interface. θ₁ is the angle in medium 1, θ₂ in medium 2, both measured from the normal.

Total Internal Reflection

For θ > θ_c: 100% reflection, no transmitted ray. Requires going from denser to less dense medium.

Mirror and Lens Equation

Sign Conventions

• xₒ: always positive for real objects (object on incoming side)
• xᵢ > 0: real image (opposite side from object for lens; same side as object for mirror)
• xᵢ < 0: virtual image
• m < 0: inverted image   m > 0: upright image
• f < 0: diverging lens or convex mirror

Single Slit Diffraction

Two-Source / Double-Slit Interference

Thin Film Interference

Where t = film thickness, n = film refractive index, λ₀ = vacuum wavelength.

Effect of Submerging in Medium