MHT-CET Physics · Optics (Ray)

Refraction, Apparent Depth and Total Internal Reflection

Light entering a denser medium slows to c/μ and bends toward the normal, sin i = μ sin r; an object under a medium appears raised to its real depth divided by μ; and light going from denser to rarer is totally reflected once its angle of incidence exceeds the critical angle sin⁻¹(1/μ).

All Optics (Ray) notes MHT-CET Physics strategy

Why this matters

26 PYQs, 7 of them HARD — the largest page in the chapter. Eight apply Snell's law — the index from a speed, the deviation on entering a slab, the angle when reflected and refracted rays are perpendicular. Nine are apparent depth through one or several layers, a bubble in a cube or a slab on an ink mark; nine are total internal reflection. Three cards.

Concept 1 of 3: Snell's Law and the Refractive Index

The refractive index μ = c/v says how much light slows; a 20% slower speed means μ = 1/0.8 = 1.25. Snell's law, sin i = μ sin r, bends the ray toward the normal; for small angles i ≈ μr, so the deviation i − r is i(1 − 1/μ). Between two media the relative index is the ratio of speeds. The optical path μd is the distance light would cover in vacuum in the same time, so equal times through two materials mean equal μd. When i = 2r, μ = sin 2r / sin r = 2 cos r.

Definition

$\mu = \dfrac{c}{v}$ ; $\sin i = \mu\sin r$ ; relative index $_1\mu_2 = \dfrac{v_1}{v_2}$ .
Small angles: deviation $i - r = i\left(1 - \dfrac{1}{\mu}\right)$ (speed down 25% ⇒ i/4).
Deviation δ at one surface: $\dfrac{1}{\mu} = \cos\delta - \dfrac{\sin\delta}{\tan i}$ .
Optical path $\mu d$ : 3 cm of 1.6 = 3.84 cm of 1.25. Same time in slab and air ⇒ $d_{\text{air}} = \mu t$ .
$i = 2r$ ⇒ $i = 2\cos^{-1}\dfrac{\mu}{2}$ .

Snell's law

\sin i = \mu\sin r, \qquad \mu = \frac{c}{v}

Worked example

Light enters water (μ = 4/3) from air at 53°. Angle of refraction? (sin 53° = 0.8)

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q28Moderate

Example 1 · Optics (Ray) · Refraction, Apparent Depth, and Total Internal Reflection

Velocity of light in diamond is

\frac{5}{12}

th times that in air. Velocity of light in water is

\frac{3}{4}

th times that in air. The angle of incidence of ray of light travelling from water to diamond is (angle of refraction

r = 30^\circ

)

[\sin 30^\circ = \frac{1}{2}]

Taking μ as the fraction of speed kept

Speed reduced by 25% means v = 0.75c and μ = 1/0.75 = 4/3 — not 0.75.

Drill 7 more on snell's law and the refractive index

Concept 2 of 3: Apparent Depth and Normal Shift

Looking straight down into a medium, an object at real depth d appears at d/μ, raised by d(1 − 1/μ). Layers add their apparent depths: d₁/μ₁ + d₂/μ₂. A slab of thickness t on an ink mark raises it by x = t(1 − 1/μ), so t = μx/(μ − 1). A bubble inside a cube, viewed from both faces, gives two equations d/μ and (L − d)/μ, and their sum fixes μ. A mirror at the bottom of a liquid puts the image of an object h above it at h below the mirror; the observer sees both through the liquid, so their apparent separation is 2h/μ.

Definition

Apparent depth $\dfrac{d}{\mu}$ ; normal shift $d\left(1 - \dfrac{1}{\mu}\right)$ (4.8 cm of 1.5 ⇒ 1.6 cm).
Layers: $\sum \dfrac{d_i}{\mu_i}$ (3, 4, 6 cm of 3/2, 4/3, 6/5 ⇒ 10 cm).
Slab on a mark: $t = \dfrac{\mu x}{\mu - 1}$ .
Bubble in a cube of side L: $\dfrac{d}{\mu} + \dfrac{L - d}{\mu} = \text{sum of the two readings}$ (24 cm, 10 + 6 ⇒ μ = 1.5, d = 15 cm).

Apparent depth

d_{\text{app}} = \frac{d}{\mu}, \qquad \text{shift} = d\left(1 - \frac{1}{\mu}\right)

Worked example

A container 30 cm tall should look half full of water (μ = 4/3) from above. Water depth?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q46Hard

Example 2 · Optics (Ray) · Refraction, Apparent Depth, and Total Internal Reflection

A transparent glass cube of length 24 cm has a small air bubble trapped inside. When seen normally through one surface from air outside, its apparent distance is 10 cm from the surface. When seen normally from opposite surface, its apparent distance is 6 cm. The distance of the air bubble from first surface is

Multiplying by μ instead of dividing

Objects in a denser medium look NEARER. Apparent depth is the real depth divided by μ.

Drill 8 more on apparent depth and normal shift

Concept 3 of 3: Total Internal Reflection

Going from denser to rarer, the ray bends away from the normal, and at the critical angle sin C = 1/μ (or μ₂/μ₁ between two media) it grazes the surface; beyond that, all the light reflects. Two conditions, then: denser to rarer, and i > C. Blue light has a higher index than red, so a smaller critical angle: when green is just totally reflected, violet, indigo and blue are too. Glass to air has the smallest critical angle among common pairs. A source at depth h is hidden by a floating disc of radius h tan C = h/√(μ² − 1). Mirages, a diamond's sparkle and optical fibres rely on it; the apparent depth of a pond does not.

Definition

Conditions: denser → rarer and i > C; $\sin C = \dfrac{1}{\mu} = \dfrac{\mu_2}{\mu_1} = \dfrac{v_1}{v_2}$ .
Speeds 1.5 and 2 × 10⁸ m/s ⇒ $C = \sin^{-1}\dfrac{3}{4}$ .
Blue has the smallest C; with green just reflected, VIB are reflected too.
Hiding disc: $r = h\tan C = \dfrac{h}{\sqrt{\mu^2 - 1}}$ .
Reflected ⟂ refracted (denser to rarer): $C = \sin^{-1}(\tan r)$ .

Critical angle

\sin C = \frac{1}{\mu}, \qquad r_{\text{disc}} = \frac{h}{\sqrt{\mu^2 - 1}}

Worked example

A lamp is 4 cm below water (μ = 4/3). Radius of the smallest disc on the surface that hides it?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q42Hard

Example 3 · Optics (Ray) · Refraction, Apparent Depth, and Total Internal Reflection

An LED is placed at a depth

h

below the water surface. An opaque disc is floating on the surface of water such that the bulb is not visible from the surface. The minimum radius of the disc will be

Allowing total internal reflection from rarer to denser

Entering a denser medium the ray bends toward the normal and always gets through. Total internal reflection needs denser to rarer.

Giving red the smaller critical angle

Red has the smallest refractive index, so the LARGEST critical angle. Blue and violet are totally reflected first.

Drill 8 more on total internal reflection

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Snell's Law and the Refractive Index
Snell's law
$\sin i = \mu\sin r, \qquad \mu = \frac{c}{v}$
Apparent Depth and Normal Shift
Apparent depth
$d_{\text{app}} = \frac{d}{\mu}, \qquad \text{shift} = d\left(1 - \frac{1}{\mu}\right)$
Total Internal Reflection
Critical angle
$\sin C = \frac{1}{\mu}, \qquad r_{\text{disc}} = \frac{h}{\sqrt{\mu^2 - 1}}$

Watch out for (4)

Taking μ as the fraction of speed kept
→ Snell's Law and the Refractive Index
Multiplying by μ instead of dividing
→ Apparent Depth and Normal Shift
Allowing total internal reflection from rarer to denser
→ Total Internal Reflection
Giving red the smaller critical angle
→ Total Internal Reflection

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