JEE Mains Physics · Ray Optics

Refraction at Plane Surfaces and Apparent Depth

At a flat boundary n₁ sin i = n₂ sin r; a parallel slab shifts a ray sideways without turning it, and an object under a liquid of index μ looks only d/μ deep.

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Why this matters

Twenty-one PYQs, eleven of them multiple choice and ten asking for a number, and one from 2026. Twelve apply Snell's law at one flat surface or through a parallel slab: an angle given from the surface instead of the normal, a ray along a vector, the sideways shift through a slab, the time spent in a slab, a shadow under water. Nine are about apparent depth: layered liquids, a bubble seen from two sides, a microscope refocused, a bird seen by a fish.

Concept 1 of 2: Snell's law and the parallel slab

Light slows down in an optically denser medium, n = c/v, and bends towards the normal. Snell's law links the angles measured from the normal on the two sides. A parallel-sided slab bends the ray in and bends it back out by the same amount, so the ray leaves parallel to its original direction, shifted sideways.

Definition

$n = c/v$ . Snell's law: $n_1\sin i = n_2\sin r$ , with both angles measured from the normal. Deviation at one surface: $i - r$ .
An angle given 'with the surface' is $90^{\circ}$ minus the angle of incidence.
A ray along a vector $\vec A$ , boundary normal $\hat n$ : $\cos i = \dfrac{|\vec A\cdot\hat n|}{|\vec A|}$ . Equivalently, $\sin i$ is the size of the part of $\vec A$ along the surface over $|\vec A|$ .
Parallel slab of thickness t: the emergent ray is parallel to the incident one, with lateral shift $d = \dfrac{t\sin(i - r)}{\cos r}$ .
Path inside the slab $t/\cos r$ ; time $\dfrac{n t}{c\cos r}$ , which is $nt/c$ at normal incidence.
Refractive index measures optical density, how much light slows down. It is not proportional to mass density.

Snell's law and lateral shift

n_1\sin i = n_2\sin r, \qquad n = \frac{c}{v}, \qquad d = \frac{t\sin(i - r)}{\cos r}

Worked example

A ray in air meets a glass slab of refractive index

\sqrt{3}

and thickness

3\sqrt{3}

cm at an angle of incidence of

60^{\circ}

. Find the angle of refraction and the lateral shift of the emergent ray.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q29Moderate

Example 1 · Ray Optics · Refraction at Plane Surfaces and Apparent Depth

The X-Y plane be taken as the boundary between two transparent media

M_{1}

and

M_{2}.M_{1}

Z \geq 0

has a refractive index of

\sqrt{2}

and

M_{2}

with

Z < 0

has a refractive index of

\sqrt{3}

. A ray of light travelling in

M_{1}

along the direction given by the vector

\overrightarrow{A}= 4\sqrt{3}\widehat{i}- 3\sqrt{3}\widehat{j}- 5\widehat{k}

, is incident on the plane of separation. The value of difference between the angle of incident in

M_{1}

and the angle of refraction in

M_{2}

will be_____ degree.

Angles are measured from the normal

A ray 'at 30° with the surface' has an angle of incidence of 60°. Putting 30° into Snell's law gives a wrong index, and that wrong value is usually one of the options.

A slab shifts a ray but does not turn it

The emergent ray is parallel to the incident one; only a sideways shift remains. The bending i − r at the first face is undone at the second.

Refractive index is not mass density

Refractive index measures how much light slows down, n = c/v. A medium with a larger mass density does not, for that reason, have a proportionally larger refractive index.

Drill 11 more on snell's law and the parallel slab

Concept 2 of 2: Apparent depth and the normal shift

Seen from straight above, an object under a liquid looks raised. Rays leaving the surface bend away from the normal, and the eye traces them back to a point higher up. For near-normal viewing the depth shrinks by the factor μ. Each layer of a stack shrinks by its own μ, and the apparent depths add.

Definition

Viewed from a rarer medium, near the normal: apparent depth $= \dfrac{d}{\mu}$ ; the object appears raised by $d\left(1 - \dfrac{1}{\mu}\right)$ .
Layers: $d_{\text{app}} = \sum \dfrac{d_i}{\mu_i}$ ; total shift $= \sum d_i\left(1 - \dfrac{1}{\mu_i}\right)$ .
Looking from the denser medium at an object in the rarer one (a fish looking up at a bird): the object appears $\mu$ times farther, $d_{\text{app}} = \mu d$ .
Moving objects: apparent distances scale the same way, so apparent speeds do too.
A bubble in a block seen from opposite faces: $x/\mu$ from one and $(t - x)/\mu$ from the other, so the two apparent distances add to $t/\mu$ .
A microscope focused on the bottom of a vessel must be raised by the shift $d(1 - 1/\mu)$ when liquid is poured in.
A glass plate of thickness t placed in a converging beam moves the image away from the lens by $t\left(1 - \dfrac{1}{\mu}\right)$ .

Apparent depth

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

Worked example

A glass block 6 cm thick (

\mu = 1.5

) lies on the bottom of a tank, under 8 cm of water (

\mu = 4/3

). A mark on the underside of the block is viewed from straight above. Find its apparent depth below the water surface and how much it appears raised.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · 2021 Paper 24 · Q12Moderate

Example 2 · Ray Optics · Refraction at Plane Surfaces and Apparent Depth

A glass tumbler having inner depth of

17.5\text{ }cm

is kept on a table. A student starts pouring water

(\mu = 4/3)

into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

The shift is not the apparent depth

A question may give the shift d(1 − 1/μ) or the apparent depth d/μ. With μ = 4/3 they are d/4 and 3d/4. Read which one is stated before solving.

Never average the indices of a stack

For layered liquids, divide each layer's own thickness by its own μ and add the results. A single averaged μ gives the wrong apparent depth.

Looking up multiplies, looking down divides

An object in water seen from air appears at d/μ. An object in air seen from under water appears at μd, farther away than it is.

Drill 8 more on apparent depth and the normal shift

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 (2)

Snell's law and the parallel slab
Snell's law and lateral shift
$n_1\sin i = n_2\sin r, \qquad n = \frac{c}{v}, \qquad d = \frac{t\sin(i - r)}{\cos r}$
Apparent depth and the normal shift
Apparent depth
$d_{\text{app}} = \frac{d}{\mu}, \qquad \text{shift} = d\left(1 - \frac{1}{\mu}\right), \qquad d_{\text{app}} = \sum \frac{d_i}{\mu_i}$

Watch out for (6)

Angles are measured from the normal
→ Snell's law and the parallel slab
A slab shifts a ray but does not turn it
→ Snell's law and the parallel slab
Refractive index is not mass density
→ Snell's law and the parallel slab
The shift is not the apparent depth
→ Apparent depth and the normal shift
Never average the indices of a stack
→ Apparent depth and the normal shift
Looking up multiplies, looking down divides
→ Apparent depth and the normal shift

Test yourself on Ray Optics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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