The mixture a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions $\ll$ vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance $d \ll h$, the height of the column.

The mixture a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions ll vertical dim

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A ray bends continuously in a medium with a vertical refractive index gradient. By treating the gradient as constant over the small horizontal distance $d$ , the deviation angle is $\theta \approx \frac{d}{n_0} \left( \frac{dn}{dy} \right)$ , where $n_0$ is the refractive index at the entry point and $\frac{dn}{dy}$ is the vertical gradient.

Why This Works: The Physics of Bending Light

When a ray of light enters a medium where the refractive index varies smoothly in a direction perpendicular to its path, the ray does not suddenly change direction at a single point — it curves continuously. This is the same principle that causes mirages and the twinkling of stars.

The key insight is Snell's law in differential form: for a ray moving through a medium with a gradient in refractive index, the local radius of curvature is determined by how fast $n$ changes across the ray's path. For small deviations, we can treat the bending as a gradual accumulation of tiny angular deflections.

Tip

Think of it like a car driving on a road that tilts sideways — the car drifts in the direction of the tilt. Here, the "tilt" is the refractive index gradient, and the "drift" is the bending of the light ray.

Step-by-Step Solution

1. Set up the geometry and the gradient

Let the vertical direction be $y$ (positive upward) and the horizontal direction be $x$ (the direction the ray initially travels). The refractive index varies only with $y$ : $n = n(y)$ . At the entry point, let $y = y_0$ and $n(y_0) = n_0$ .

The ray enters horizontally, so initially the wavefront is vertical. As the ray moves a small horizontal distance $dx$ , it encounters slightly different refractive indices at different heights.

2. Apply the differential form of Snell's law

For a ray in a medium with a gradient, the local curvature is given by:

\frac{1}{R} = \frac{1}{n} \frac{dn}{dy} \cos \theta

where $\theta$ is the angle the ray makes with the horizontal. For small deviations, $\theta$ is small, so $\cos \theta \approx 1$ . The radius of curvature $R$ relates to the change in direction $d\theta$ over a path length $ds$ by $d\theta = ds / R$ .

Since the ray is nearly horizontal, $ds \approx dx$ . Therefore:

d\theta = \frac{dx}{R} = \frac{1}{n} \frac{dn}{dy} \, dx

3. Integrate over the horizontal distance

The ray travels a horizontal distance $d$ . Over this short distance, $n$ and $\frac{dn}{dy}$ change very little from their values at the entry point. So we can treat them as constants:

\theta = \int_0^d \frac{1}{n_0} \left( \frac{dn}{dy} \right)_{y_0} dx = \frac{d}{n_0} \left( \frac{dn}{dy} \right)_{y_0}

This $\theta$ is the total deviation angle — the angle between the final direction of the ray and its original horizontal direction. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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CBSE 2026Set A1 markMCQ
$Q.The refractive index of water is 1.33. What will be the speed of light in water? (A) 1.33 × 10^8 m/s (B) 4 × 10^8 m/s (C) 2.25 × 10^8 m/s (D) 3 × 10^8 m/s$
›Reveal solutionSolution

Speed in a medium = c/n = (3 × 10⁸)/1.33 ≈ 2.25 × 10⁸ m/s.

The refractive index of a medium is the ratio of the speed of light in vacuum to that in the medium:

$n=\frac{c}{v}\quad\Rightarrow\quad v=\frac{c}{n}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The refractive indices of glass and water with respect to air are 3/2 and 4/3, respectively. The refractive index of glass w.r.t. water will be$
(a) 8/9
(b) 9/8
(c) 7/6
(d) 6/7
›Reveal solutionSolution

$n_{g/w} = n_{g/a}/n_{w/a} = (3/2)/(4/3) = 9/8$ .

Refractive index of glass w.r.t. water can be obtained by combining the refractive indices w.r.t. air:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.Refraction takes place due to$
(a) change in the speed of light
(b) no change in the speed of light
(c) change in the colour of light
(d) polarization
›Reveal solutionSolution

Light bends when it crosses into a medium where its speed is different; that speed change is the fundamental cause of refraction.

When light travels from one medium into another (e.g. air into glass), its speed changes because the two media have different optical densities (different refractive indices). According to Snell's law, n1sin(theta1) = n2sin(theta2), and refractive index n = c/v (speed of light in vacuum divided by speed in the medium). Because the speed v changes at the boundary, the direction of the light ray bends - this bending is refraction. If the speed did no …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. Which of the following four rays incident on turpentine in figure, the path shown is correct?$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution
Since turpentine is optically the densest of the three (n(turpentine) > n(water) > n(air)), a ray must bend toward the normal at the air–turpentine surface, then bend away from the normal (but not all the way back) at the turpentine–water surface.

The question states the optical density (refractive index) order is turpentine > water > air (even though turpentine's mass density is lower than water's, which is why it floats — optical density and mass density are unrelated). Refraction at each interface follows Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ :

Air → Turpentine: going into an optically denser medium, the ray bends towards the normal (angle decreases).

Turpentine → Water: water is rarer than turpentine, so going into it the ray bends away from the normal (angle increases) compared to its path inside turpentine. …
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CBSE 2026Set ANNUAL1 mark
$Q.In phenomenon of refraction of light, which property of it remains unchanged ?$
›Reveal solutionSolution

In refraction, frequency stays constant; speed and wavelength change.

When light travels from one medium into another, its speed changes (v = c/n) and consequently its wavelength changes (λ = v/f). However, the frequency f is determined by the source of light and does not change when the wave crosses the boundary — the number of wavefronts arriving per second must equal the number leaving per second ( …

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CBSE 2025Set ANNUAL1 mark
$Q.Write Snell's law of refraction.$
›Reveal solutionSolution

Snell's law relates the angle of incidence and angle of refraction at a boundary between two media through a constant called the refractive index.

Snell's law of refraction states that when light passes from one medium into another, the ratio of the sine of the angle of incidence $i$ to the sine of the angle of refraction $r$ is a constant for the given pair of media, called the refractive index of the second medium with respect to the first:

$n_{21} = \dfrac{\sin i}{\sin r} = \dfrac{n_2}{n_1}$

Equivalently, in the symmetric form: $n_1 \sin i = n_2 \sin r$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q.The time taken by the light ray to pass through a glass slab of thickness 5 mm and refractive index μ = 1.5 will be$
(a) 0.25×10⁻¹¹ s
(b) 0.2×10⁻¹⁰ s
(c) 2.5×10⁻¹¹ s
(d) 15×10⁻¹¹ s
›Reveal solutionSolution

Speed of light inside the slab is c/μ; time = thickness ÷ (c/μ).

Speed of light in the glass slab: $v = c/\mu$ . Time taken to cross thickness $d = 5\,mm = 5\times10^{-3}\,m$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$Q.For light incident from air on a slab of refractive index 2, the maximum possible angle of refraction is :$
(a) 60°
(b) 30°
(c) 90°
(d) 45°
›Reveal solutionSolution

At grazing incidence ( $i=90°$ ), Snell's law with $n=2$ gives the maximum possible angle of refraction, $r=30°$ .

Working

For light entering a medium of refractive index $n=2$ from air, Snell's law gives

$\sin i = n\sin r \ \Rightarrow\ \sin r = \dfrac{\sin i}{n}$

…

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CBSE 2025Set ANNUAL1 mark
$Q.The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. What will be the refractive index of glass with respect to water ?$
›Reveal solutionSolution

Using $^{a}n_g$ and $^{a}n_w$ , the refractive index of glass relative to water is their ratio: $^{w}n_g = {}^{a}n_g/{}^{a}n_w = 9/8$ .

Given: refractive index of glass w.r.t. air, $^{a}n_g = 3/2$ ; refractive index of water w.r.t. air, $^{a}n_w = 4/3$ .

Refractive index of medium 2 with respect to medium 1 can be written as

$^{1}n_2 = \dfrac{{}^{a}n_2}{{}^{a}n_1}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q.The velocity of light in a glass block of refractive index 1.5 is -$
(a) $1.5 \times 10^8\ ms^{-1}$
(b) $2 \times 10^8\ ms^{-1}$
(c) $3 \times 10^8\ ms^{-1}$
(d) $2 \times 10^9\ ms^{-1}$
›Reveal solutionSolution

$v=c/n$ .

The speed of light in a medium of refractive index $n$ is $v = \dfrac{c}{n}$ :

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. When a ray of light enters a glass slab, then$
(a) its frequency and velocity change.
(b) only frequency changes
(c) its frequency and wavelength changes
(d) its frequency does not change.
›Reveal solutionSolution

When light passes from one medium to another, only its speed and wavelength change; the frequency stays fixed because it is determined by the source that produced the light.

When a light ray crosses from air into a glass slab, the medium's refractive index $n$ slows the wave down: $v = c/n$ . Since $v=f\lambda$ and $f$ cannot change (frequency corresponds to the number of oscillations per second of the source, which is unaffected by what medium the wave later travels through — this is a fundamental result of the wave's boundary conditions at an interface), the wavelength must adjust: $\lambda = v/f = (c/n)/f$ , i.e. wavelength decreases in the denser medium (glass) since $v$ decreases.

…

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CBSE 2023Set B1 mark
$Q. Fill in the blank: The velocity of light ______ when it goes from rare medium to denser medium.$
›Reveal solutionSolution

Speed of light is inversely related to the refractive index of the medium; a denser medium has a higher refractive index and hence a lower light speed.

The refractive index of a medium is defined as $n = c/v$ , where $c$ is the speed of light in vacuum and $v$ is its speed in the medium. An optically denser medium has a higher refractive index, which by this relation means light travels slower in it. So as light passes from a rarer medium (e.g., air) into a denser medium (e.g., glass or water), its spee …

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Concept understanding — Refraction and Snell's Law

Refraction and Snell's Law

Why This Works: The Physics of Bending Light

Step-by-Step Solution

Method: Deviation of a Ray Through a Medium With a Refractive-Index Gradient

Steps

Why This Works: The Physics of Bending Light

Step-by-Step Solution

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