A ray of light incident at an angle $\theta$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^{\circ}$ and the prism is made of a material of refractive index 1.5, the angle of incidence is (a) 7.5°. (b) 5°. (c) 15°. (d) 2.5°.

A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If

NCERT Exemplar · Q1

\theta

(a) 7.5°.

(b) 5°.

(d) 2.5°.

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Concept understanding — Refraction and Snell's Law

Refraction and Snell's Law

When light crosses the boundary between two transparent media its speed changes, so the ray bends at the surface. The bending is governed by Snell's law, which relates the angle of incidence $i$ (measured from the normal) in medium 1 to the angle of refraction $r$ in medium 2:

n_1 \sin i = n_2 \sin r

Here $n = c/v$ is the (absolute) refractive index of a medium, always $\geq 1$ for ordinary matter.

Which way does it bend?

Going into a denser medium ( $n_2 > n_1$ ): light slows down, $\sin r < \sin i$ , so the ray bends toward the normal.
Going into a rarer medium ( $n_2 < n_1$ ): the ray bends away from the normal.
At normal incidence ( $i=0$ ) the ray passes straight through, and a ray never bends past the normal for ordinary media.

Across a stack of layers (e.g. air $\to$ turpentine $\to$ water), apply Snell's law at each interface in turn. Since $n_{air} < n_{water} < n_{turpentine}$ , a ray descending from air bends toward the normal on entering the turpentine and then away from the normal on entering the (less dense) water.

Worked idea: light in air ( $n_1=1$ ) hits glass ( $n_2=1.5$ ) at $i=30^\circ$ . Then $\sin r = \frac{1}{1.5}\sin 30^\circ = 0.333$ , so $r \approx 19.5^\circ$ — smaller than $i$ , confirming the bend toward the normal.

Refraction and Snell's law form a core part of the NCERT Class 12 Physics chapter on Ray Optics and Optical Instruments, and "Snell's law formula and derivation" or "refraction of light class 12 important questions" are searched heavily by CBSE board and NEET/JEE Main physics aspirants. The multi-layer refraction idea shown above — applying Snell's law at each interface in turn — is exactly the style of numerical that appears in JEE Main optics questions built on this NCERT Class 12 topic.

This is a plain prism-refraction problem (no total internal reflection anywhere) - normal emergence at the second face forces the internal ray to hit the second face at $0^\circ$ , so Snell's law at the first face gives the angle of incidence $\theta \approx 7.5^\circ$ , matching option (a).

Setting up the prism geometry

A ray enters the first face of the prism at angle of incidence $\theta$ , refracts to angle $r_1$ inside, travels to the second face, hits it at internal angle $r_2$ , and emerges there at angle $e$ . For any prism, the two internal angles are related to the prism (apex) angle $A$ by

r_1 + r_2 = A.

Here $A = 5^\circ$ and the refractive index is $n = 1.5$ .

Using the "emerges normally" condition

"Emerges normally" means the ray leaves the second face along the normal, so the emergence angle $e = 0^\circ$ . By Snell's law at the second face,

n\sin r_2 = 1\cdot\sin e = 0 \quad\Rightarrow\quad \sin r_2 = 0 \quad\Rightarrow\quad r_2 = 0^\circ.

(This makes sense physically too: a ray travelling exactly along the normal inside the glass hits the exit face perpendicularly and passes straight through with no bending.)

Finding $r_1$

From $r_1 + r_2 = A$ :

r_1 = A - r_2 = 5^\circ - 0^\circ = 5^\circ.

Applying Snell's law at the first face

1\cdot\sin\theta = n\sin r_1 = 1.5\sin(5^\circ).

Using $\sin 5^\circ \approx 0.0872$ :

\sin\theta \approx 1.5\times0.0872 = 0.1308 \quad\Rightarrow\quad \theta \approx \sin^{-1}(0.1308) \approx 7.5^\circ.

Watch out

A common mistake is to reach for minimum-deviation results ( $r_1=A/2$ ) - that applies only to the symmetric passage through a prism, not to this case, where the exit condition ( $r_2=0$ ) is given directly. Always work from $r_1+r_2=A$ using whatever the problem actually specifies.

Tip

For a small apex angle $A$ (in radians), the approximation $\sin\theta\approx nA$ gives a fast sanity check: $\theta \approx 1.5\times5^\circ = 7.5^\circ$ - matching the exact calculation.

✓Final answer

\theta \approx 7.5^\circ

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

Showing the 12 most recent of 29 on this concept.

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:

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

Substituting $c=3\times10^{8}\,\text{m/s}$ and $n=1.33$ :

$v=\frac{3\times10^{8}}{1.33}\approx 2.25\times10^{8}\,\text{m/s}$

✓Final answer
$(C) 2.25 \times 10⁸ m/s.$
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:

$n_{glass/water} = \dfrac{n_{glass/air}}{n_{water/air}} = \dfrac{3/2}{4/3} = \dfrac{3}{2}\times\dfrac{3}{4} = \dfrac{9}{8}$

✓Final answer
$(b) 9/8.$
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 not change, the ray would pass straight through undeviated. Refraction does not, by itself, change the colour/frequency of light (that is unchanged across the interface); only the wavelength and speed change.

✓Final answer
$(a) change in the speed of light.$
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.

A useful check: because the two interfaces are parallel (horizontal), applying Snell's law straight through — $n_{air}\sin\theta_{air} = n_{turp}\sin\theta_{turp} = n_{water}\sin\theta_{water}$ — shows the final angle in water depends only on $n_{air}$ and $n_{water}$ (turpentine's index only affects the sideways kink of the path inside the middle layer, not the final direction). Since $n_{water} > n_{air}$ , the ray in water is still bent closer to the normal than the original incident ray in air, even though it bent away from the normal at the second surface relative to its steeper bend inside turpentine.

So the correct path is a ray that: (i) bends sharply toward the normal on entering turpentine, (ii) bends back away from the normal (but only partially) on entering water, and (iii) is overall still closer to the normal than the incident ray was in air. Only one of the four drawn rays can show this exact 'bend-in-then-partly-out' pattern; based on the figure that is ray 2.

✓Final answer
$Ray 2 (option b) shows the physically correct path: it bends toward the normal entering turpentine and partially back away from the normal entering water, remaining net closer to the normal than in air.$
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 (otherwise waves would pile up at the boundary).

Hence the property that remains unchanged in refraction is the frequency of light.

✓Final answer
$Frequency remains unchanged.$
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: $n_1 \sin i = n_2 \sin r$ .

(The incident ray, refracted ray, and the normal at the point of incidence all lie in the same plane — this is the accompanying first law of refraction.)

✓Final answer
$\dfrac{\sin i}{\sin r} = n_{21} =$
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 $d = 5\,mm = 5\times10^{-3}\,m$ :

$t=\dfrac{d}{v}=\dfrac{d\mu}{c}=\dfrac{5\times10^{-3}\times1.5}{3\times10^{8}}=\dfrac{7.5\times10^{-3}}{3\times10^{8}}=2.5\times10^{-11}\ s$

✓Final answer
$t = 2.5\times10⁻¹¹ s — option (c).$
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

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

$\sin r$ is greatest when $\sin i$ is greatest, i.e., at grazing incidence $i=90°$ ( $\sin i=1$ ):

$\sin r_{max} = \dfrac{1}{2} \ \Rightarrow\ r_{max} = 30°$

✓Final answer
$30°$
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

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

So refractive index of glass with respect to water:

$^{w}n_g = \dfrac{{}^{a}n_g}{{}^{a}n_w} = \dfrac{3/2}{4/3} = \dfrac{3}{2}\times\dfrac{3}{4} = \dfrac{9}{8} = 1.125$

✓Final answer
$^{w}n_g = 9/8 = 1.125$
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 $v = \dfrac{c}{n}$ :

$v = \frac{3\times10^8}{1.5} = 2\times10^8\ \text{m/s}$

✓Final answer
$2\times10^8\ \text{ms}^{-1}$
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.

Checking against the options: (a) is wrong because frequency does not change; (b) is wrong because frequency alone changing is false — it doesn't change at all; (c) is wrong for the same reason (frequency does change is false); only (d), 'its frequency does not change', correctly describes what happens (velocity and wavelength both decrease, frequency stays constant).

✓Final answer
$Its frequency does not change (option d).$
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 speed decreases — this reduction in speed (along with the wave bending toward the normal) is precisely what causes refraction.

✓Final answer
$The velocity of light decreases when it goes from a rarer medium to a denser medium.$

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Q.A ray of light incident at an angle $\theta$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^{\circ}$ and the prism is made of a material of refractive index 1.5, the angle of incidence is

Concept understanding — Refraction and Snell's Law

Refraction and Snell's Law

Setting up the prism geometry

Using the "emerges normally" condition

Finding $r_1$

Applying Snell's law at the first face

Method: Prism Refraction Using the Angle-Sum Relation $r_1+r_2=A$

Steps

Q.A ray of light incident at an angle θ\thetaθ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5∘5^{\circ}5∘ and the prism is made of a material of refractive index 1.5, the angle of incidence is

Setting up the prism geometry

Using the "emerges normally" condition

Finding r1r_1r1​

Applying Snell's law at the first face

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Q.A ray of light incident at an angle $\theta$ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is $5^{\circ}$ and the prism is made of a material of refractive index 1.5, the angle of incidence is

Finding $r_1$

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