(a) Using Huygens' principle, draw a ray diagram showing the propagation of a plane wave refracting at a plane

Question

Q.(a) Using Huygens' principle, draw a ray diagram showing the propagation of a plane wave refracting at a plane surface separating two media. Also verify Snell's law of refraction.

(OR)

(b) Why is a reflecting telescope preferred over a refracting telescope? Justify your answer giving two reasons.

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Huygens' Principle and Wavefronts

A wavefront is the locus of all points that are in the same phase (state) of vibration at a given instant; because points on a wavefront are all in step, a wavefront is always perpendicular to the direction of the wave's propagation (the ray). A point source at a finite distance produces spherical wavefronts, a line (extended) source produces cylindrical wavefronts, and any source located effectively at infinity produces plane wavefronts. Huygens' principle is a geometrical construction rule stating that every point on a wavefront acts as a source of secondary spherical wavelets, spreading outward with the wave's own speed; the new wavefront at a later time $t$ is the common forward tangent (the 'envelope') of all these secondary wavelets, each having grown to radius $ct$ from its originating point. This construction correctly reproduces spherical wavefronts from a nearby point source and plane wavefronts from a distant source, and it explains how a wavefront propagates without needing to track individual rays. Its one acknowledged shortcoming is that it …

Part (b)Concept understanding — Resolving Power of Telescope

Resolving Power of a Telescope – From Intuition to Precision

Imagine you are looking at two stars that are very close together in the night sky. Through a small telescope, they might appear as a single blurry blob. Through a larger telescope, you see them as two distinct points. Why? The answer is not about magnification — it is about resolution, the ability to distinguish fine detail.

The Core Intuition

Light behaves like a wave. When it enters a telescope, it passes through the circular opening (the aperture) and diffracts — it spreads out slightly. Even a perfect lens cannot focus light to an infinitely sharp point. Instead, it produces a small central bright spot surrounded by faint rings, called an Airy pattern. This is not a flaw in the lens; it is a fundamental property of wave optics.

If two stars are extremely close, their Airy patterns overlap. When the overlap is too great, your eye cannot tell that there are two sources — they merge into one. The resolving power of a telescope is its ability to show two closely spaced objects as separate.

The Rayleigh Criterion

There is a precise condition for when two point sources are just barely resolvable. It is called the Rayleigh criterion:

Two sources are resolved when the central maximum of one falls on the first minimum of the other.

For a circular aperture, the angular separation $\Delta\theta$ (in radians) at which this happens is:

\Delta\theta = 1.22 \frac{\lambda}{D}

where:

$\lambda$ is the wavelength of light
$D$ is the diameter of the telescope's objective (the main lens or mirror)

Important

A smaller $\Delta\theta$ means better resolving power — you can distinguish finer details. So a larger $D$ gives a smaller $\Delta\theta$ , which is why big telescopes see more detail.

Why 1.22?

The factor 1.22 comes from the mathematics of diffraction through a circular opening. It is the first zero of the Bessel function that describes the Airy pattern. For a slit (a rectangular opening), the factor would be 1.00 — but telescopes have circular apertures, so 1.22 is the number you need.

A Concrete Example

Suppose you use a telescope with $D = 10$ cm and observe yellow light with $\lambda = 550$ nm ( $5.5 \times 10^{-7}$ m). Then:

\Delta\theta = 1.22 \times \frac{5.5 \times 10^{-7}}{0.1} = 6.71 \times 10^{-6} \text{ radians}

That is about 1.4 arcseconds. Two stars closer than this would appear as one.

Now double the aperture to $D = 20$ cm. The resolving angle halves to $3.35 \times 10^{-6}$ radians (about 0.7 arcseconds). You can now separate stars that are twice as close.

Tip

| Aperture $D$ | Resolving angle $\Delta\theta$ (for $\lambda = 550$ nm) |

|--------------|--------------------------------------------------------|

| 5 cm | 2.8 arcseconds |

| 10 cm | 1.4 arcseconds |

| 20 cm | 0.7 arcseconds |

| 1 m | 0.14 arcseconds |

What Resolving Power Is NOT …

Start your 14-day free trial to unlock the full solution →

Part (a): Huygens' wavelet construction gives $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=n_{21}$ (Snell's law).

Part (b): a reflecting telescope beats a refracting one—no chromatic aberration and a larger aperture (more light, higher resolving power).

Diagram of a plane wavefront refracting at a plane interface between Medium 1 (speed v1) and Medium 2 (speed v2 greater than v1), used to derive Snell's law from Huygens' construction with incident wavefront AB, angle of incidence i, angle of refraction r, and points A, B, C, E.

Part (a)

Let a plane wavefront $AB$ travel in medium 1 (speed $v_1$ ) and strike the plane boundary $XY$ at angle of incidence $i$ . Each point of the boundary becomes a source of secondary wavelets that spread into medium 2 with speed $v_2$ . In the time $t$ taken by the end $B$ to reach the surface at $C$ ( $BC=v_1t$ ), the wavelet from $A$ advances a distance $AD=v_2t$ into medium 2. Drawing the tangent $CD$ gives the refracted wavefront, making angle $r$ with the surface.

In right triangles $ABC$ and $ADC$ (common hypotenuse $AC$ ):

\sin i=\frac{BC}{AC}=\frac{v_1t}{AC},\qquad \sin r=\frac{AD}{AC}=\frac{v_2t}{AC}.

Dividing,

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{n_2}{n_1}=n_{21}=\text{constant},$ …

Start your 14-day free trial to unlock the full solution →

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 30 on this concept.

CBSE 2026Set V11 markMCQ
$Q. According to Huygen's principle, speed of the secondary wavelets is :$
(a) twice that of the wave
(b) zero
(c) same as that of the wave
(d) infinite
›Reveal solutionSolution

(c) same as that of the wave …

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. The wavefront due to a point source at a finite distance from the source is (A) plane (B) circular (C) cylindrical (D) spherical$
›Reveal solutionSolution

A point source radiates uniformly outward, giving spherical wavefronts (they become nearly plane only very far away).

A wavefront is the surface joining points of the same phase. A point source sends out disturbances equally in every direction, so at any finite distance the set of in-phase points forms a sphere centred on the source — a ** …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The phase difference between any two points on a wavefront is$
(a) 0°
(b) 45°
(c) 90°
(d) 120°
›Reveal solutionSolution

A wavefront is defined as the locus of points that are in the same phase of oscillation, so the phase difference between any two points on it is zero.

By definition, a wavefront is a surface of constant phase - every point on it has travelled the same optical path length from the source and is therefore oscillating in exactly the same …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The concept of secondary wavelets was given by$
(a) Fresnel
(b) Newton
(c) Huygens
(d) Maxwell
›Reveal solutionSolution

Huygens' principle says every point of a wavefront is itself a source of secondary wavelets, and the new wavefront is their common tangent surface.

Christiaan Huygens proposed that every point on an existing wavefront can be regarded as a source of new, secondary spherical wavelets that spread out in the forward direction with the speed of the wave in that medium. The surface that is tangential to all these secondary wavelets at any later instant gives the new position of the wavefront. This principle (Huygens' principle) is used to explain reflection, refraction, and diffraction of light p …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The phase difference between two points located on the same wavefront is:$
(a) 2π
(b) 0
(c) π
(d) π/2
›Reveal solutionSolution

A wavefront is, by definition, the locus of points that are all at the same stage (phase) of their vibration, so the phase difference between any two of its points is zero.

A wavefront is defined as the continuous locus of all points in a medium that are oscillating in the same phase, having been disturbed by the wave at the same instant relative to the source (e.g. a sphere for a point source, or a plane far from the source). Since every point on a wavefront is, by construction, at the identical stage of its oscillation cycle, the phase difference between any two such points …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Draw the diagram of wave front produced by a point source of light.$
›Reveal solutionSolution
Figure — Explicit 'Draw the diagram of wave front produced by a point source' hard gate. Catalog fig-10-1 is exactly a

A point source produces spherical wavefronts.

A point source of light emits waves equally in all directions, so at any given instant, all the points at the same distance from the source are vibrating in the same phase. The locus of these points — the wavefront — is therefore a sphere centred on the source, and as time progresses, this spherical wavefront expands outward with radius increasing at t …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): A light wave can be considered to travel from one point to another along a straight line. Reason (R): Wavelength of light is very small compared to size of ordinary objects in daily life.$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is incorrect.
(d) A and R both are incorrect.
›Reveal solutionSolution

Ray (straight-line) propagation of light is a good approximation exactly because light's wavelength is far smaller than the objects/apertures it interacts with in everyday life.

Assertion (A) is true: over everyday distances and apertures, light can be treated as travelling in straight lines (rays) — this is the basis of ray/geometrical optics. Reason (R) is also true: the wavelength of visible light (~400–700 nm) is indeed extremely small compared to the size of ordinary objects and apertures encountered in daily life. R correctly explains A: diffraction effects (bending of waves around obstacles/edges) become significant only when the size of an obstacle or aperture is com …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. What is the shape of the wave front when light is diverging from a point source?$
›Reveal solutionSolution

A point source emits waves equally in all directions, so surfaces of constant phase are spheres centred on the source.

A wavefront is the locus of all points that a wave disturbance reaches at the same instant (a surface of constant phase). A point source radiates light uniformly in all directions in three-dimensional space, so at any given instant, all points at the same distance from the source have travelled for the same time and are in the same phase. The locus of such points is a sphere centred at the source. …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. According to Huygens, light waves are$
(a) Mechanical, longitudinal
(b) Mechanical, transverse
(c) Electromagnetic
(d) Mechanical, spherical
›Reveal solutionSolution

Huygens proposed light as a wave phenomenon needing a material medium (the luminiferous ether) to travel through, and modelled it, like sound, as a longitudinal mechanical wave.

Huygens' wave theory explained reflection, refraction and the (then-mysterious) phenomenon of double refraction using the idea of wavefronts built from secondary wavelets. However, since it treated light as a mechanical disturbance in a medium (analogous to sound waves in air), the vibrations were taken to be longitudinal.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Hugen's wave theory allows us to know about :$
(a) the wavelength of the wave.
(b) the velocity of the wave.
(c) the amplitude of the wave.
(d) the propagation of wavefronts.
›Reveal solutionSolution

Huygens' principle is a geometrical construction that predicts how a wavefront advances with time, i.e., the propagation of wavefronts.

Huygens' wave theory states that every point on a wavefront acts as a source of secondary wavelets, which spread out in all directions with the wave's speed. The new wavefront, at a later time, is the envelope (tangent surface) of these secondary wavelets. This construction is a geometric tool specifically for tracking how a wavefront propagates and changes shape …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A11 markMCQ
$Q. According to Huygen's principle, the speed of the secondary wavelets is$
(a) twice that of the wave
(b) zero
(c) same as the wave
(d) infinite
›Reveal solutionSolution

(c) same as the wave …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. According to Huygen's principle, the direction of energy propagation of a wave is -$
(a) along the surface of wave front
(b) perpendicular to wave front
(c) at an angle of 45 $^\circ$ to wave front
(d) None of these
›Reveal solutionSolution

Energy flows along the ray direction, and rays are always normal to the wavefront.

A wavefront is the locus of points having the same phase of vibration. The direction of propagation of energy (the ray) is always perpendicular (normal) to the …

Start your 14-day free trial to see the full solution →

Q.(a) Using Huygens' principle, draw a ray diagram showing the propagation of a plane wave refracting at a plane surface separating two media. Also verify Snell's law of refraction.

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days