Nature of Light & Huygens' Principle — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Nature of Light & Huygens' Principle MCQs with step-by-step solutions (21 questions). Part of Wave Optics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Nature of Light & Huygens' Principle · easy · theory
According to Huygens' principle, every point on a wavefront acts as:
A. A point of zero amplitude
B. A source of longitudinal waves only
C. A perfect reflector of light
D. A source of secondary spherical wavelets ✓ Correct
Solution: The new wavefront at a later instant is the forward envelope of all these secondary wavelets — the construction that explains reflection and refraction.
Q2 — Nature of Light & Huygens' Principle · easy · theory
A wavefront is defined as the locus of all points which are:
A. At the same distance from the observer
B. In the same phase of vibration ✓ Correct
C. Of equal intensity only
D. Moving with the same speed
Solution: Every point on a given wavefront has been reached by the disturbance at the same instant, so they all vibrate in step.
Q3 — Nature of Light & Huygens' Principle · easy · theory
The wavefront produced by a point source of light at a finite distance is:
A. Plane
B. Spherical ✓ Correct
C. Cylindrical
D. Parabolic
Solution: The disturbance spreads out equally in all directions, so surfaces of constant phase are concentric spheres. Far from the source a small patch approximates a plane wavefront.
Q4 — Nature of Light & Huygens' Principle · medium · theory
The relationship between a ray of light and the wavefront is that the ray is:
A. Inclined at $45^\circ$ to the wavefront
B. Always along the wavefront
C. Perpendicular to the wavefront ✓ Correct
D. Parallel to the wavefront
Solution: A ray simply marks the direction in which the wavefront advances, which is along the normal to the surface of constant phase.
Q5 — Nature of Light & Huygens' Principle · easy · theory
Light waves are:
A. Longitudinal electromagnetic waves
B. Transverse mechanical waves
C. Transverse electromagnetic waves ✓ Correct
D. Longitudinal mechanical waves
Solution: The electric and magnetic fields oscillate perpendicular to the direction of propagation, which is why light can be polarised.
Q6 — Nature of Light & Huygens' Principle · easy · theory
The speed of light in vacuum is approximately:
A. $3 \times 10^8\text{ m/s}$ ✓ Correct
B. $3 \times 10^6\text{ m/s}$
C. $3 \times 10^{10}\text{ m/s}$
D. $3 \times 10^5\text{ m/s}$
Solution: The accepted value is $2.998 \times 10^8\text{ m/s}$, usually rounded to $3 \times 10^8\text{ m/s}$.
Q7 — Nature of Light & Huygens' Principle · medium · theory
When light passes from air into a denser medium, the quantity that remains unchanged is its:
A. Frequency ✓ Correct
B. Amplitude
C. Wavelength
D. Speed
Solution: Frequency is fixed by the source. The speed falls to $\dfrac{c}{\mu}$ and the wavelength shortens in the same proportion.
Q8 — Nature of Light & Huygens' Principle · medium · theory
Huygens' wave construction successfully explains:
A. The photoelectric effect
B. Compton scattering
C. The emission spectrum of hydrogen
D. Reflection and refraction of light ✓ Correct
Solution: The secondary-wavelet construction reproduces the laws of reflection and refraction. Photoelectric and Compton effects need the particle picture instead.
Q9 — Nature of Light & Huygens' Principle · easy · numerical
The refractive index of glass is $1.5$. The speed of light in the glass is:
A. $1.5 \times 10^8\text{ m/s}$
B. $3 \times 10^8\text{ m/s}$
C. $4.5 \times 10^8\text{ m/s}$
D. $2 \times 10^8\text{ m/s}$ ✓ Correct
Solution: $v = \dfrac{c}{\mu} = \dfrac{3 \times 10^8}{1.5} = 2 \times 10^8\text{ m/s}$.
Q10 — Nature of Light & Huygens' Principle · medium · numerical
Light of wavelength $600\text{ nm}$ in vacuum enters a medium of refractive index $1.5$. Its wavelength there is:
A. $300\text{ nm}$
B. $900\text{ nm}$
C. $400\text{ nm}$ ✓ Correct
D. $600\text{ nm}$
Solution: $\lambda_{medium} = \dfrac{\lambda_0}{\mu} = \dfrac{600}{1.5} = 400\text{ nm}$.
Q11 — Nature of Light & Huygens' Principle · medium · numerical
The refractive index of water is $\dfrac{4}{3}$. The speed of light in water is:
A. $2 \times 10^8\text{ m/s}$
B. $1.33 \times 10^8\text{ m/s}$
C. $2.25 \times 10^8\text{ m/s}$ ✓ Correct
D. $4 \times 10^8\text{ m/s}$
Solution: $v = \dfrac{c}{\mu} = 3 \times 10^8 \times \dfrac{3}{4} = 2.25 \times 10^8\text{ m/s}$.
Q12 — Nature of Light & Huygens' Principle · medium · numerical
The frequency of light of wavelength $500\text{ nm}$ in vacuum is:
A. $6 \times 10^{12}\text{ Hz}$
B. $5 \times 10^{14}\text{ Hz}$
C. $1.5 \times 10^{15}\text{ Hz}$
D. $6 \times 10^{14}\text{ Hz}$ ✓ Correct
Solution: $n = \dfrac{c}{\lambda} = \dfrac{3 \times 10^8}{5 \times 10^{-7}} = 6 \times 10^{14}\text{ Hz}$.
Q13 — Nature of Light & Huygens' Principle · easy · numerical
Light of wavelength $600\text{ nm}$ enters a medium of refractive index $1.2$. Its wavelength in the medium is:
A. $720\text{ nm}$
B. $300\text{ nm}$
C. $500\text{ nm}$ ✓ Correct
D. $400\text{ nm}$
Solution: $\lambda_{medium} = \dfrac{600}{1.2} = 500\text{ nm}$.
Q14 — Nature of Light & Huygens' Principle · medium · numerical
The optical path corresponding to a geometric path of $2\text{ cm}$ in a medium of refractive index $1.5$ is:
A. $2\text{ cm}$
B. $0.5\text{ cm}$
C. $3\text{ cm}$ ✓ Correct
D. $1.33\text{ cm}$
Solution: Optical path $= \mu \times$ geometric path $= 1.5 \times 2 = 3\text{ cm}$.
Q15 — Nature of Light & Huygens' Principle · hard · numerical
Light travels at $2 \times 10^8\text{ m/s}$ in medium $1$ and at $1.5 \times 10^8\text{ m/s}$ in medium $2$. The refractive index of medium $2$ relative to medium $1$ is:
A. $0.75$
B. $2.0$
C. $1.33$ ✓ Correct
D. $1.5$
Solution: $_1\mu_2 = \dfrac{v_1}{v_2} = \dfrac{2 \times 10^8}{1.5 \times 10^8} \approx 1.33$.
Q16 — Nature of Light & Huygens' Principle · hard · numerical
The time taken by light to cross a glass plate $3\text{ mm}$ thick of refractive index $1.5$ is:
A. $2 \times 10^{-11}\text{ s}$
B. $4.5 \times 10^{-11}\text{ s}$
C. $1 \times 10^{-11}\text{ s}$
D. $1.5 \times 10^{-11}\text{ s}$ ✓ Correct
Solution: Speed in glass $= 2 \times 10^8\text{ m/s}$, so $t = \dfrac{3 \times 10^{-3}}{2 \times 10^8} = 1.5 \times 10^{-11}\text{ s}$.
Q17 — Nature of Light & Huygens' Principle · easy · numerical
Light of wavelength $400\text{ nm}$ enters a medium of refractive index $2$. Its wavelength there becomes:
A. $800\text{ nm}$
B. $400\text{ nm}$
C. $200\text{ nm}$ ✓ Correct
D. $100\text{ nm}$
Solution: $\lambda_{medium} = \dfrac{400}{2} = 200\text{ nm}$.
Q18 — Nature of Light & Huygens' Principle · hard · numerical
Light of vacuum wavelength $600\text{ nm}$ travels through glass. Its frequency in the glass is:
A. $3.3 \times 10^{14}\text{ Hz}$
B. $6 \times 10^{14}\text{ Hz}$
C. $7.5 \times 10^{14}\text{ Hz}$
D. $5 \times 10^{14}\text{ Hz}$ ✓ Correct
Solution: Frequency does not change on entering a medium: $n = \dfrac{3 \times 10^8}{6 \times 10^{-7}} = 5 \times 10^{14}\text{ Hz}$.
Q19 — Nature of Light & Huygens' Principle · hard · numerical
A glass slab of thickness $4\,\mu\text{m}$ and refractive index $1.5$ is placed in the path of light. The optical path difference introduced is:
A. $1\,\mu\text{m}$
B. $6\,\mu\text{m}$
C. $2\,\mu\text{m}$ ✓ Correct
D. $4\,\mu\text{m}$
Solution: The extra optical path is $(\mu - 1)t = (1.5 - 1) \times 4 = 2\,\mu\text{m}$.
Q20 — Nature of Light & Huygens' Principle · medium · numerical
Light of frequency $6 \times 10^{14}\text{ Hz}$ travels in vacuum. Its wavelength is:
A. $400\text{ nm}$
B. $300\text{ nm}$
C. $500\text{ nm}$ ✓ Correct
D. $600\text{ nm}$
Solution: $\lambda = \dfrac{c}{n} = \dfrac{3 \times 10^8}{6 \times 10^{14}} = 5 \times 10^{-7}\text{ m} = 500\text{ nm}$.
Q21 — Nature of Light & Huygens' Principle · medium · numerical
A point source emits light in vacuum. The radius of its spherical wavefront after $1\text{ second}$ is:
A. $3 \times 10^8\text{ m}$ ✓ Correct
B. $3 \times 10^{10}\text{ m}$
C. $1\text{ m}$
D. $3 \times 10^6\text{ m}$
Solution: The wavefront expands at the speed of light, so its radius is $ct = 3 \times 10^8 \times 1 = 3 \times 10^8\text{ m}$.