Q.(a)

Showing the 12 most recent of 46 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity $\left(\dfrac{3}{2}\right)$ and relative magnetic permeability $\left(\dfrac{8}{3}\right)$. Then, its (A) wavelength is doubled and frequency remains unchanged. (B) wavelength is doubled and frequency is halved. (C) wavelength is halved and frequency remains unchanged. (D) wavelength and frequency both will remain unchanged.

   ›Reveal solutionSolution

   The key idea is that frequency is determined by the source and never changes when a wave enters a new medium, while wavelength scales with the wave speed. Here the speed reduces by a factor of 2, so the wavelength is halved — making option (C) correct.

   When a wave crosses from one medium into another, the frequency never changes — it is set by the source and cannot be altered by the medium. What does change is the wave speed, and with it the wavelength, because $v = f \lambda$ must hold in every medium.

   The question gives us the relative permittivity $\varepsilon_r = \frac{3}{2}$ and relative permeability $\mu_r = \frac{8}{3}$. These determine the refractive index of the dielectric, which tells us how much the speed changes.

   1. Find the refractive index. For any medium, the refractive index is $n = \sqrt{\varepsilon_r \mu_r}$ (for a non-magnetic medium $\mu_r \approx 1$ and this reduces to $n = \sqrt{\varepsilon_r}$; here $\mu_r = \frac{8}{3}$, so we keep both factors).

   $n = \sqrt{\frac{3}{2} \times \frac{8}{3}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2.$

   So the dielectric has refractive index $n = 2$.

   2. Relate speed to wavelength. In vacuum, speed is $c$, wavelength is $\lambda_0$, and $c = f \lambda_0$. In the medium, speed is $v = \frac{c}{n} = \frac{c}{2}$, and $v = f \lambda$. Since $f$ is unchanged,

   $\frac{c}{2} = f \lambda \quad \Rightarrow \quad \lambda = \frac{c}{2f} = \frac{\lambda_0}{2}.$

   The wavelength is halved.

   3. Check the options.
      • (A) says wavelength doubled — wrong. …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.[FIGURE: Ray diagram at a plane interface between medium $n_1$ (rarer) and medium $n_2$ (denser), $n_2>n_1$, showing an incident ray, a reflected ray and a refracted ray.] When monochromatic light is incident on a surface separating two transparent media (first medium is rarer and second medium is denser) then some light is reflected back into first medium and remaining light is refracted in second medium. In this case -

   (i) Frequency of incident, refracted and reflected light is same.

   (ii) Frequency of incident and reflected light is same but frequency of refracted wave is decreased.

   (iii) Frequency of incident and reflected light is same but frequency of refracted light is increased.

   (iv) Frequency of incident light and refracted light is same but frequency of reflected light is changed.

   ›Reveal solutionSolution

   Frequency of light is fixed by the source and never changes on reflection or refraction.

   The frequency of a light wave is determined by the source that produces it, not by the medium it travels through. On reflection and refraction only the wavelength and speed change (since $v=f\lambda$ and $v$ depends on the medium); the frequency …

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3. CBSE 2026Set A1 markMCQ
   Q.The ratio of the refractive index of red light to blue light in air is (A) less than unity (B) greater than unity (C) equal to unity (D) none of these

   ›Reveal solutionSolution

   Because refractive index rises from red to blue (dispersion), n_red < n_blue, so their ratio is less than 1.

   Dispersion arises because the refractive index of a medium depends on wavelength: it is larger for shorter wavelengths (blue) and smaller for longer wavelengths (red). Thus $n_{red}<n_{blue}$, and their ratio is

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The focal length of a lens is minimum for which colour?

   (a) Red

   (b) Yellow

   (c) Violet

   (d) Blue

   ›Reveal solutionSolution

   Since refractive index decreases from violet to red (dispersion), and $f\propto 1/(\mu-1)$, the colour with the highest $\mu$ — violet — bends most and has the smallest focal length.

   For a thin lens, the lens-maker's formula gives $\dfrac{1}{f}=(\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so $f\propto\dfrac{1}{\mu-1}$: a larger refractive index means a smaller focal length. Because the refractive index of ordinary glass is greater for shorter wavelengths, the order from lowest to highest $\mu$ across the visible spectrum is red < orange < yellow < green < blue < …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.If a wave gets refracted into a denser medium, then which of the following is true?

   (a) wavelength, speed and frequency decrease.

   (b) wavelength increases, speed decreases and frequency remain constant.

   (c) wavelength and speed decrease but frequency remains constant.

   (d) wavelength, speed and frequency increase.

   ›Reveal solutionSolution

   Frequency is fixed by the source and never changes on refraction; since $v = f\lambda$ and $v$ decreases in a denser medium, $\lambda$ must decrease too.

   When a wave crosses from a rarer to a denser medium:

   • Frequency is determined only by the source that generates the wave and stays the same in every medium — this is why frequency is used to define colour of light, unlike wavelength.
   • Speed decreases in a denser medium because the refractive index $n = c/v$ is larger there, so $v = c/n$ is smaller. …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The speed of light in an isotropic medium depends on :

   (a) the nature of propagation

   (b) its intensity

   (c) the motion of the source with respect to medium

   (d) its wavelength

   ›Reveal solutionSolution

   A medium's refractive index (and hence the speed of light within it) varies with wavelength -- the phenomenon of dispersion.

   Working

   The speed of light in a medium is $v=c/n$. For any real (non-vacuum) isotropic medium, the refractive index $n$ is not fixed but depends on the wavelength of the light (e.g., violet light travels slightly slower than red light in glass) -- this dependence is the cause of dispersion (splitting of white light by a prism).

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.In glass, the velocity of the light is minimum for :

   (a) red

   (b) violet

   (c) yellow

   (d) green

   ›Reveal solutionSolution

   Violet light bends the most and travels slowest in glass because glass has its highest refractive index for violet light.

   Speed of light in a medium is $v = c/n$. Due to dispersion, the refractive index of glass is different for different colours and increases from red to violet (violet has the shortest wavelength and is refracted most). Since $n_{violet}$ is the largest …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.If red light and violet rays are of focal lengths f_r and f_v, then which one of the following is true ?

   (a) lambda_r <= lambda_v

   (b) lambda_r = lambda_v

   (c) mu_r > mu_v

   (d) mu_r < mu_v

   ›Reveal solutionSolution

   Violet light has a shorter wavelength and is refracted more, so its refractive index is higher than red light's.

   By Cauchy's dispersion relation, the refractive index of a medium decreases as wavelength increases: $\mu = A + \dfrac{B}{\lambda^2}$ (approximately). Since red light has a longer wavelength than violet light ($\lambda_r > \lambda_v$), red light is refracted less and has a smaller refractive index than violet light.

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.If the refractive index of a material of equilateral prism is sqrt(3), then the angle of minimum deviation is

   (i) 30 degrees

   (ii) 45 degrees

   (iii) 60 degrees

   (iv) 75 degrees

   ›Reveal solutionSolution

   Dm = 60 degrees for an equilateral prism with mu = sqrt(3).

   The prism formula is $\mu = \frac{\sin\!\left(\frac{A + D_m}{2}\right)}{\sin\!\left(\frac{A}{2}\right)}$. For an equilateral prism $A = 60^\circ$, so $\sin(A/2) = \sin 30^\circ = 0.5$. Then …

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10. CBSE 2024Set IMPROVEMENT1 markMCQ
    Q.When light enters from one medium to other medium then its —

    (a) Frequency is changed, wavelength remains unchanged.

    (b) Frequency remains unchanged, wavelength is changed.

    (c) Frequency and wavelength both are changed.

    (d) Frequency and wavelength both remain unchanged.

    ›Reveal solutionSolution

    On crossing a boundary between media, frequency (fixed by the source) stays the same; wavelength changes because speed changes.

    When light passes from one medium to another, its frequency $\nu$ is determined by the source and does not change. However, its speed $v$ changes because it depends on the refractive index of the medium ($v=c/n$). Since $v=\nu\lambda$, a change in speed at constant frequency means the wavelength …

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11. CBSE 2024Set FS1 markMCQ
    Q.The maximum focal length of convex lens is for:

    (i) blue light

    (ii) green light

    (iii) red light

    (iv) yellow light

    ›Reveal solutionSolution

    $\dfrac{1}{f}\propto(n-1)$; red light has the least refractive index in glass, so it gives the greatest focal length — option (iii).

    Concept. For a lens, $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so $f$ is largest when the refractive index $n$ is smallest.

    …

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12. CBSE 2024Set A1 markMCQ
    Q.The wavelength of which colour is minimum? (A) Violet (B) Yellow (C) Blue (D) Red

    ›Reveal solutionSolution

    Violet has the minimum wavelength in the visible spectrum.

    In the visible spectrum (VIBGYOR), wavelength increases steadily from violet to red:

    $\text{Violet} (\sim 400\,\text{nm}) < \text{Blue} < \text{Yellow} < \text{Red} (\sim 700\,\text{nm}).$

    …

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