Q.A small bulb (a point source) is placed at the bottom of a tank containing water to a depth of 80 cm. What is the radius of the circular surface of water through which light emerge out? Refractive index of water is 1.33.

Showing the 12 most recent of 37 on this concept.

1. CBSE 2026Set 55/3/11 markMCQ
   Q.In which of the following does total internal reflection NOT occur ? (A) Twinkling of stars (B) Brilliance of diamonds (C) Optical fibre (D) Reflecting prism

   ›Reveal solutionSolution

   Total internal reflection requires light to travel from a denser to a rarer medium at an angle greater than the critical angle. Twinkling of stars involves refraction through atmospheric layers of varying density, not TIR. The answer is (A).

   Total internal reflection is a phenomenon that occurs when light traveling in a denser medium strikes the boundary with a rarer medium at an angle greater than the critical angle. At this point, instead of refracting into the second medium, all the light reflects back into the denser medium. The key conditions are: light must go from higher to lower refractive index, and the angle of incidence must exceed the critical angle $\theta_c = \sin^{-1}(n_2/n_1)$ where $n_1 > n_2$.

   Let's examine each option to see where this principle applies:

   1. Twinkling of stars: Starlight enters Earth's atmosphere and passes through layers of air with slightly different temperatures and densities. As light moves through these layers, it undergoes refraction — bending at each boundary because the refractive index changes gradually. Sometimes light bends toward regions of higher density, sometimes toward lower density, but it continues to propagate through the atmosphere. The random fluctuations in atmospheric density cause the apparent position and brightness of stars to change rapidly, creating the twinkling effect. This is purely refraction, not total internal reflection.

   2. Brilliance of diamonds: A diamond has a very high refractive index ($n \approx 2.42$). When light enters a diamond and strikes the internal faces, it often does so at angles greater than the critical angle (which is quite small, around $24.4°$ for diamond-air interface). The light then undergoes total internal reflection multiple times inside the diamond before emerging, creating the characteristic sparkle and brilliance. …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set V11 markMCQ
   Q.For total internal reflection of light :

   (a) light should be travelling from rarer medium to denser medium

   (b) light should be travelling from denser medium to rarer medium

   (c) light should be incident along the normal

   (d) angle of incidence should be equal to $90^\circ$

   ›Reveal solutionSolution

   (b) light should be travelling from denser medium to rarer medium …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set DS1 markMCQ
   Q.If critical angle in a medium be $\alpha$, then for total internal reflection, the angle of incidence $\beta$ should be:

   i) $\beta < \alpha$

   ii) $\beta > \alpha$

   iii) $\beta = \alpha$

   iv) $\beta \le \alpha$

   ›Reveal solutionSolution

   TIR occurs only for angles of incidence larger than the critical angle, so $\beta > \alpha$.

   Concept. When light travels from a denser to a rarer medium, at the critical angle $\alpha$ the refracted ray just grazes the surface (angle of refraction $=90^\circ$). If the angle of incidence is increased beyond this critical angle, refraction is no longer possible and the light is completely reflected back into the denser medium — this is total internal reflection.

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 mark
   Q.When light passes from a denser medium to a rarer medium, and the angle of incidence equals the critical angle, what is the angle of refraction?

   ›Reveal solutionSolution

   The critical angle is DEFINED as the angle of incidence at which the refracted ray just grazes the interface, making the angle of refraction exactly 90 degrees.

   When light travels from a denser medium towards a rarer medium, it bends away from the normal. As the angle of incidence is increased, the angle of refraction increases faster and eventually reaches 90 degrees (the refracted ray travels right along the boundary between the two media) - the angle of incidence at which this happens is called the critical angle (theta_c). By …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 mark
   Q.Answer in one word/sentence: What is the phenomena that causes a bubble in water to shine brightly?

   ›Reveal solutionSolution

   Air bubbles in water appear to shine brightly (like silver) due to total internal reflection of light at the water-air interface of the bubble.

   Light travelling in the denser medium (water) strikes the surface of the air bubble (a rarer medium) at angles greater than the critical angle for the water-air interface. Since it cannot refract out, it is totally internal …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set JS1 mark
   Q.What do you mean by total internal reflection? Show it by drawing ray diagram.

   ›Reveal solutionSolution

   When light travelling in a denser medium hits a rarer medium at an angle greater than the critical angle, it is fully reflected back — this is total internal reflection.

   Definition. When light passes from an optically denser medium to an optically rarer medium, it bends away from the normal. As the angle of incidence increases, the refracted ray bends more, until at the critical angle $\theta_c$ the refracted ray grazes the surface ($r=90^\circ$). For any angle of incidence greater than $\theta_c$, no refraction occurs and the light is completely reflected back into the denser medium. This is called total internal reflection (TIR).

   Conditions.

   1. Light must travel from a denser to a rarer medium.
   2. The angle of incidence must exceed the critical angle: $i>\theta_c$, where $\sin\theta_c = \dfrac{1}{n}$ ($n$ = refractive index of the denser medium relative to the rarer).

   Ray diagram (description). For a ray striking the denser–rarer boundary at three increasing angles:

   • $i<\theta_c$: refracts into the rarer medium (bends away from normal).
   • $i=\theta_c$: refracted ray travels along the surface ($90^\circ$). …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set D1 markMCQ
   Q.If the critical angle for total internal reflection from any medium to vacuum is 30°, then the velocity of light in the medium is (A) 3 × 10^8 m/sec (B) 1.5 × 10^8 m/sec (C) 6 × 10^8 m/sec (D) 4.5 × 10^8 m/sec

   ›Reveal solutionSolution

   From the critical angle, refractive index n = 1/sin C = 2; the speed of light in the medium is c/n = 1.5×10⁸ m/s.

   The critical angle C for total internal reflection from a medium into vacuum is related to the refractive index n by

   $\sin C = \frac{1}{n}$

   With C = 30°:

   $n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2$

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set A1 mark
   Q.Fill in the blank with appropriate word: In total internal reflection, refractive angle is ______, when incidence angle is equal to the critical angle.

   ›Reveal solutionSolution

   At the critical angle, the refracted ray just grazes the boundary, so the angle of refraction is 90°.

   The critical angle C is defined, for light travelling from a denser to a rarer medium, as that particular angle of incidence for which the angle of refraction becomes exactly 90°, i.e. the refracted ray travels along the boundary/interface separating the two media. By Snell's law, n1 sin C = n2 sin 90° = n2, which gives sin C = n2/n1. For any …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2025Set ANNUAL1 mark
   Q.How is the critical angle related to the refractive index?

   ›Reveal solutionSolution

   The critical angle is the angle of incidence (in the denser medium) at which the refracted ray just grazes the interface at 90 degrees, which by Snell's law gives sin(C) = 1/n.

   Consider light travelling from a denser medium (refractive index n, relative to a rarer medium like air) toward the interface. By Snell's law, for the boundary case where the angle of refraction is exactly 90 degrees (refracted ray grazes along the surface):

   n sin(C) = 1 x sin(90 degrees) = 1

   sin(C) = 1/n

   Equivalently, n = 1/sin(C) (or n = cosec C).

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2025Set ANNUAL1 mark
    Q.Write the reason for the shining of air bubbles in the water.

    ›Reveal solutionSolution

    Light travelling from the denser medium (water) towards the rarer medium (air bubble) strikes the curved surface at an angle greater than the critical angle and undergoes total internal reflection, making the bubble appear bright/silvery.

    Water is optically denser than air. When light travelling inside water falls on the curved surface of an air bubble, it is going from a denser medium (water) to a rarer medium (air).

    For total internal reflection to occur, two conditions must be met:

    1. Light must travel from a denser medium to a rarer medium.
    2. The angle of incidence at the interface must be greater than the critical angle $\theta_c$ for the water–air pair (where $\sin\theta_c = 1/n_{water}$). …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2025Set ANNUAL1 markMCQ
    Q.If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in medium is (velocity of light = 3×10⁸ ms⁻¹).

    (a) 1.5 × 10⁸ ms⁻¹

    (b) 0.75 × 10⁸ ms⁻¹

    (c) 3 × 10⁸ ms⁻¹

    (d) 2 × 10⁸ ms⁻¹.

    ›Reveal solutionSolution

    Using $\sin C = 1/n$ with C = 30° gives n = 2, so the speed of light in the medium is c/2 = 1.5×10⁸ m/s.

    For total internal reflection from a medium into vacuum, the critical angle C satisfies

    $\sin C = \frac{1}{n}$ where n is the refractive index of the medium (with respect to vacuum).

    $\sin 30° = 0.5 = \frac{1}{n} \Rightarrow n = 2$

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2025Set ANNUAL1 mark
    Q.In the case of refraction, write down the relation between critical angle and refractive index of the denser medium. OR What will be the change in focal length f of a concave mirror when immersed in a liquid of refractive index μ?

    ›Reveal solutionSolution

    The refractive index of the denser medium (with respect to the rarer medium) is the reciprocal of the sine of the critical angle.

    At the critical angle C, light travelling from a denser to a rarer medium refracts exactly along the interface (angle of refraction = 90°). By Snell's law, applied to the ray going from the denser medium (refrac …

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