Q.Let us list some of the factors, which could possibly influence the speed of wave propagation:

Showing the 12 most recent of 29 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The angle of prism is equal to the angle of minimum deviation for a prism of refractive index 1.5. Then the value of angle of the prism is (A) $\cos^{-1}\left(\dfrac{3}{4}\right)$ (B) $2\cos^{-1}\left(\dfrac{3}{4}\right)$ (C) $\cos^{-1}\left(\dfrac{3}{2}\right)$ (D) $2\cos^{-1}\left(\dfrac{3}{2}\right)$

   ›Reveal solutionSolution

   This tests the prism minimum-deviation formula under the special condition $D_m = A$; solving gives $A = 2\cos^{-1}(3/4)$.

   Concept and Intuition

   The standard minimum-deviation relation links the prism angle $A$, the minimum deviation $D_m$, and refractive index $\mu$. When the problem imposes the special condition that the minimum deviation equals the prism angle itself, the formula collapses into a much simpler direct relation between $\mu$ and $A/2$ using a half-angle identity — this is a common trick to test whether the student can simplify $\sin A$ in terms of $\sin(A/2)$ and $\cos(A/2)$.

   Step-by-Step Solution

   1. Start from $\mu = \dfrac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$.
   2. Substitute $D_m = A$: $\mu = \dfrac{\sin\left(\frac{2A}{2}\right)}{\sin(A/2)} = \dfrac{\sin A}{\sin(A/2)}$.
   3. Use $\sin A = 2\sin(A/2)\cos(A/2)$: $\mu = \dfrac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The refractive index of water is $\dfrac{4}{3}$ and that of glass is $\dfrac{3}{2}$. The refractive index of glass with respect to water is (A) 9/8 (B) 8/9 (C) 6/4 (D) 2.0

   ›Reveal solutionSolution

   Tests the relative-refractive-index relation between two media given their absolute refractive indices (both relative to vacuum/air). Answer: 9/8.

   Concept and Intuition

   The refractive index of medium 2 with respect to medium 1 is simply the ratio of their absolute refractive indices (each measured relative to vacuum): $_1n_2=\dfrac{n_2}{n_1}=\dfrac{v_1}{v_2}$. This follows directly from combining $n=c/v$ for each medium.

   Step-by-Step Solution

   1. Absolute refractive indices: $n_{water}=4/3$, $n_{glass}=3/2$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.For the same angle of incidence, the angles of refraction of light ray in different media A, B, C are $35°, 25°, 15°$. If $V_A, V_B, V_C$ are velocities of light in A, B, C media respectively, then (A) $V_A = V_B = V_C = 0$ (B) $V_A = V_B = V_C$ (C) $V_A > V_B > V_C$ (D) $V_A < V_B < V_C$

   ›Reveal solutionSolution

   Tests reading off relative refractive indices (and hence relative speeds of light) from refraction angles at fixed incidence angle. Answer: $V_A>V_B>V_C$.

   Concept and Intuition

   Snell's law, $n=\dfrac{\sin i}{\sin r}$, shows that for a fixed angle of incidence, a medium that bends the ray more (smaller refraction angle $r$, hence larger $\sin i/\sin r$) has a higher refractive index. Since a higher refractive index physically means light travels slower in that medium ($v=c/n$), the medium bending light the most is the one where light moves slowest.

   Step-by-Step Solution

   1. Snell's law with common $i$: $n_A=\dfrac{\sin i}{\sin35^\circ}$, $n_B=\dfrac{\sin i}{\sin25^\circ}$, $n_C=\dfrac{\sin i}{\sin15^\circ}$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.For a prism of angle $5^\circ$, the angle of minimum deviation ($\delta$) varies with the refractive index ($\mu$) as shown in the graph (graph: $\delta$ on the y-axis, $\mu$ on the x-axis, a straight line rising from point P near the origin on the $\mu$-axis to point Q). The slope of the graph is (A) $5^\circ$ (B) $5\ rad$ (C) $0.5^\circ$ (D) $0.5\ rad$

   ›Reveal solutionSolution

   This tests the thin-prism deviation formula $\delta=(\mu-1)A$ and reading a slope off a $\delta$ vs $\mu$ graph. Answer: $5^\circ$.

   Concept and Intuition

   For a prism of small angle $A$ (here $5^\circ$, small enough for the thin-prism approximation), the minimum deviation is related to the refractive index by the simple linear law $\delta = (\mu - 1)A$. Plotting $\delta$ against $\mu$ gives a straight line $\delta = A\mu - A$, whose slope (rate of change of $\delta$ with $\mu$) is exactly the prism angle $A$, since $\mu$ itself carries no units.

   Step-by-Step Solution

   1. Thin prism relation: $\delta = (\mu - 1)A = A\mu - A$.
   2. This is a straight line in the $(\mu,\delta)$ plane, matching the graph described (a line rising from a point on the $\mu$-axis).
   3. Slope $= \dfrac{d\delta}{d\mu} = A$.
   4. Given $A = 5^\circ$, the slope is $5^\circ$.

   Common Mistakes …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A Vessel of depth x is half filled with oil of refractive index $\mu_1$ and the other half is filled with water of refractive index $\mu_2$. The apparent depth of the vessel when viewed from above is (A) $\frac{x(\mu_1+\mu_2)}{2\mu_1\mu_2}$ (B) $\frac{x\mu_1\mu_2}{2(\mu_1+\mu_2)}$ (C) $\frac{2x\mu_1\mu_2}{(\mu_1+\mu_2)}$ (D) $\frac{2x(\mu_1+\mu_2)}{\mu_1\mu_2}$

   ›Reveal solutionSolution

   Apparent depth of each half-layer (real depth / refractive index) simply adds for the two stacked liquids, giving $\frac{x(\mu_1+\mu_2)}{2\mu_1\mu_2}$.

   Concept and Intuition

   For near-normal viewing, a slab of transparent medium of real thickness $t$ and refractive index $\mu$ appears to have thickness (apparent depth) $t/\mu$ when viewed from a medium of lower refractive index (air) above it. When several such slabs are stacked, the total apparent depth as seen from above is (to this same paraxial approximation) just the sum of each slab's own apparent depth — the layers don't "interact" in this formula, each contributes independently.

   Step-by-Step Solution

   1. Vessel depth $x$, split into two equal layers, each of real thickness $x/2$: one of oil (index $\mu_1$), one of water (index $\mu_2$).
   2. Apparent depth contributed by the oil layer: $\dfrac{x/2}{\mu_1}$.
   3. Apparent depth contributed by the water layer: $\dfrac{x/2}{\mu_2}$.
   4. Total apparent depth (they add, since one is stacked directly under the other when viewed from directly above): …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A rectangular glass block of thickness 10 cm and refractive index 1.5 is placed over a small coin. A beaker is filled with water of refractive index $\dfrac{4}{3}$ to a height of 10 cm and placed over glass block. The apparent depth of coin when viewed at near normal incidence (A) 3.3 cm (B) 5.8 cm (C) 12.0 cm (D) 14.2 cm

   ›Reveal solutionSolution

   Each refracting layer shifts the apparent position of the coin by (real thickness)/(refractive index); adding the glass and water contributions gives an apparent depth of about 14.2 cm.

   Concept and Intuition

   When you look down through several stacked transparent media at an object, each layer independently "raises" the apparent position of what's beneath it by a factor of its own refractive index (for near-normal viewing, apparent depth $=\dfrac{\text{real depth}}{n}$). Because refraction at each interface is treated layer by layer, the total apparent depth as seen from above is simply the sum of each layer's real thickness divided by its own refractive index.

   Step-by-Step Solution

   1. Glass layer: thickness 10 cm, $n=1.5$. Its contribution to apparent depth: $10/1.5=6.667$ cm.
   2. Water layer: thickness 10 cm, $n=4/3$. Its contribution: $10/(4/3)=10\times3/4=7.5$ cm. …

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7. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.When a light ray incidents on an equilateral prism of material of refractive index $\sqrt{2}$, the angle of minimum deviation is D. If the light ray incidents on another equilateral prism of material of refractive index $\sqrt{3}$, then the angle of minimum deviation is (A) $\sqrt{1.5}\ D$ (B) $\sqrt{3}\ D$ (C) $0.5\ D$ (D) $2\ D$

   ›Reveal solutionSolution

   Tests the prism formula $n=\dfrac{\sin\frac{A+D}{2}}{\sin\frac{A}{2}}$ for two different refractive indices on the same (equilateral) prism geometry, comparing their minimum deviations.

   Concept and Intuition

   For a given prism angle, a higher refractive index bends light more, producing a larger minimum-deviation angle. The prism-angle formula relates $n$, $A$ (fixed at $60^\circ$ for an equilateral prism) and $D$ directly through a sine relation, so each refractive index maps to one specific deviation angle, and we just solve the equation twice.

   Step-by-Step Solution

   1. Prism-angle formula for minimum deviation: $n = \dfrac{\sin\left(\dfrac{A+D}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$. For an equilateral prism, $A=60^\circ$, so $\sin(A/2)=\sin30^\circ = 0.5$.
   2. First prism, $n_1=\sqrt2$: $\sin\left(\dfrac{60+D_1}{2}\right) = \sqrt2\times0.5 = \dfrac{\sqrt2}{2} = \sin45^\circ$.
   3. So $\dfrac{60+D_1}{2}=45^\circ \Rightarrow 60+D_1=90^\circ \Rightarrow D_1 = 30^\circ$. This is the "$D$" referred to in the question. …

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8. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A thin prism with angle $6°$, has refractive index $\mu_v = 1.532, \mu_r = 1.514$ for violet and red light respectively. The angular dispersion produced by the prism is (A) $0.210°$ (B) $0.108°$ (C) $0.153°$ (D) $0.151°$

   ›Reveal solutionSolution

   Angular dispersion of a thin prism is $(\mu_v-\mu_r)A$; substituting gives $0.108°$.

   Concept and Intuition

   A prism disperses white light because refractive index depends on wavelength (violet bends more than red). For a thin prism, the deviation of each colour is $\delta = (\mu-1)A$, and the angular dispersion (spread between violet and red) is the difference of these deviations, which simplifies to $(\mu_v-\mu_r)A$.

   Step-by-Step Solution

   1. Angular dispersion $= \delta_v - \delta_r = (\mu_v-1)A - (\mu_r-1)A = (\mu_v-\mu_r)A$.
   2. $\mu_v - \mu_r = 1.532 - 1.514 = 0.018$. …

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9. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A ray of light incident on a glass plate of refractive index $\sqrt{3}$. If the angle between refracted ray and reflected ray is $90^0$, then the angle of incidence is (A) $30^0$ (B) $45^0$ (C) $60^0$ (D) $90^0$

   ›Reveal solutionSolution

   The condition that reflected and refracted rays are perpendicular defines Brewster's angle, $\tan\theta_B=n$; with $n=\sqrt3$, $\theta_B=60^\circ$.

   Concept and Intuition

   Brewster's angle is the special angle of incidence at which the reflected ray is completely polarized because the reflected and refracted rays are exactly perpendicular to each other. This geometric condition, combined with Snell's law, gives the simple relation $\tan\theta_B=n$ (the refractive index of the second medium relative to the first).

   Step-by-Step Solution

   1. Let $\theta_i$ be the angle of incidence (=angle of reflection) and $\theta_r$ the angle of refraction.
   2. Given: reflected ray ⊥ refracted ray, so $\theta_i+\theta_r=90^\circ\Rightarrow\theta_r=90^\circ-\theta_i$.
   3. Snell's law: $n=\dfrac{\sin\theta_i}{\sin\theta_r}=\dfrac{\sin\theta_i}{\sin(90^\circ-\theta_i)}=\dfrac{\sin\theta_i}{\cos\theta_i}=\tan\theta_i$. …

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10. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.When a light ray is incident on a small angle prism of material of refractive index 1.5, the angle of minimum deviation is $7^0$. If the prism is immersed in a liquid of refractive index 1.2, then the angle of minimum deviation is (A) $10.5^0$ (B) $1.75^0$ (C) $3.5^0$ (D) $14^0$

    ›Reveal solutionSolution

    For a thin prism, deviation depends on the refractive index relative to the surrounding medium; immersing the prism in a liquid of index 1.2 reduces the relative index and hence the deviation to 3.5°. Answer: (C).

    Concept and Intuition

    For a small-angle (thin) prism, the minimum deviation formula simplifies to $\delta = (n_{rel}-1)A$, where $n_{rel}$ is the refractive index of the prism material relative to whatever it is immersed in (not its absolute index). Surrounding the prism with a denser medium (a liquid with index closer to the prism's own index) reduces the relative refractive index, which directly reduces the deviation — this is the same principle behind why a glass lens becomes less powerful underwater.

    Step-by-Step Solution

    1. In air, relative index $= 1.5/1 = 1.5$. Given $\delta_{air}=(1.5-1)A = 0.5A = 7^\circ \Rightarrow A=14^\circ$.
    2. In liquid, relative index $n_{rel} = \dfrac{n_{prism}}{n_{liquid}} = \dfrac{1.5}{1.2}=1.25$. …

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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.A light ray incidents on an equilateral prism made of material of refractive index $\sqrt{3}$. Inside the prism, if the light ray moves parallel to the base of the prism, then the angle of incidence of the light ray is (A) $30°$ (B) $45°$ (C) $75°$ (D) $60°$

    ›Reveal solutionSolution

    A ray travelling parallel to the base inside a prism is in the symmetric path, giving $r_1=r_2=A/2$; applying Snell's law gives an angle of incidence of $60°$.

    Concept and Intuition

    For a prism, the ray path becomes symmetric about the prism's axis (i.e. $r_1=r_2$ and $i=e$) precisely when it travels parallel to the base inside the prism — this is exactly the condition for minimum deviation. For an equilateral prism ($A=60°$), symmetry gives $r_1=r_2=A/2=30°$, and Snell's law at the first surface then fixes the incidence angle.

    Step-by-Step Solution

    1. Equilateral prism: apex angle $A=60°$.
    2. Ray parallel to base inside $\Rightarrow$ symmetric path $\Rightarrow r_1=r_2=\dfrac{A}{2}=30°$. …

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.A ray of light incidents at an angle of 60° on the first face of a prism. The angle of the prism is 30° and its second face is silvered. If the light ray inside the prism retraces its path after reflection from the second face, then the refractive index of the material of the prism is (A) $\dfrac{2}{\sqrt{3}}$ (B) $\dfrac{3}{2}$ (C) $\sqrt{2}$ (D) $\sqrt{3}$

    ›Reveal solutionSolution

    This tests the "auto-collimation" condition for a silvered prism (the ray hits the silvered face normally to retrace its path); the answer is (D) $\sqrt{3}$.

    Concept and Intuition

    When a prism has its second face silvered and a ray retraces its exact original path after reflection, the physical requirement is that the ray must hit the mirrored surface perpendicularly (angle of incidence at that face = 0°), because only a normal-incidence reflection sends the ray straight back the way it came. This converts an otherwise two-refraction-and-reflection problem into a simple one-face refraction problem, since the geometry inside the prism is now fixed by $r_1 + r_2 = A$ with $r_2 = 0$.

    Step-by-Step Solution

    1. Let $r_1$ = angle of refraction at the first face, $r_2$ = angle of incidence at the second (silvered) face. Prism geometry gives $r_1 + r_2 = A = 30°$.
    2. For the ray to retrace its path after reflecting from the silvered face, it must strike that face normally: $r_2 = 0°$. …

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