Q.You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is $90^{\circ}$. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A flag on a boat at rest is fluttering in the south-east direction when the wind is blowing at a speed of $72$ kmph. If the boat starts moving towards south with a speed of $36\sqrt{2}$ kmph, then the direction of the flag on the boat is (A) South (B) West (C) North (D) East

   ›Reveal solutionSolution

   The flag points opposite to the relative wind direction. Initially the wind is from the northwest (southeast flutter). When the boat moves south, the relative wind shifts to the west, so the flag points east. The correct option is (D).

   Concept & Intuition

   A flag flutters in the direction the wind is blowing toward — that is, opposite to the direction the wind is coming from. When the boat is at rest, the flag flutters southeast, meaning the wind is blowing toward the southeast. Therefore the wind is coming from the northwest.

   When the boat moves, the relative wind (wind relative to the boat) changes. The flag will flutter in the direction of this relative wind. So we need to find the vector sum: (wind velocity relative to ground) minus (boat velocity relative to ground). That gives the wind velocity as experienced by the boat.

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   Step-by-step solution

   1. Interpret the initial condition

      • Flag flutters southeast → wind blows toward southeast.
      • So wind velocity vector $\vec{v}_w$ points southeast.
      • Southeast is exactly halfway between south and east, so its direction is $45^\circ$ south of east (or equivalently $45^\circ$ east of south).
      • Speed: $72$ km/h.

   2. Write wind velocity in components

      Take east as positive $x$, north as positive $y$.

      Southeast means: east component positive, south component negative.

   $\vec{v}_w = 72 \left( \cos 45^\circ \,\hat{i} - \sin 45^\circ \,\hat{j} \right) = 72 \left( \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \right) = 36\sqrt{2} \, \hat{i} - 36\sqrt{2} \, \hat{j} \quad \text{(km/h)}.$

   3. Boat velocity Boat moves south at $36\sqrt{2}$ km/h. South is negative $y$ direction:

   $\vec{v}_b = -36\sqrt{2} \, \hat{j}.$

   4. Relative wind velocity (as seen from the boat) The relative wind is the wind velocity minus the boat’s velocity:

   $\vec{v}_{\text{rel}} = \vec{v}_w - \vec{v}_b.$

   Substitute:

   $$$$

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.When an object is placed at a distance of 40 cm from a convex lens, a real image is formed at a distance V from the lens. If the convex lens is replaced with a concave lens of same focal length, then the change in the position of the image is (A) $\dfrac{V(V-40)}{(V-20)}$ cm (B) $\dfrac{V(V+40)}{(V-20)}$ cm (C) $\dfrac{V(V-40)}{(V+20)}$ cm (D) $\dfrac{V(V+40)}{(V+20)}$ cm

   ›Reveal solutionSolution

   Find $f$ of the convex lens from the given real image, then locate the image made by a concave lens of the same $|f|$; the shift is $\dfrac{V(V+40)}{(V+20)}$ cm (D).

   Step 1 — Focal length of the convex lens.

   With $u=-40$ cm and image distance $v=+V$ (real):

   $\frac{1}{V}-\frac{1}{-40}=\frac{1}{f}\;\Rightarrow\;\frac{1}{f}=\frac{40+V}{40V}\;\Rightarrow\;f=\frac{40V}{40+V}$

   Step 2 — Image with the concave lens.

   A concave lens of the same focal length has $f'=-f=-\dfrac{40V}{40+V}$. With the same object ($u=-40$):

   $\frac{1}{v'}=\frac{1}{f'}+\frac{1}{u}=-\frac{40+V}{40V}-\frac{1}{40}=-\frac{40+2V}{40V}$

   $v'=-\frac{40V}{40+2V}=-\frac{20V}{20+V}\quad(\text{virtual, on the object side})$ …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.In hydrogen spectrum, if the longest wavelength of the spectral line in Balmer series is $\lambda$, then the shortest wavelength of the spectral line in Paschen series is (A) $5\lambda$ (B) $3.75\lambda$ (C) $2.5\lambda$ (D) $1.25\lambda$

   ›Reveal solutionSolution

   The key idea is to relate the longest Balmer wavelength (transition $3\to2$) to the shortest Paschen wavelength (transition $\infty\to3$) using the Rydberg formula. The shortest Paschen wavelength is $\frac{5}{4}\lambda = 1.25\lambda$, so option (D) is correct.

   The Balmer series corresponds to transitions ending at $n=2$, and the Paschen series to transitions ending at $n=3$. The longest wavelength in any series comes from the smallest energy gap — that is, the transition from the next higher level to the series limit. For Balmer, that's $3\to2$. The shortest wavelength in any series comes from the largest energy gap — the transition from infinity (ionization) to the series limit. For Paschen, that's $\infty\to3$.

   The Rydberg formula gives the wavenumber (inverse wavelength) for hydrogen:

   $\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

   where $R$ is the Rydberg constant. We'll apply it to both cases and relate the two wavelengths.

   1. Longest wavelength in Balmer series ($n_f = 2$, $n_i = 3$):

   $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36}$

   So $\lambda = \frac{36}{5R}$.

   2. Shortest wavelength in Paschen series ($n_f = 3$, $n_i = \infty$):

   $\frac{1}{\lambda_{\text{Paschen,min}}} = R\left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = R\left(\frac{1}{9} - 0\right) = \frac{R}{9}$

   Hence $\lambda_{\text{Paschen,min}} = \frac{9}{R}$.

   3. Find the ratio: …

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If the height of a transmitting antenna is 45 m and the height of the receiving antenna is H, then the maximum line-of-sight distance between the two antennas is 1% of the radius of the earth. If the radius of the earth is 6400 km, then the value of H is (A) 125 m (B) 75 m (C) 90 m (D) 150 m

   ›Reveal solutionSolution

   The maximum line-of-sight distance between two antennas is limited by the Earth's curvature. By using the formula $d_M = \sqrt{2R_E h_T} + \sqrt{2R_E h_R}$ and the given total distance, the height of the receiving antenna is found to be $\boxed{125 \text{ m}}$.

   The ability to communicate directly between two antennas, known as line-of-sight (LOS) communication, is limited by the curvature of the Earth. Even if there are no physical obstructions, the signal cannot travel infinitely far because the Earth curves away from the direct path. The maximum distance a signal can travel from an antenna before it dips below the horizon is determined by the antenna's height and the Earth's radius.

   Consider an antenna of height $h$ above the Earth's surface. A signal transmitted from this antenna travels in a straight line, tangent to the Earth's surface at the maximum line-of-sight point.

   Let $R_E$ be the radius of the Earth and $d$ be the maximum line-of-sight distance. We can form a right-angled triangle with vertices at the center of the Earth, the base of the antenna (or the point of tangency on the Earth's surface), and the top of the antenna.

   ›Proof

   Derivation of maximum line-of-sight distance for a single antenna

   Let $O$ be the center of the Earth, $A$ be the top of the antenna, and $T$ be the point on the Earth's surface where the line of sight from $A$ is tangent to the Earth.

   The radius $OT$ is perpendicular to the tangent line $AT$.

   In the right-angled triangle $\triangle OAT$:

   $OA = R_E + h$ (radius of Earth plus antenna height)

   $OT = R_E$ (radius of Earth)

   $AT = d$ (maximum line-of-sight distance)

   By the Pythagorean theorem:

   $(R_E + h)^2 = R_E^2 + d^2$

   Expanding the left side:

   $R_E^2 + 2R_E h + h^2 = R_E^2 + d^2$

   Subtracting $R_E^2$ from both sides:

   $2R_E h + h^2 = d^2$

   Since the antenna height $h$ is typically much smaller than the Earth's radius $R_E$ ($h \ll R_E$), the term $h^2$ is negligible compared to $2R_E h$.

   Therefore, we can approximate:

   $d^2 \approx 2R_E h$

   $d \approx \sqrt{2R_E h}$

   When there are two antennas, one transmitting at height $h_T$ and one receiving at height $h_R$, the total maximum line-of-sight distance ($d_M$) between them is the sum of the individual maximum line-of-sight distances from each antenna.

   The maximum line-of-sight distance $d_M$ between two antennas of heights $h_T$ and $h_R$ is given by:

   $d_M = \sqrt{2R_E h_T} + \sqrt{2R_E h_R}$

   where $R_E$ is the radius of the Earth.

   Now, let's apply this concept to solve the problem.

   1. Identify given values and ensure consistent units:

      • Height of transmitting antenna, $h_T = 45 \text{ m}$.
      • Height of receiving antenna, $h_R = H$.
      • Radius of the Earth, $R_E = 6400 \text{ km} = 6400 \times 10^3 \text{ m} = 6.4 \times 10^6 \text{ m}$.
      • Maximum line-of-sight distance between the two antennas, $d_M = 1\% \text{ of } R_E$. $d_M = 0.01 \times 6400 \text{ km} = 64 \text{ km} = 64 \times 10^3 \text{ m}$.

   2. Calculate the maximum line-of-sight distance from the transmitting antenna ($d_T$):

      Using the formula $d = \sqrt{2R_E h}$: …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If A, B and C are three different physical quantities with different dimensional formulae, then the combination which can never give a proper physical quantity is (A) $\dfrac{A}{BC}$ (B) $\dfrac{AB - C^2}{BC}$ (C) $\dfrac{A - C}{B}$ (D) $AC - B$

   ›Reveal solutionSolution

   The key idea is that only terms with identical dimensions can be added or subtracted; the combination that forces an impossible subtraction of unlike dimensions is the one that can never represent a proper physical quantity. The answer is option (C).

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   Concept and Intuition

   In physics, when we write an equation that represents a real quantity, every term that is added to or subtracted from another must have the same dimensions. This is the principle of dimensional homogeneity. Multiplication and division, however, can combine different dimensions freely (e.g., speed = distance/time). So the danger lies in expressions that contain a sum or difference of terms with different dimensional formulae. If such a sum/difference appears and cannot be simplified away, the expression is dimensionally illegal and cannot represent a proper physical quantity.

   Here, A, B, and C all have different dimensional formulae. So any expression that forces us to add or subtract A, B, or C directly (without them first being multiplied/divided into the same dimensions) is invalid.

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   Step-by-step analysis

   1. Option (A): $\dfrac{A}{BC}$

      This is a pure product/quotient. No addition or subtraction. The dimensions are simply $[A]/([B][C])$, which is some valid combination. Possible.

   2. Option (B): $\dfrac{AB - C^2}{BC}$

      The numerator is $AB - C^2$. For this subtraction to be allowed, $AB$ and $C^2$ must have the same dimensions.

      • $AB$ has dimensions $[A][B]$.
      • $C^2$ has dimensions $[C]^2$. Since A, B, C all have different dimensional formulae, it is possible that $[A][B] = [C]^2$ (e.g., if A is length, B is time, C is sqrt(length×time) — though unusual, dimensionally it could match). The problem only says they are different, not that they are dimensionally unrelated in products. So this equality is not ruled out. Hence this expression could be valid for some specific choices of A, B, C. Not necessarily impossible.

   3. Option (C): $\dfrac{A - C}{B}$ …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The angle of a prism made of a material of refractive index $\sqrt{2}$ is $90^\circ$. The angle of incidence for a light ray on the first face of the prism such that the light ray suffers total internal reflection at the second face is (A) $0^\circ$ (B) $90^\circ$ (C) $60^\circ$ (D) $45^\circ$

   ›Reveal solutionSolution

   The critical angle is $45^\circ$; for a $90^\circ$ prism the ray just reaches total internal reflection when $r_2 = 45^\circ$, forcing $r_1 = 45^\circ$ and hence a grazing incidence $i = 90^\circ$ — option (B).

   ---

   Concept

   For refractive index $\sqrt{2}$, the critical angle satisfies $\sin C = \dfrac{1}{\sqrt{2}}$, so $C = 45^\circ$. In a prism, $r_1 + r_2 = A$. Total internal reflection at the second face requires $r_2 \ge C$; the limiting ray has $r_2 = C = 45^\circ$.

   Solution

   Prism angle $A = 90^\circ = r_1 + r_2$. With $r_2 = 45^\circ$: …

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If you are using eye glasses of power 2D, your near point is (A) 25 cm (B) 50 cm (C) 43 cm (D) 32 cm

   ›Reveal solutionSolution

   The power of a lens is the reciprocal of its focal length in metres. A +2 D lens has a focal length of 50 cm, which becomes the corrected near point for a hypermetropic eye. The correct option is (B).

   The key concept here is corrective lenses for hypermetropia (farsightedness). A person with hypermetropia cannot see nearby objects clearly because the eye’s lens focuses the image behind the retina. The near point (the closest distance at which the eye can see clearly) is farther than the normal 25 cm. A convex lens of appropriate power is used to bring the image of an object placed at 25 cm to the person’s actual near point. The power of the lens tells us its focal length, and that focal length is the corrected near point when the lens is placed close to the eye (as in glasses).

   Let’s work through it:

   1. Interpret the given power. The power of the glasses is $P = +2 \, \text{D}$ (the “+” indicates a convex lens, used for hypermetropia). The focal length $f$ in metres is given by

   $f = \frac{1}{P} = \frac{1}{2} = 0.5 \, \text{m} = 50 \, \text{cm}.$

   2. Understand what this focal length means for the near point. …

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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The Brewster angle for air to glass transition of light is (Refractive index of glass = 1.5) (A) $\sin^{-1}\left(\dfrac{3}{2}\right)$ (B) $\cos^{-1}\left(\dfrac{3}{2}\right)$ (C) $\tan^{-1}\left(\dfrac{3}{2}\right)$ (D) $\cos^{-1}\left(\dfrac{2}{3}\right)$

   ›Reveal solutionSolution

   The Brewster angle is the angle of incidence at which reflected light is completely polarized, given by $\tan \theta_B = n_2/n_1$. For air ($n_1=1$) to glass ($n_2=1.5$), $\theta_B = \tan^{-1}(1.5) = \tan^{-1}(3/2)$, so the correct option is (C).

   Concept & Intuition

   The Brewster angle (also called the polarizing angle) is the special angle of incidence where light reflecting off a surface becomes perfectly polarized perpendicular to the plane of incidence. This happens because at that angle, the reflected and refracted rays are perpendicular to each other. Using Snell’s law and the condition $90^\circ$ between reflected and transmitted rays, we derive $\tan \theta_B = n_2/n_1$. No sine or cosine appears — only tangent. A common mistake is to confuse Brewster’s law with Snell’s law or critical angle formulas.

   Step-by-step reasoning

   1. Recall Brewster’s law When unpolarized light strikes a boundary between two media, the reflected light is completely polarized if the angle of incidence $\theta_B$ satisfies

   $\tan \theta_B = \frac{n_2}{n_1}$

   where $n_1$ is the refractive index of the incident medium and $n_2$ of the transmitting medium.

   2. Identify the media Here light goes from air ($n_1 = 1$) to glass ($n_2 = 1.5$). So

   $\tan \theta_B = \frac{1.5}{1} = 1.5 = \frac{3}{2}.$

   3. Solve for the angle Taking the inverse tangent of both sides:

   $\theta_B = \tan^{-1}\left(\frac{3}{2}\right).$

   4. Match with the options …

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9. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
   Q.Arrange the following electromagnetic waves in the order of increasing frequency A. Microwaves B. Infrared waves C. Ultraviolet rays D. X-rays (A) A, B, C, D (B) D, C, B, A (C) B, C, A, D (D) D, A, C, B

   ›Reveal solutionSolution

   The electromagnetic spectrum is ordered by frequency (or equivalently, by energy): radio/microwaves are lowest, then infrared, visible, ultraviolet, X-rays, and gamma rays are highest. Arranging the given waves in increasing frequency gives A, B, C, D.

   Why frequency orders the electromagnetic spectrum

   All electromagnetic waves travel at the speed of light $c$ in vacuum, related to frequency $\nu$ and wavelength $\lambda$ by

   $c = \nu \lambda.$

   Higher frequency means shorter wavelength and higher photon energy ($E = h\nu$). The electromagnetic spectrum spans an enormous range—more than 20 orders of magnitude—and different regions interact with matter in characteristic ways. Microwaves excite molecular rotations, infrared excites vibrations, ultraviolet can ionize atoms, and X-rays penetrate deep into materials. These physical distinctions arise directly from the photon energy, which tracks frequency.

   Ordering the four regions

   Let me place each wave type on the spectrum:

   1. Microwaves (A): Frequency roughly $10^9$ to $10^{12}\,\text{Hz}$ (wavelengths from millimeters to centimeters). Used in radar, mobile phones, and microwave ovens. This is the lowest-frequency region among the four.

   2. Infrared waves (B): Frequency roughly $10^{12}$ to $4 \times 10^{14}\,\text{Hz}$ (wavelengths from about $750\,\text{nm}$ to a few millimeters). Felt as heat; emitted by warm objects. Sits just below visible light. …

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10. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
    Q.The figure shows the variation of photocurrent (I) with anode potential (V) for a photo sensitive surfaces for three different radiations. If $I_a$, $I_b$ and $I_c$ are the intensities and $v_a$, $v_b$, and $v_c$ are the frequencies of lights for the curves a, b and c respectively, then (A) $v_a = v_b$ and $I_a \neq I_b$ (B) $v_a = v_b$ and $I_a = I_b$ (C) $v_a = v_c$ and $I_a = I_c$ (D) $v_b = v_c$ and $I_b = I_c$

    ›Reveal solutionSolution

    The stopping potential depends only on frequency, while the saturation current depends only on intensity. Curves with the same stopping potential have equal frequency; curves with the same saturation current have equal intensity.

    The photoelectric effect gives us a clean, direct relationship between the graph of photocurrent vs. anode potential and the properties of the incident light. The key is to read two features from each curve: the stopping potential (where the current just becomes zero) and the saturation current (the flat, maximum current at large positive voltage).

    The stopping potential $V_0$ is related to the frequency $v$ of the incident light by Einstein’s photoelectric equation:

    $eV_0 = h v - \phi$

    where $\phi$ is the work function of the metal surface (same for all three curves here). So a larger stopping potential means a higher frequency. If two curves have the same stopping potential, they have the same frequency.

    The saturation current is proportional to the number of photoelectrons emitted per second, which is directly proportional to the intensity of the incident light (for a fixed frequency). So a larger saturation current means a higher intensity. If two curves have the same saturation current, they have the same intensity.

    Now look at the figure (described in the problem). Curves a and b cut the voltage axis at the same point — they have the same stopping potential. Therefore $v_a = v_b$. But their saturation currents are different: curve a saturates at a higher current than curve b. Therefore $I_a \neq I_b$.

    Curve c has a different (more negative) stopping potential, so its frequency is different from both a and b. Its saturation current happens to match that of curve a, but that is not one of the options that fits all the data. …

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11. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.Critical angle of light passing from glass to air is minimum for light with wavelength (A) 670 nm (B) 550 nm (C) 590 nm (D) 460 nm

    ›Reveal solutionSolution

    The critical angle depends on the refractive index, which varies with wavelength due to dispersion; since the refractive index is highest for shorter wavelengths, the critical angle is smallest for the shortest wavelength, 460 nm.

    The key concept here is dispersion — the variation of a material's refractive index with the wavelength of light. For most transparent materials like glass, the refractive index is higher for shorter wavelengths (blue light) and lower for longer wavelengths (red light). This is why a prism splits white light into a rainbow.

    The critical angle $\theta_c$ for light going from a medium (glass) to a less dense medium (air) is given by:

    $\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{glass}}}$

    Since $n_{\text{air}} \approx 1$, we have:

    $\theta_c = \sin^{-1}\left( \frac{1}{n_{\text{glass}}} \right)$

    Thus, a larger refractive index $n_{\text{glass}}$ gives a smaller critical angle (because the inverse sine of a larger fraction is smaller). So to find the wavelength that gives the minimum critical angle, we need the wavelength for which glass has the maximum refractive index.

    1. Recall the dispersion trend: In normal dispersion (which applies to glass in the visible range), the refractive index decreases as wavelength increases. So:

       • Short wavelength (violet/blue) → high $n$
       • Long wavelength (red) → low $n$

    2. List the given wavelengths:

       • (A) 670 nm (red)
       • (B) 550 nm (green)
       • (C) 590 nm (yellow)
       • (D) 460 nm (blue)

    3. Identify the shortest wavelength: Among these, 460 nm is the shortest (blue region). Therefore, it corresponds to the highest refractive index for glass. …

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12. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.If the frequency of the electromagnetic wave is $3 \times 10^{17}$ Hz, then it corresponds to which part of electromagnetic spectrum (A) Visible (B) X-ray (C) Microwave (D) Gamma ray

    ›Reveal solutionSolution

    The frequency $3 \times 10^{17}$ Hz lies in the X-ray region of the electromagnetic spectrum, so the correct option is (B).

    The electromagnetic spectrum is arranged by frequency (or wavelength). Higher frequency means shorter wavelength and more energetic photons. The key is to recall the approximate frequency ranges for each region: visible light is around $10^{14}$–$10^{15}$ Hz, X-rays are around $10^{16}$–$10^{19}$ Hz, gamma rays are above $10^{19}$ Hz, and microwaves are much lower (around $10^9$–$10^{12}$ Hz). Given $3 \times 10^{17}$ Hz, we can directly compare.

    1. Recall the frequency ranges

       • Microwaves: $10^9$ to $10^{12}$ Hz
       • Visible light: $4 \times 10^{14}$ to $7.5 \times 10^{14}$ Hz
       • X-rays: $10^{16}$ to $10^{19}$ Hz
       • Gamma rays: above $10^{19}$ Hz

    2. Locate the given frequency

       $3 \times 10^{17}$ Hz is between $10^{16}$ and $10^{19}$ Hz, squarely in the X-ray band. It is far above visible light (which is $\sim 10^{14}$ Hz) and below typical gamma rays (which start around $10^{19}$–$10^{20}$ Hz).

    3. Eliminate other options

       • (A) Visible: too low by a factor of about 1000.
       • (C) Microwave: too low by a factor of about $10^8$.
       • (D) Gamma ray: too high; gamma rays usually begin above $10^{19}$ Hz, though there is some overlap, $3 \times 10^{17}$ Hz is conventionally X-ray. …

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