Q.Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Showing the 12 most recent of 18 on this concept.

1. COMEDK 2025Set 2025-A1 markMCQ
   Q.A beam of light parallel to the principal axis of a concave and convex lens, first passes through the concave lens of focal length 0.5 m and then through the convex lens of focal length 1.75 m . If the lenses are placed 1.25 apart, which of the given statement is true? (A) The emergent beam will focus at 1.75 m on the principal axis. (B) The emergent beam will focus at 1.25 m on the principal axis. (C) The emergent beam will pass parallel to the principal axis. (D) The emergent beam will focus at 0.75 m on the principal axis.

   ›Reveal solutionSolution

   The parallel beam diverges through the concave lens as if from its focus; that virtual image sits exactly at the focus of the convex lens, so the emergent beam leaves parallel to the axis — option (C).

   Concave lens ($f_1=-0.5\ \text{m}$): a beam parallel to the axis forms a virtual image at its focal point,

   $v_1=f_1=-0.5\ \text{m},$

   i.e. $0.5\ \text{m}$ in front of the concave lens (on the incoming side).

   Convex lens ($f_2=+1.75\ \text{m}$), placed $1.25\ \text{m}$ beyond the concave lens: the virtual image acts as the object. Its distance from the convex lens is

   $u=1.25+0.5=1.75\ \text{m}.$

   The object lies exactly at the focal distance of the convex lens ($|u|=f_2=1.75\ \text{m}$). Rays coming from the focal point of a converging lens emerge parallel to the principal axis.

   ✓Final answer

   The emergent beam passes parallel to the principal axis — option (C).

2. COMEDK 2025Set 2025-E1 markMCQ
   Q.A beam of incident parallel light falls on a diverging lens of focal length 20 cm in magnitude. If a converging lens of focal length 15 cm in magnitude is placed at a distance of 10 cm to the right of the diverging lens on the other side, then, the final image formed is: (A) Virtual and is at a distance of 30 cm to the right of the converging lens. (B) Virtual and is at a distance of 40 cm to the right of the diverging lens. (C) Real and is at a distance of 30 cm to the right of the converging lens. (D) Real and is at a distance of 40 cm to the right of the converging lens.

   ›Reveal solutionSolution

   The problem is a two-lens system where parallel light first hits a diverging lens, creating a virtual object for the converging lens. Using the lens formula step by step, the final image is real and located 30 cm to the right of the converging lens, matching option (C).

   Concept & Intuition

   When parallel light (rays from infinity) enters a diverging lens, the lens makes them appear to come from its virtual focus. That virtual image then acts as an object for the second lens. The trick is that if this object lies behind the second lens (on the opposite side from where light is traveling), it’s a virtual object for that lens — and virtual objects can produce real images. We apply the lens formula carefully, keeping sign conventions consistent.

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   1. First lens (diverging, focal length $f_1 = -20$ cm) Parallel light means object distance $u_1 = \infty$. Lens formula: $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$. With $u_1 = \infty$, $\frac{1}{u_1} = 0$, so

   $\frac{1}{v_1} = \frac{1}{f_1} = \frac{1}{-20} \quad \Rightarrow \quad v_1 = -20 \text{ cm}.$

   The negative sign means the image is 20 cm to the left of the diverging lens (virtual, on the same side as the incoming light).

   2. Position relative to the second lens

      The converging lens is 10 cm to the right of the diverging lens.

      So the first image is $20$ cm left of the diverging lens, which is $20 + 10 = 30$ cm left of the converging lens.

      For the converging lens, object distance $u_2 = -30$ cm (negative because the object is on the opposite side from the incoming light — a virtual object).

   3. Second lens (converging, focal length $f_2 = +15$ cm)

      Use the lens formula again:

   $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}.$

   Substitute $u_2 = -30$ cm, $f_2 = +15$ cm:

   $\frac{1}{v_2} - \frac{1}{-30} = \frac{1}{15} \quad \Rightarrow \quad \frac{1}{v_2} + \frac{1}{30} = \frac{1}{15}.$

   $\frac{1}{v_2} = \frac{1}{15} - \frac{1}{30} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30}.$

   So $v_2 = +30$ cm. The positive sign means the final image is real and 30 cm to the right of the converging lens.

   Tip

   A virtual object (negative $u$) often leads to a real image with a converging lens — this is a classic trick in two-lens problems.

   Watch out

   A common mistake is to treat the first image as a real object for the second lens, giving $u_2 = +30$ cm. That would yield a virtual image — but the correct sign is negative because the light from the first image never actually reaches the second lens; it’s a virtual image.

   4. Check the options The final image is real, 30 cm to the right of the converging lens. This matches option (C).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. COMEDK 2025Set 2025-E1 markMCQ
   Q.An object of height $h$ is placed midway between $f$ and $2 f$ in front of a biconvex lens. A real inverted image is captured on a screen placed a little beyond 2 f on the other side of the lens. If the whole arrangement is immersed in water without disturbing the object, lens and the screen positions, then the new image formed will be (A) Magnified virtual erect image on the same side of the lens where the object is placed (B) Magnified real inverted image on the same side of the lens where the object is placed (C) Diminished real inverted image on the other side of the lens between $f$ and $2 f$ (D) Magnified real inverted image on the other side of the lens beyond 2f

   ›Reveal solutionSolution

   In air the object at $1.5f$ gives a real, inverted, magnified image beyond $2f$. Immersing the glass lens in water makes its focal length roughly $4$ times larger, so the fixed object now lies inside the new focal length — producing a magnified, virtual, erect image on the same side as the object — option (A).

   Concept

   The focal length of a biconvex lens depends on the surrounding medium through the lens-maker relation

   $\frac{1}{f}=\left(\frac{n_{\text{lens}}}{n_{\text{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

   Going from air to water sharply reduces the effective refractive-index contrast, so $f$ increases.

   Step-by-step solution

   1. Original situation (in air). Object is midway between $f$ and $2f$, i.e. at $u=1.5f$. Since $f<u<2f$, the image is real, inverted and magnified, formed just beyond $2f$ — consistent with the given screen position.

   2. Focal length in water. Taking $n_{\text{glass}}=1.5$ and $n_{\text{water}}=1.33$,

   $\frac{f_{\text{water}}}{f_{\text{air}}}=\frac{n_{g}-1}{\dfrac{n_{g}}{n_{w}}-1}=\frac{0.5}{\dfrac{1.5}{1.33}-1}\approx\frac{0.5}{0.128}\approx 3.9.$

   So the new focal length is $f_{\text{water}}\approx 3.9\,f$.

   3. New object position relative to the new focus. The object stays fixed at $u=1.5f$, but now $f_{\text{water}}\approx 3.9f > 1.5f$. The object therefore lies inside the focal length of the water-immersed lens.

   4. Nature of the image. For a real object placed within the focal length of a converging lens, the image is magnified, virtual and erect, and lies on the same side as the object. No real image can form on the screen anymore.

   ✓Final answer

   A magnified, virtual, erect image on the same side of the lens as the object. The correct option is (A).

   ANSWER: A

4. COMEDK 2025Set 2025-E1 markMCQ
   Q.A plano-convex lens made of refractive index 1.5 and having radius of curvature $\mathrm{R}=4 \mathrm{~cm}$ fits exactly into a planoconcave lens made of refractive index 1.3 and having the same radius of curvature $R=4 \mathrm{~cm}$ such that their plane surfaces are parallel to each other. The focal length of the combination is : (A) 5 cm (B) 50 cm (C) 2 cm (D) 20 cm

   ›Reveal solutionSolution

   The combination behaves as a single lens whose net power is the sum of the individual lens powers. Using the lens maker’s formula for each lens and adding them gives a focal length of 20 cm, so the correct option is (D).

   We have two lenses placed in contact: a plano-convex lens (refractive index 1.5) and a plano-concave lens (refractive index 1.3), both with the same radius of curvature $R = 4 \, \text{cm}$. Their plane surfaces are parallel and facing each other. The key is to treat the combination as two thin lenses in contact, so the net power is simply the algebraic sum of their individual powers. The lens maker’s formula gives the power of each lens from its geometry and refractive index.

   1. Lens maker’s formula For a thin lens in air, the power $P$ (in diopters if $R$ is in meters) is

   $P = \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

   where $R_1$ and $R_2$ are the radii of curvature of the two surfaces, with sign convention: convex toward the incident light is positive, concave is negative.

   2. First lens: plano-convex (refractive index $n_1 = 1.5$)
      • One surface is plane: $R_1 = \infty$ (radius infinite, so $1/R_1 = 0$).
      • The other surface is convex toward the incident light (say light comes from the plane side first, but the sign is consistent). Typically, for a plano-convex lens, the convex surface has $R_2 = -R$ if the plane is on the incident side? Let’s be careful: We can choose the orientation. Since the two lenses fit together, their curved surfaces are in contact. For the plano-convex lens, the curved surface is convex and faces the other lens. If we take light incident from the plane side, then the first surface is plane ($R_1 = \infty$) and the second surface is convex toward the light, so $R_2 = +R = +4 \, \text{cm}$ (convex surface has positive radius when the center of curvature is on the outgoing side). Then

   $P_1 = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{4} \right) = 0.5 \times \left(0 - \frac{1}{4}\right) = -0.125 \, \text{cm}^{-1}$

    That gives a negative power — but a plano-convex lens is converging! This sign error is a classic pitfall.
   

   Watch out

   The sign convention must be applied consistently: For a lens in air, the standard Cartesian sign convention is:

   • Radius is positive if the center of curvature lies to the right of the surface (for light traveling left to right).
   • For a plano-convex lens with the plane side facing the incident light, the second surface is convex, so its center of curvature is to the right → $R_2 = +R$. But then the formula gives negative power? That can’t be right. Actually, the correct form of the lens maker’s formula is $\frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ with the sign convention: $R$ positive if the surface is convex toward the incident light. For a plano-convex lens with the plane side first, the first surface is plane ($R_1 = \infty$), the second surface is convex toward the incident light? No — the light hits the plane surface first, then the convex surface. The convex surface is curved away from the incident light (its center is on the far side), so it is concave toward the incident light. Therefore $R_2$ should be negative. Let’s redo properly.

   3. Correct sign convention for lens maker’s formula Use the standard: For each surface, $R$ is positive if the center of curvature lies on the side of the outgoing light (i.e., the surface is convex toward the incident light).
      • For the plano-convex lens: Light enters from the plane side.
        • First surface (plane): $R_1 = \infty$.
        • Second surface (curved): The center of curvature is on the outgoing side (to the right), so the surface is convex toward the incident light? Actually, if light is going left to right, and the convex surface bulges to the right, then the incident light sees a convex surface — yes, because the surface curves away from the incoming light. So $R_2 = +R = +4 \, \text{cm}$. Then

   $\frac{1}{f_1} = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{+4} \right) = 0.5 \times (0 - 0.25) = -0.125 \, \text{cm}^{-1}$

    This gives a negative focal length, meaning diverging — but a plano-convex lens is converging. The error: The formula is actually  
   

   $\frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

    where $ R_1 $ is the radius of the first surface (incident side) and $ R_2 $ of the second surface (exit side), with the sign: positive if the center of curvature is on the *right* (for light traveling left to right). For a plano-convex lens with the plane side left, the first surface is plane ($ R_1 = \infty $), the second surface is convex, so its center is to the right → $ R_2 = +4 $. Then the formula gives negative power. That suggests the formula is actually  
   

   $\frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

    but with $ R_1 $ and $ R_2 $ signed such that convex toward the incident light is positive. For a plano-convex lens with the plane side first, the second surface is convex *away* from the incident light, so it is concave toward the incident light → $ R_2 $ should be negative. Let’s adopt that:  
    - For the plano-convex lens: first surface plane ($ R_1 = \infty $), second surface convex away from incident light → $ R_2 = -R = -4 \, \text{cm} $.  
    Then  
   

   $\frac{1}{f_1} = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{-4} \right) = 0.5 \times (0 + 0.25) = +0.125 \, \text{cm}^{-1}$

    So $ f_1 = \frac{1}{0.125} = 8 \, \text{cm} $ (converging). That matches expectation.
   

   4. Second lens: plano-concave (refractive index $n_2 = 1.3$)

   This lens has one plane surface and one concave surface. It fits exactly with the first lens, so its curved surface is concave and mates with the convex surface of the first lens. Light now passes from the first lens into the second. For the second lens, we consider light incident from its curved side (since it’s in contact).

   • First surface (curved, concave toward incident light): center of curvature is on the incident side → $R_1 = -R = -4 \, \text{cm}$.
   • Second surface (plane): $R_2 = \infty$. Then

   $\frac{1}{f_2} = (1.3 - 1)\left( \frac{1}{-4} - \frac{1}{\infty} \right) = 0.3 \times (-0.25 - 0) = -0.075 \, \text{cm}^{-1}$

   So $f_2 = -\frac{1}{0.075} \approx -13.33 \, \text{cm}$ (diverging).

   5. Net power of combination For lenses in contact, the net power is the sum:

   $P_{\text{net}} = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = 0.125 - 0.075 = 0.05 \, \text{cm}^{-1}$

   Therefore the net focal length is

   $f_{\text{net}} = \frac{1}{0.05} = 20 \, \text{cm}.$

   Tip

   Notice that the curved surfaces are in contact and have the same radius but different refractive indices. The net power depends only on the difference of the refractive indices:

   $P_{\text{net}} = (n_1 - n_2)/R$. Here $(1.5 - 1.3)/4 = 0.2/4 = 0.05 \, \text{cm}^{-1}$, giving $f = 20 \, \text{cm}$. This shortcut works because the plane surfaces contribute nothing and the curved surfaces cancel the $(n-1)$ terms partially.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2025Set 2025-M1 markMCQ
   Q.A symmetric double convex lens is cut into two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is 4D, the difference between the powers of the original lens and the cut lens is (A) zero (B) 3D (C) D (D) 2D

   ›Reveal solutionSolution

   When a symmetric double convex lens is cut perpendicular to the principal axis, each half becomes a plano-convex lens with the same radius of curvature on the curved side. The power of each half is half the original power, so the difference between the original power (4D) and the power of one cut lens (2D) is 2D.

   Concept and Intuition

   The power of a lens depends on its curvature and the refractive index of the material. For a symmetric double convex lens, both surfaces have the same radius of curvature $R$. The lens maker’s formula gives the power as $P = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. When you cut the lens perpendicular to the principal axis, you split it into two identical plano-convex lenses. Each half has one flat surface (radius of curvature $\infty$) and one curved surface with the original radius $R$. The flat surface contributes zero power, so the power of each half is exactly half the original power. The question asks for the difference between the original power and the power of one cut lens.

   Step-by-step solution

   1. Original lens power For a symmetric double convex lens, both surfaces have the same radius of curvature $R$ but opposite signs (convex toward the incoming light). Using the lens maker’s formula:

   $P_{\text{original}} = (n-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n-1)\left(\frac{2}{R}\right)$

   Given $P_{\text{original}} = 4\,\text{D}$, we have:

   $(n-1)\frac{2}{R} = 4 \quad \Rightarrow \quad (n-1)\frac{1}{R} = 2$

   2. Power of one cut lens After cutting, each half is a plano-convex lens. The flat side has $R_1 = \infty$ (so $1/R_1 = 0$), and the curved side has $R_2 = R$ (positive if the convex side faces the incoming light). The power is:

   $P_{\text{half}} = (n-1)\left(\frac{1}{\infty} - \frac{1}{R}\right) = (n-1)\left(0 - \frac{1}{R}\right) = -(n-1)\frac{1}{R}$

   The negative sign just indicates the orientation; the magnitude of the power is:

   $|P_{\text{half}}| = (n-1)\frac{1}{R} = 2\,\text{D}$

   So each cut lens has power $2\,\text{D}$.

   3. Difference in powers The problem asks for the difference between the original power and the power of the cut lens. Taking magnitudes:

   $\text{Difference} = P_{\text{original}} - P_{\text{half}} = 4\,\text{D} - 2\,\text{D} = 2\,\text{D}$

   Watch out

   A common mistake is to think the cut lens still has two curved surfaces, but cutting perpendicular to the axis creates a flat surface. Another pitfall is forgetting that the power of a plano-convex lens is half that of an equivalent double convex lens with the same radii.

   Tip

   You can also think of it this way: the original lens is like two plano-convex lenses placed together. Cutting it simply separates them, so each half has half the original power.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. COMEDK 2025Set 2025-M1 markMCQ
   Q.One surface of a lens is convex and the other is concave. If the radii of curvatures are $R$ and $r$ respectively, the lens will be convex if (A) $R >r$ (B) $R< r$ (C) $R=\frac{1}{r}$ (D) $R=r$

   ›Reveal solutionSolution

   The lens is convex when the net curvature is positive; for a convex-concave lens, this happens when the convex side has a smaller radius of curvature than the concave side, i.e., $R < r$. The correct option is (B).

   The key idea is the lens maker’s formula and the sign convention for radii of curvature. A lens is convex (converging) if its net focal length is positive. For a lens with one convex and one concave surface, the sign of the focal length depends on which radius is larger.

   Why this works:

   In the lens maker’s formula, $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, the sign of each radius is taken as positive if the surface is convex toward the incoming light, and negative if concave. For a convex-concave lens, the convex side has $R_1 > 0$ and the concave side has $R_2 < 0$ (if light enters from the convex side). The term $\frac{1}{R_1} - \frac{1}{R_2}$ becomes $\frac{1}{R} - \frac{1}{(-r)} = \frac{1}{R} + \frac{1}{r}$. For the lens to be convex (converging), we need this sum to be positive — which it always is — but the thickness and shape matter: a convex-concave lens is actually convex only if the convex surface is more curved (smaller radius) than the concave surface.

   Let’s work through it step by step.

   1. Set up the sign convention.

      Assume light enters from the convex side. Then the first surface (convex) has radius $R_1 = +R$. The second surface (concave) has radius $R_2 = -r$ (since the center of curvature lies on the opposite side of the incoming light).

   2. Apply the lens maker’s formula.

      For a thin lens in air:

   $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

   Substitute $R_1 = R$, $R_2 = -r$:

   $\frac{1}{f} = (n-1)\left(\frac{1}{R} - \frac{1}{-r}\right) = (n-1)\left(\frac{1}{R} + \frac{1}{r}\right)$

   3. Interpret the sign of $f$.

      Since $n>1$ and both $R$ and $r$ are positive lengths, $\frac{1}{R} + \frac{1}{r} > 0$, so $f > 0$. This suggests the lens is always converging — but that’s misleading. The formula assumes the lens is thin and the radii are measured from the same direction. In reality, a convex-concave lens can be either converging or diverging depending on which surface is more curved.

   4. Think physically: which side bulges more?

      A lens is called “convex” if it is thicker at the center than at the edges. For a lens with one convex and one concave surface, the center thickness depends on the curvatures.

      • If the convex side is more curved (smaller radius $R$) than the concave side (larger radius $r$), the center bulges out — the lens is convex (converging).
      • If the convex side is less curved (larger $R$) than the concave side (smaller $r$), the center is thinner — the lens is actually concave (diverging), despite having one convex surface.

   5. Translate to inequality.

      Convex lens condition: convex surface more curved → smaller radius → $R < r$.

      So the lens is convex when $R < r$.

   Watch out

   A common mistake is to think that because one surface is convex, the lens is automatically convex. But a convex-concave lens can be diverging if the concave side is more curved. Always compare the radii.

   Tip

   A neat shortcut: For a convex-concave lens, the lens is converging (convex) if the convex side has the smaller radius. Remember: “Convex wins when its curve is tighter.”

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. KCET 2024Set D-21 markMCQ
   Q.The final image formed by an astronomical telescope is (A) real, erect and diminished (B) virtual, inverted and diminished (C) real, inverted and magnified (D) virtual, inverted and magnified

   ›Reveal solutionSolution

   Objective ⇒ real inverted image; eyepiece used as a magnifier ⇒ that image is re-imaged as a virtual, magnified image, still inverted with respect to the object.

   Step 1 — Stage 1: the objective

   The object (a star, a planet) is effectively at infinity. The objective is a converging lens of large focal length $f_o$, so it forms a real, inverted, diminished image $A'B'$ in its focal plane — i.e. at a distance $f_o$ behind the objective.

   This intermediate image is real, and that is essential: it becomes the object for the eyepiece.

   Step 2 — Stage 2: the eyepiece

   The eyepiece is a converging lens of short focal length $f_e$, placed so that the intermediate image $A'B'$ lies at or just inside its first focal point. A converging lens with an object inside its focal length behaves as a simple microscope (magnifying glass), and such a lens always produces a virtual, erect, magnified image of its own object.

   • "Erect with respect to $A'B'$" — but $A'B'$ was itself inverted with respect to the original object.
   • Hence the final image is inverted with respect to the original object.

   Step 3 — Why "magnified"

   For a telescope, magnification means angular magnification, not linear size. In normal adjustment,

   $m=-\frac{f_o}{f_e}$

   Since $f_o\gg f_e$ for a telescope, $|m|\gg1$: the object subtends a much larger angle at the eye than it did unaided. So the final image is magnified, and the minus sign again records the inversion.

   Step 4 — Assemble the three attributes

   Attribute	Value	Reason
   Real/virtual	Virtual	The eyepiece works as a magnifier — rays only appear to diverge from the final image; it cannot be caught on a screen
   Erect/inverted	Inverted	Objective inverts; eyepiece does not re-invert
   Size	Magnified	$

   This inversion is why an astronomical telescope is unsuitable for terrestrial use (stars look the same either way up) — a terrestrial telescope adds an erecting lens.

   ✓Final answer

   The correct option is (D) — virtual, inverted and magnified.

   ANSWER: D

8. COMEDK 2024Set 2024-A1 markMCQ
   Q.An object is placed at a distance of $12 \mathrm{~cm}$ from a convex lens on its principal axis and a virtual image of certain size is formed. If the object is moved $4 \mathrm{~cm}$ away from the lens, a real image of the same size as that of the virtual image is formed. The focal length of the lens in $\mathrm{cm}$ is (A) 12 (B) 14 (C) 18 (D) 16

   ›Reveal solutionSolution

   The problem uses the lens formula and magnification to relate two object positions that produce images of equal size (one virtual, one real). Solving the resulting equations gives the focal length as 16 cm.

   The key insight is that the magnitude of magnification is the same in both cases, but the sign differs because one image is virtual (upright, positive magnification) and the other is real (inverted, negative magnification). The lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (with sign convention: object distance $u$ is negative for real objects) lets us express image distances in terms of $u$ and $f$, then equate the absolute magnifications.

   1. Set up the sign convention and first case. For a convex lens, we use the Cartesian sign convention: object distance $u$ is negative. Let the focal length be $f$ (positive for convex). First object distance: $u_1 = -12$ cm. The image is virtual, so $v_1$ is negative. Lens formula:

   $\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{v_1} - \frac{1}{-12} = \frac{1}{v_1} + \frac{1}{12}.$

   Magnification $m_1 = \frac{v_1}{u_1} = \frac{v_1}{-12}$. Since the image is virtual and upright, $m_1 > 0$, so $v_1$ is negative (consistent). The size (absolute value) is $|m_1| = \frac{|v_1|}{12}$.

   2. Second case. Object moved 4 cm away: $u_2 = -(12+4) = -16$ cm. Now a real image is formed, so $v_2 > 0$. Lens formula:

   $\frac{1}{f} = \frac{1}{v_2} - \frac{1}{-16} = \frac{1}{v_2} + \frac{1}{16}.$

   Magnification $m_2 = \frac{v_2}{u_2} = \frac{v_2}{-16}$, which is negative (real, inverted). The absolute size is $|m_2| = \frac{v_2}{16}$.

   3. Equal image size condition. The problem says the size (absolute value) is the same:

   $|m_1| = |m_2| \quad \Rightarrow \quad \frac{|v_1|}{12} = \frac{v_2}{16}.$

   Since $v_1$ is negative, $|v_1| = -v_1$. So:

   $\frac{-v_1}{12} = \frac{v_2}{16} \quad \Rightarrow \quad v_2 = -\frac{4}{3} v_1. \qquad(1)$

   4. Use the lens formula for both cases. From case 1:

   $\frac{1}{f} = \frac{1}{v_1} + \frac{1}{12}. \qquad(2)$

   From case 2:

   $\frac{1}{f} = \frac{1}{v_2} + \frac{1}{16}. \qquad(3)$

   Equate (2) and (3):

   $\frac{1}{v_1} + \frac{1}{12} = \frac{1}{v_2} + \frac{1}{16}.$

   Substitute $v_2$ from (1):

   $\frac{1}{v_1} + \frac{1}{12} = \frac{1}{-\frac{4}{3}v_1} + \frac{1}{16} = -\frac{3}{4v_1} + \frac{1}{16}.$

   Bring terms with $v_1$ together:

   $\frac{1}{v_1} + \frac{3}{4v_1} = \frac{1}{16} - \frac{1}{12}.$

   Left side: $\frac{4}{4v_1} + \frac{3}{4v_1} = \frac{7}{4v_1}$.

   Right side: common denominator 48: $\frac{3}{48} - \frac{4}{48} = -\frac{1}{48}$.

   So:

   $\frac{7}{4v_1} = -\frac{1}{48} \quad \Rightarrow \quad v_1 = -\frac{7 \times 48}{4} = -84 \text{ cm}.$

   5. Find focal length. Use (2):

   $\frac{1}{f} = \frac{1}{-84} + \frac{1}{12} = -\frac{1}{84} + \frac{7}{84} = \frac{6}{84} = \frac{1}{14}.$

   Hence $f = 14$ cm.

   Watch out

   A common mistake is to forget the sign of $v_1$ when equating magnifications. Since the virtual image gives a positive magnification but $v_1$ is negative, you must use absolute values correctly.

   Tip

   Notice that the two object distances (12 cm and 16 cm) are symmetric around 14 cm? That’s not a coincidence — when object distances are $f \pm d$, the magnifications are reciprocals in magnitude. Here $f = 14$ fits perfectly.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. COMEDK 2024Set 2024-A1 markMCQ
   Q.To a fish under water, viewing obliquely, a fisherman standing on the bank of a lake looks (A) Same height as he actually is (B) Shorter than what he actually is (C) Taller than what he actually is (D) Taller or shorter depending on the refractive index of water

   ›Reveal solutionSolution

   When light travels from air (rarer) into water (denser), the apparent position of an object is shifted upward. For a fisherman on the bank viewed from underwater, his image appears taller than his actual height because the rays from his feet and head bend differently, stretching the perceived vertical size.

   The key concept here is apparent depth and apparent height due to refraction. When you view an object from a denser medium (water) into a rarer medium (air), the light rays bend away from the normal as they enter the water. This makes the object appear farther away and larger in the vertical direction. The classic example is a stick partly immersed in water appearing bent — but here, the observer is underwater, looking up at the fisherman.

   Let’s work through the reasoning step by step.

   1. Set up the geometry

      Imagine the fisherman standing on the bank, with his feet at the water’s edge and his head above. An underwater fish looks obliquely upward at him. Light from the fisherman’s head and feet travels from air (refractive index $n_1 \approx 1$) into water (refractive index $n_2 \approx 1.33$). At the water surface, the rays bend toward the normal (since they go from rarer to denser medium).

   2. Apply Snell’s law for a small angle approximation

      For a ray coming from a point on the fisherman at height $h$ above the water, the angle of incidence in air is $i$ and the angle of refraction in water is $r$, with $n_1 \sin i = n_2 \sin r$. For small angles (oblique but not extreme), $\sin \theta \approx \tan \theta \approx \theta$. The apparent height $h'$ as seen from underwater is given by the formula for apparent depth in reverse:

   $\text{Apparent height} = \frac{n_2}{n_1} \times \text{real height}$

   Since $n_2 > n_1$, the apparent height is larger than the real height.

   3. Why this formula works Consider a point at real height $h$ above the water. The ray from that point to the fish’s eye makes an angle $i$ in air and $r$ in water. The apparent position is where the backward extension of the refracted ray in water meets the vertical line through the object. Using geometry and Snell’s law, the apparent height $h'$ satisfies

   $h' \tan r = h \tan i$

   For small angles, $\tan i \approx i$ and $\tan r \approx r$, and Snell’s law gives $i = n_2 r / n_1$. Substituting yields

   $h' \cdot r = h \cdot \frac{n_2}{n_1} r \quad \Rightarrow \quad h' = \frac{n_2}{n_1} h$

   So the fisherman appears taller by a factor of about 1.33.

   4. Consider the whole body

      Both the head and feet are shifted upward by the same factor, but because the feet are at the water surface (real height zero) and the head is at a positive height, the entire image is stretched vertically. The fish sees the fisherman’s head higher than it really is, while his feet appear at the same level (since they are at the water surface). This makes him look taller.

   5. Check the options

      • (A) Same height: No, refraction changes apparent size.
      • (B) Shorter: No, the factor is >1.
      • (C) Taller: Yes, as derived.
      • (D) Depends on refractive index: While the exact factor depends on $n$, for any $n>1$ the fisherman appears taller, not shorter. So the qualitative answer is fixed.

   Watch out

   A common mistake is to think the fisherman appears shorter because water magnifies objects. But magnification in water usually makes objects appear larger and closer when viewed from above; from below, the effect is opposite for depth but same for height — the vertical dimension is stretched.

   Tip

   You can remember this by thinking of a straw in a glass of water: from above, the straw appears bent upward (the submerged part looks shallower). If you were a fish looking at the part above water, that part would appear higher and thus the straw longer.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. COMEDK 2024Set 2024-A1 markMCQ
    Q.In a container of height $21 \mathrm{~cm}$, certain transparent liquid is taken to a height of $12 \mathrm{~cm}$. When seen from above, it appears half filled. The refractive index of the liquid is (A) $\frac{7}{6}$ (B) $\frac{5}{4}$ (C) $\frac{4}{3}$ (D) $\frac{3}{2}$

    ›Reveal solutionSolution

    The liquid's real depth of 12 cm appears reduced to the 9 cm empty column above it, so $n = \dfrac{\text{real depth}}{\text{apparent depth}} = \dfrac{12}{9} = \dfrac{4}{3}$ — option (C).

    The container is 21 cm tall with liquid to a real depth of 12 cm, leaving a $21 - 12 = 9$ cm air column above. Viewing from above raises the apparent bottom, so the liquid's apparent depth is

    $d_{\text{app}} = \frac{d_{\text{real}}}{n}.$

    "Appears half filled" means the liquid's apparent depth equals the 9 cm air column above it, so the liquid appears to span the same amount as the empty part:

    $\frac{12}{n} = 9 \;\Rightarrow\; n = \frac{12}{9} = \frac{4}{3}.$

    Check: apparent depth $= 12 \div \tfrac{4}{3} = 9$ cm, matching the 9 cm air gap.

    ✓Final answer

    Refractive index $n = \dfrac{4}{3}$ — option (C).

11. COMEDK 2024Set 2024-E1 markMCQ
    Q.Figure below shows a lens of refractive index, $\mu=1.4$. $C_1$ and $C_2$ are the centres of curvature of the two faces of the lens of radii of curvature $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$ respectively. The lens behaves as a (A) diverging lens of focal length $20 \mathrm{~cm}$ (B) converging lens of focal length $20 \mathrm{~cm}$ (C) converging lens of focal length $12 \mathrm{~cm}$ (D) diverging lens of focal length $12 \mathrm{~cm}$

    ›Reveal solutionSolution

    The lens is a meniscus with both centres of curvature on the same side; using the lens maker’s formula with the correct sign convention shows it is a converging lens of focal length 20 cm. The correct option is (B).

    ---

    Concept & Intuition

    The lens maker’s formula relates the focal length of a thin lens to its refractive index and the radii of curvature of its two surfaces. The crucial step is assigning the correct sign to each radius according to the Cartesian sign convention: distances measured from the lens toward the incident light are positive, and distances measured opposite to the incident light are negative.

    Here the incident light comes from the left. Both centres of curvature $C_1$ and $C_2$ lie to the right of the lens. That means:

    • For the left surface (first surface the light hits), the centre of curvature is on the opposite side from the incident light → radius is negative.
    • For the right surface, the centre of curvature is also on the opposite side from the incident light → radius is also negative.

    A negative radius for a surface means it is concave toward the incident light. So both surfaces are concave from the left side, making the lens a meniscus shape that is thicker at the centre than at the edges? Actually, if both surfaces curve away from the incident light, the lens is thinner in the middle — that would be a diverging lens in air. But wait: the refractive index is 1.4 (greater than air), and the shape is a meniscus with both centres on the same side. The sign of the focal length depends on the combination. Let’s compute carefully.

    ---

    Step-by-step solution

    1. Identify the radii with sign

       • Left surface: radius $R_1 = -4\ \text{cm}$ (centre $C_1$ is to the right, opposite to incident light).
       • Right surface: radius $R_2 = -8\ \text{cm}$ (centre $C_2$ is also to the right).

    2. Lens maker’s formula (thin lens in air):

    $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

    Here $\mu = 1.4$, so $\mu - 1 = 0.4$.

    3. Substitute the signed radii:

    $\frac{1}{f} = 0.4\left(\frac{1}{-4} - \frac{1}{-8}\right)$

    Simplify inside parentheses:

    $\frac{1}{-4} - \frac{1}{-8} = -\frac{1}{4} + \frac{1}{8} = -\frac{2}{8} + \frac{1}{8} = -\frac{1}{8}$

    4. Compute focal length:

    $\frac{1}{f} = 0.4 \times \left(-\frac{1}{8}\right) = -\frac{0.4}{8} = -\frac{1}{20}$

    Hence

    $f = -20\ \text{cm}$

    A negative focal length means the lens is diverging. But wait — the options include a diverging lens of focal length 20 cm (option A) and a converging lens of 20 cm (option B). Our calculation gives $f = -20\ \text{cm}$, which is diverging. However, we must double-check the sign convention: many textbooks define $R$ positive if the centre of curvature is on the same side as the outgoing light. Let’s re-evaluate with the standard convention used in most Indian exams (the “new Cartesian” convention).

    Watch out

    A common pitfall: mixing sign conventions. In the standard lens maker’s formula used in many textbooks (especially for competitive exams), the sign convention is:

    • For a convex surface (bulging toward the incident light), $R$ is positive.
    • For a concave surface (curving away from incident light), $R$ is negative. Here, both surfaces curve away from the incident light (since centres are on the right), so both $R_1$ and $R_2$ are negative. That is what we used. But the result $f = -20$ cm suggests a diverging lens. Yet the figure shows a meniscus that is thicker at the centre? Let’s check the shape: if both centres are on the right, the left surface is convex toward the left? Actually, if the centre is to the right, the surface bulges toward the left — that is convex to the left, meaning it is converging for the left surface. Wait, we need to be careful:
    • A surface with centre on the right: the surface curves toward the left (like a dome facing left). That is a convex surface for light coming from the left.
    • So the left surface is convex (bulging toward incident light) → $R_1$ should be positive in the standard convention.
    • The right surface also has its centre on the right, so it curves away from the incident light (concave) → $R_2$ is negative.

    Let’s correct the sign assignment:

    • Left surface: centre $C_1$ is on the right, so the surface bulges toward the left (convex to incident light) → $R_1 = +4\ \text{cm}$.
    • Right surface: centre $C_2$ is on the right, so the surface bulges away from incident light (concave) → $R_2 = -8\ \text{cm}$.

    Now apply the lens maker’s formula:

    $\frac{1}{f} = (1.4 - 1)\left(\frac{1}{+4} - \frac{1}{-8}\right) = 0.4\left(\frac{1}{4} + \frac{1}{8}\right) = 0.4 \times \frac{3}{8} = \frac{1.2}{8} = \frac{3}{20}$

    Thus

    $f = \frac{20}{3} \approx 6.67\ \text{cm}$

    That is not among the options. So this sign assignment is also not matching. Let’s use the convention that is standard in the problem’s context (likely the one where both radii are taken as positive if the centre is on the same side as the outgoing light). In many textbooks, the formula is written as:

    $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

    with the rule: $R$ is positive if the centre of curvature is on the same side as the outgoing light (i.e., the side where the image is formed). For a lens in air, light travels left to right, so outgoing light is to the right.

    • Left surface: centre $C_1$ is on the right (same side as outgoing light) → $R_1 = +4\ \text{cm}$.
    • Right surface: centre $C_2$ is on the right (same side as outgoing light) → $R_2 = +8\ \text{cm}$.

    Then:

    $\frac{1}{f} = 0.4\left(\frac{1}{4} - \frac{1}{8}\right) = 0.4 \times \frac{1}{8} = \frac{0.4}{8} = \frac{1}{20}$

    So $f = +20\ \text{cm}$, a converging lens.

    This matches option (B).

    Tip

    The key insight: when both centres of curvature lie on the same side of the lens, the lens is a meniscus. If that side is the side where the light exits (the right side), then the lens is converging if the refractive index is greater than the surrounding medium. The sign convention that gives $R$ positive for centres on the outgoing side yields the correct physical behaviour.

    ---

    Final Answer

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. KCET 2023Set A-31 markMCQ
    Q.An equiconvex lens made of glass of refractive index $\dfrac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\dfrac{4}{3}$. The percentage change in the focal length is (A) 300% decrease (B) 400% decrease (C) 300% increase (D) 400% increase

    ›Reveal solutionSolution

    Apply the lens-maker's formula twice — once with $n = 3/2$ in air, once with the relative index $n_{rel} = (3/2)/(4/3)$ in water — and compare.

    1. The concept — the lens-maker's formula.

    A lens focuses light because of refraction at its two surfaces. The focal length depends on the index of the lens relative to its surroundings:

    $\frac{1}{f} = \left(\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

    That relative index is the whole point of this question: put the lens in a medium closer to its own index, and it bends light less.

    2. Geometry of an equiconvex lens.

    For an equiconvex lens, $R_1 = +R$ and $R_2 = -R$, so

    $\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{R} - \left(-\frac{1}{R}\right) = \frac{2}{R}$

    3. In air ($n_{\text{medium}} = 1$).

    $\frac{1}{f} = \left(\frac{3}{2} - 1\right)\frac{2}{R} = \frac12 \cdot \frac{2}{R} = \frac{1}{R} \quad\Longrightarrow\quad f = R$

    4. In water ($n_{\text{medium}} = 4/3$).

    The relative index is

    $n_{rel} = \frac{3/2}{4/3} = \frac{3}{2}\times\frac{3}{4} = \frac{9}{8}$

    $\frac{1}{f_w} = \left(\frac98 - 1\right)\frac{2}{R} = \frac18\cdot\frac{2}{R} = \frac{1}{4R} \quad\Longrightarrow\quad f_w = 4R = 4f$

    5. Percentage change.

    $\%\text{ change} = \frac{f_w - f}{f}\times 100 = \frac{4f - f}{f}\times 100 = 300\%$

    Since $f_w > f$, the focal length has increased by 300% (the lens becomes four times weaker — its converging power drops, exactly as expected when the surrounding medium's index approaches the lens's own).

    ✓Final answer

    The correct option is (C) — 300% increase.

    ANSWER: C