Q.The photograph of a house occupies an area of $1.75\ \text{cm}^2$ on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55\ \text{m}^2$. What is the linear magnification of the projector-screen arrangement?

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A convex lens forms a real image and a virtual image of same size of an object placed separately at distances $u_1$ and $u_2$ respectively from the lens. Then the focal length of the lens is (A) $\dfrac{\sqrt{u_1 u_2}}{2}$ (B) $\left(\dfrac{u_1+u_2}{2}\right)$ (C) $\sqrt{u_1 u_2}$ (D) $2(u_1+u_2)$

   ›Reveal solutionSolution

   When a convex lens produces a real image and a virtual image of the same size from two different object positions $u_1, u_2$, its focal length is simply the average of the two object distances: $f = (u_1+u_2)/2$.

   Concept and Intuition

   A convex lens forms a real, diminished-or-magnified, inverted image when the object is beyond the focal length, and a virtual, magnified, erect image when the object is placed between the lens and the focus. "Same size" here means the two magnifications have equal magnitude but opposite sign (one inverted, one erect). Writing the lens equation for both cases in terms of a common magnification magnitude $k$ and eliminating $k$ reveals a clean linear relationship between $u_1$, $u_2$ and $f$ — this is a standard, elegant optics result worth remembering.

   Step-by-Step Solution

   1. Real image case (object at distance $u_1$, using Cartesian sign convention so $u_1<0$): magnification $m_1 = -k$ (negative, inverted), with $v_1 = -k\,u_1$. Substituting into $\dfrac{1}{v_1}-\dfrac{1}{u_1}=\dfrac{1}{f}$ and simplifying gives $|u_1| = \dfrac{f(1+k)}{k}$.
   2. Virtual image case (object at distance $u_2$, also $u_2<0$, but nearer the lens than $f$): magnification $m_2=+k$ (positive, erect, and here $k>1$ since a convex lens's virtual image is always magnified), with $v_2 = k\,u_2$. Substituting similarly gives $|u_2| = \dfrac{f(k-1)}{k}$.
   3. Add the two object-distance magnitudes: $|u_1|+|u_2| = \frac{f(1+k)}{k}+\frac{f(k-1)}{k} = \frac{f\big[(1+k)+(k-1)\big]}{k} = \frac{f(2k)}{k} = 2f.$ …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A Convex lens forms a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object, again a real image is formed on the screen which is 16 cm tall. The length of the object must be (A) 4 cm (B) 8 cm (C) 12 cm (D) 20 cm

   ›Reveal solutionSolution

   Two conjugate lens positions give magnifications that are reciprocals of each other, so the object length is the geometric mean of the two image lengths — $\sqrt{4\times16}=8$ cm.

   Concept and Intuition

   When a convex lens is displaced between object and screen (keeping the object–screen separation fixed) and a sharp real image is found at two different lens positions, those two positions are conjugate: the object distance in one case equals the image distance in the other, and vice versa. Since magnification $m=v/u$, the two magnifications are reciprocals: $m_1 = v_1/u_1$ and $m_2 = u_1/v_1 = 1/m_1$, so $m_1 m_2 = 1$.

   Step-by-Step Solution

   1. Let the object height be $h_0$. The two image heights are $h_1 = m_1 h_0 = 4$ cm and $h_2 = m_2 h_0 = 16$ cm.
   2. Multiply: $h_1 h_2 = m_1 m_2\, h_0^2$.
   3. Since $m_1 m_2 = 1$ (conjugate positions): $h_1 h_2 = h_0^2$.
   4. $h_0 = \sqrt{h_1 h_2} = \sqrt{4\times16} = \sqrt{64} = 8$ cm. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.An object is placed at 10 cm from a lens and a real image is formed with magnification of 0.5. The lens is (A) Concave with focal length of $\dfrac{10}{3}$ cm (B) Convex with focal length of $\dfrac{10}{3}$ cm (C) Concave with focal length of 10 cm (D) Convex with focal length of 10 cm

   ›Reveal solutionSolution

   This tests using magnification and the lens formula together, with correct sign convention, to identify both the type of lens and its focal length from an object distance and magnification.

   Concept and Intuition

   Using the Cartesian sign convention for lenses (distances measured from the optical centre; distances in the direction of incident light are positive), the magnification is $m = v/u$. A real image formed by a lens is always inverted relative to the object, which is captured by $m$ being negative (since a real image has $v$ with a sign opposite to what an erect, virtual image would have relative to $u$). Since the object distance $u$ is always negative for a real object, a real image requires $v$ and $u$ to have a specific relation making $m<0$ overall — concretely, real images have $v>0$ (for a converging/convex lens producing a real image) while $u<0$, so $m=v/u<0$ automatically.

   Step-by-Step Solution

   1. Object distance: $u = -10$ cm (object on the incoming side, standard convention).
   2. The image is real, so $m$ is negative; magnitude given is $0.5$, so $m = -0.5$.
   3. From $m = v/u$: $v = m\,u = (-0.5)\times(-10) = +5$ cm.
   4. Since $v>0$, the image forms on the side opposite to the object — consistent with a real image formed by a converging (convex) lens (a diverging/concave lens can only form virtual, diminished, upright images for a real object, never a real image).
   5. Apply the lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$: …

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4. COMEDK 2026Set 2026-A1 markMCQ
   Q.An object placed 40 cm in front of a thin convex lens is moved to 60 cm from the lens. If the focal length of the lens is 30 cm the ratio of magnification of the image at the initial position to the final position is: (A) $3: 2$ (B) $2: 3$ (C) $1: 3$ (D) $3: 1$

   ›Reveal solutionSolution

   Magnifications are $|m_1|=3$ at $u=40$ cm and $|m_2|=1$ at $u=60$ cm, so the ratio is $3:1$ — option (D).

   Lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ with $f=+30$ cm (object distances taken negative).

   Initial position, $u=-40$ cm:

   $\frac{1}{v}=\frac{1}{30}-\frac{1}{40}=\frac{1}{120}\ \Rightarrow\ v=120\text{ cm},\quad m_1=\frac{v}{u}=\frac{120}{-40}=-3$

   Final position, $u=-60$ cm: …

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5. KCET 2025Set D-41 markMCQ
   Q.The image formed by an objective lens of a compound microscope is (A) Real and diminished (B) Real and enlarged (C) Virtual and enlarged (D) Virtual and diminished

   ›Reveal solutionSolution

   The object sits between $f_o$ and $2f_o$ of the objective, which by the lens rules gives a real, inverted, enlarged image — the intermediate image that the eyepiece then acts on.

   Step 1 — How a compound microscope is arranged.

   It uses two converging lenses:

   • The objective — short focal length $f_o$, placed close to the specimen.
   • The eyepiece — placed at the other end of the tube, used as a simple magnifier.

   The object is deliberately placed just beyond the objective's focus, i.e.

   $f_o < u < 2f_o$

   Step 2 — What such an object position gives.

   Apply the standard convex-lens image table for an object between $f$ and $2f$: the image forms beyond $2f$ on the far side, and it is

   • real (rays actually converge and cross — it can be caught on a screen),
   • inverted,
   • enlarged ($|m| > 1$).

   We can verify with the lens formula and the magnification relation. With $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_o}$ and, say, $f_o = 1\ \mathrm{cm}$, $u = -1.2\ \mathrm{cm}$:

   $\frac{1}{v} = \frac{1}{1} + \frac{1}{(-1.2)} = 1 - 0.833 = 0.167 \ \Rightarrow\ v = +6\ \mathrm{cm}$

   Positive $v$ $\Rightarrow$ real image on the far side. Magnification:

   $m_o = \frac{v}{u} = \frac{6}{-1.2} = -5$

   $|m_o| = 5 > 1$ (enlarged), and the minus sign says inverted.

   Step 3 — Why it must be real. …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The total magnification produced by a compound microscope is 24 when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is 5 cm, the magnification produced by the objective is (A) 4 (B) 4.8 (C) 120 (D) 6

   ›Reveal solutionSolution

   The total magnification of a compound microscope is the product of the objective magnification and the eyepiece magnification. For the eyepiece used at the least distance of distinct vision (25 cm), its magnification is $1 + \frac{D}{f_e}$. Given total magnification 24 and $f_e = 5$ cm, the objective magnification is 4.

   Concept & Intuition

   A compound microscope magnifies in two stages: the objective lens produces a real, enlarged image of the object, and the eyepiece then magnifies that image further. The total magnification $M$ is simply the product:

   $M = m_o \times m_e$

   where $m_o$ is the linear magnification of the objective and $m_e$ is the angular magnification of the eyepiece.

   The eyepiece acts like a simple magnifier. When the final image is formed at the least distance of distinct vision (typically $D = 25$ cm), the eyepiece’s magnification is given by:

   $m_e = 1 + \frac{D}{f_e}$

   This formula comes from the fact that the eye sees the image at the near point, so the angular size is maximized.

   We are told $M = 24$ and $f_e = 5$ cm. So we can find $m_e$ first, then solve for $m_o$.

   Step-by-step solution

   1. Identify the eyepiece magnification formula For a simple magnifier (or eyepiece) used with the final image at the near point $D = 25$ cm:

   $m_e = 1 + \frac{D}{f_e}$

   This is a standard result — the “1” accounts for the relaxed eye case being $D/f_e$, and adding 1 gives the near-point case.

   2. Plug in the given focal length

   $m_e = 1 + \frac{25}{5} = 1 + 5 = 6$

   So the eyepiece alone gives a magnification of 6.

   3. Use the total magnification relation $$M = m_o \times m_e \quad \Rightarrow \quad 24 = m_o \times 6 …$$

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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Images of same size are formed by a convex lens when an object is placed either at 20 cm or 10 cm distance from the lens. The focal length of the lens is (A) 12 cm (B) 40 cm (C) 18 cm (D) 15 cm

   ›Reveal solutionSolution

   Two object distances giving equal-sized images on a convex lens means one image is real & magnified while the other is virtual & magnified by the same factor — this symmetry pins the focal length at 15 cm.

   Concept and Intuition

   A convex lens can form images of different character depending on where the object sits relative to $f$ and $2f$: beyond $2f$ the image is real, inverted, and diminished; between $f$ and $2f$ it's real, inverted, and magnified; inside $f$ it's virtual, erect, and magnified. Here, two DIFFERENT object distances (20 cm and 10 cm) give images of the SAME SIZE. Since these are close to the lens, it's plausible one produces a magnified real image and the other a magnified virtual image, of equal size but opposite orientation. Algebraically, requiring $|m_1|=|m_2|$ with $m=f/(u+f)$ forces the denominators to be equal in magnitude but opposite in sign, which directly gives $f$ in terms of $u_1,u_2$.

   Step-by-Step Solution

   1. Use the Cartesian sign convention: real object distances are negative. So $u_1 = -20$ cm, $u_2=-10$ cm.
   2. Lens formula: $\dfrac1v - \dfrac1u = \dfrac1f \Rightarrow v = \dfrac{uf}{u+f}$.
   3. Magnification: $m = \dfrac{v}{u} = \dfrac{f}{u+f}$.
   4. Equal image sizes means $|m_1| = |m_2|$: $\left|\dfrac{f}{u_1+f}\right| = \left|\dfrac{f}{u_2+f}\right| \Rightarrow |u_1+f| = |u_2+f|$.
   5. Since $u_1 \ne u_2$, the only solution (other than the trivial equal case) is $u_1+f = -(u_2+f) \Rightarrow u_1+u_2+2f = 0$. …

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8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.When an object of height 12 cm is placed at a distance from a convex lens, an image of height 18 cm is formed on a screen. Without changing the positions of the object and the screen, if the lens is moved towards the screen, another clear image is formed on the screen. The height of this image is (A) 4 cm (B) 6 cm (C) 8 cm (D) 10 cm

   ›Reveal solutionSolution

   This is the classic “lens displacement” problem: for fixed object and screen distances, two lens positions produce a sharp image, and the product of the two image heights equals the square of the object height. The second image height is 8 cm.

   ---

   The key idea is that when the object and screen are fixed, there are two positions of a convex lens that form a real image on the screen (provided the distance between object and screen is greater than 4 times the focal length). This is known as the displacement method for finding focal length. The two images are conjugate: one is magnified, the other diminished, and their heights multiply to give the square of the object height.

   Why this works:

   For a thin lens, the lens equation is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, where $u$ is object distance and $v$ is image distance. If the total distance $D = u + v$ is fixed, then $u$ and $v$ are the two roots of a quadratic. Swapping $u$ and $v$ gives the second lens position. Magnification $m = v/u = h_i / h_o$. So if the first image height is $h_1$ and the second is $h_2$, then $h_1 h_2 = h_o^2$.

   ---

   1. Set up the given data.

      Object height $h_o = 12$ cm.

      First image height $h_1 = 18$ cm (magnified, so $m_1 > 1$).

      The second image is formed when the lens is moved toward the screen — this swaps object and image distances, giving a diminished image.

   2. Relate magnifications.

      For the first position: $m_1 = \frac{v_1}{u_1} = \frac{h_1}{h_o} = \frac{18}{12} = 1.5$. …

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9. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.If $m_1$ and $m_2$ ($m_1 > m_2$) are the magnification for two position of the lens between the object and the screen, and 'd' is the distance between the two positions of the lens, the focal length of the lens is (A) $\dfrac{m_1-m_2}{d}$ (B) $\dfrac{m_1m_2}{d}$ (C) $(m_1-m_2)d$ (D) $\dfrac{d}{(m_1-m_2)}$

   ›Reveal solutionSolution

   This tests the displacement method for measuring focal length: two lens positions between a fixed object and screen give reciprocal magnifications, and the focal length can be expressed purely via the magnifications and the displacement $d$ between positions.

   Concept and Intuition

   In the displacement method, the object and screen (a fixed distance $D$ apart) stay put while the lens is moved to two different positions, both giving a sharp image on the screen. By the principle of reversibility of light paths, the object and image distances simply swap between the two positions — so if the magnification is $m_1$ at the first position, it becomes $m_2=1/m_1$ at the second. Combining this reciprocal relation with the thin lens equation lets us eliminate the object/image distances entirely and express $f$ in terms of just $d$ (the lens displacement) and the two magnifications.

   Step-by-Step Solution

   1. At position 1: object distance $u_1$, image distance $v_1$, magnification $m_1=v_1/u_1$.
   2. By reversibility, at position 2 the distances swap: $u_2=v_1$, $v_2=u_1$, so $m_2=v_2/u_2=u_1/v_1=1/m_1$.
   3. Displacement between positions: $d=u_2-u_1=v_1-u_1=u_1(m_1-1)$, giving $u_1=\dfrac{d}{m_1-1}$. …

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10. MHT-CET 2024Set pcm-2024-05-02-M1 markMCQ
    Q.A plane mirror produces a magnification of (A) $-1$ (B) zero (C) $+1$ (D) $+2$

    ›Reveal solutionSolution

    +1.

    A plane mirror forms an image of the same size, erect: m=+1. …

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11. MHT-CET 2024Set pcm-2024-05-03-E1 markMCQ
    Q.A convex lens of focal length ' $f$ ' produces a real image whose size is ' $n$ ' times the size of an object. The distance of the object from the lens is (A) $\frac{\mathrm{n}+1}{\mathrm{nf}}$ (B) $\mathrm{f}\left(1-\frac{1}{\mathrm{n}}\right)$ (C) $\frac{\mathrm{nf}}{\mathrm{n}+1}$ (D) $f\left(1+\frac{1}{n}\right)$

    ›Reveal solutionSolution

    Real image means $|v|=n|u|$; substituting into the lens equation gives $u=f(1+1/n)$.

    Let object distance magnitude be $u$; for a real image of magnitude $n$ times, image distance $v=nu$.

    Lens formula (magnitudes, real object–real image): $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ …

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12. MHT-CET 2024Set pcm-2024-05-03-M1 markMCQ
    Q.A convex lens of focal length ' $f$ ' $m$ forms a real, inverted image twice in size of the object. The object distance from the lens in metre is (A) 0.5 f (B) 0.66 f (C) f (D) 1.5 f

    ›Reveal solutionSolution

    Magnification 2 with real image. …

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