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?

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: $|u_1|+|u_2| = \frac{f(1+k)}{k}+\frac{f(k-1)}{k} = \frac{f\big[(1+k)+(k-1)\big]}{k} = \frac{f(2k)}{k} = 2f.$ …

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. …

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 $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$: …

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 $u_1+f = -(u_2+f) \Rightarrow u_1+u_2+2f = 0$. …

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 $u_1=\dfrac{d}{m_1-1}$. …

Concept and Intuition

For any thin lens, magnification $m = v/u$ and the lens formula is $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f_{signed}}$. For a concave (diverging) lens of magnitude focal length $f$, the signed focal length is $-f$ under the standard Cartesian sign convention.

Step-by-Step Solution

1. From $m = \dfrac{v}{u} = \dfrac{1}{x}$, we get $v = \dfrac{u}{x}$.
2. Lens formula for a diverging lens (signed focal length $=-f$):

3. Substitute $v = u/x$:

4. Then $v = \dfrac{u}{x} = \dfrac{-f(x-1)}{x}$.
5. The magnitude of the image distance from the lens is therefore $|v| = \dfrac{x-1}{x}f$. …

Concept and Intuition

The usual thin-lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ measures distances from the lens. Newton's alternative form measures the object's distance $x$ from the FIRST focal point (beyond it, so the object is farther than $f$ from the lens) and the image's distance $x'$ from the SECOND focal point. These satisfy the elegant relation $x\,x' = f^2$, and the magnification works out to $m=-f/x=-x'/f$ — a compact, symmetric result that is often faster than working with $u$ and $v$ directly.

Step-by-Step Solution

1. Object distance from the lens is $u = f+x$ (it sits a distance $x$ beyond the first focal point). …