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 & 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:

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:

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:

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

So the eyepiece alone gives a magnification of 6.

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

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

The heart of this problem is the mirror formula and the sign convention. A concave mirror has its focus and centre of curvature on the same side as the incoming light (the object side), so by the Cartesian sign convention both $f$ and $R$ are taken as negative. The object distance $u$ is also negative because the object is placed in front of the mirror. Once you plug these into $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, you get the image distance $v$, and then magnification follows directly from $m = -\frac{v}{u}$.

Let’s walk through it step by step.

1. Identify the given data with proper signs.

   Object distance: $u = -30$ cm (negative because object is in front of the mirror).

   Radius of curvature: $R = -20$ cm (negative for a concave mirror).

   Focal length: $f = \frac{R}{2} = -10$ cm.

2. Apply the mirror formula.

   The mirror formula is

Substitute $u = -30$ and $f = -10$:

This simplifies to

3. Solve for $v$. Bring $\frac{1}{30}$ to the right:

The right-hand side is $-\frac{3}{30} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15}$.

Hence

The negative sign means the image is formed on the same side as the object — a real image, as expected for an object beyond the focus of a concave mirror.

4. Calculate magnification. Magnification is given by

Substitute $v = -15$ and $u = -30$:

$m = -\frac{(-15)}{(-30)} = -\frac{15}{30} = -\frac{1}{2}.$ …

The key to this question is understanding the role of the objective lens in a compound microscope. A compound microscope uses two lenses: an objective (short focal length) and an eyepiece (or ocular). The objective does the heavy lifting of magnification — it creates a large, real image of the tiny object placed just beyond its focal point. The eyepiece then acts as a simple magnifier to view that real image.

Let’s walk through the ray diagram logic step by step.

1. Object placement relative to the objective.

   The object is placed just outside the focal point of the objective lens, i.e., at a distance $f_o < u_o < 2f_o$ (where $f_o$ is the focal length of the objective). This is a deliberate choice: if the object were inside the focal point, the objective would produce a virtual image, which cannot be projected or further magnified by the eyepiece effectively.

2. Nature of the image formed by the objective.

   For a convex lens, when the object is placed beyond the focal point ($u > f$), the image formed is real (rays actually converge). Since the object is also between $f$ and $2f$, the image is inverted relative to the object and magnified (larger than the object). This is standard lens behaviour — you can verify it with the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and the magnification $m = v/u$ (with sign conventions giving $m < 0$ for inversion and $|m| > 1$ for magnification).

3. Why this intermediate image must be real.

   The eyepiece needs a real object to work with — it cannot magnify a virtual image from the objective. The real, inverted, magnified image formed by the objective sits at a distance just inside the focal point of the eyepiece, so the eyepiece then produces a final virtual, erect, and highly magnified image for the eye. But the question asks only about the objective’s image.

4. Eliminating the wrong options.

   • (B) real, erect and magnified — A single convex lens never gives an erect real image; real images are always inverted for a convex lens. …

The key idea is line-of-sight horizon: a transmitting antenna at height $h$ can only be seen by receivers up to the point where the line from the antenna just grazes the Earth’s surface. Because the Earth is curved, the visible region is a spherical cap. For a tower that is tiny compared to Earth’s radius, the cap’s area simplifies beautifully to $2\pi R h$ — no trigonometry needed.

1. Set up the geometry

   Imagine a tower of height $h$ at point $T$ on Earth’s surface. A ray from the top of the tower to a receiver at ground level just touches the Earth at the horizon point $H$. The line $TH$ is tangent to the Earth’s surface at $H$.

   Let $R$ be Earth’s radius. The distance from Earth’s center $O$ to the tower top is $R + h$. The tangent line $TH$ is perpendicular to the radius $OH$. So triangle $OTH$ is right-angled at $H$.

2. Find the central angle $\theta$

   In right triangle $OTH$:

For small $h/R$ (here $h=40$ m, $R=6.4\times10^6$ m, so $h/R \approx 6.25\times10^{-6}$), we can use the approximation $\cos\theta \approx 1 - \frac{\theta^2}{2}$. Equating:

Hence $\theta^2 \approx \frac{2h}{R}$, so

3. Area of the spherical cap The service area is the spherical cap on Earth’s surface with angular radius $\theta$. The exact formula for the area of a spherical cap of height $H_{\text{cap}}$ (the vertical distance from the cap’s base to the pole) is $A = 2\pi R H_{\text{cap}}$. From the geometry, the cap height is $H_{\text{cap}} = R(1 - \cos\theta)$. Using $\cos\theta \approx 1 - \frac{\theta^2}{2}$:

Substitute $\theta^2 = 2h/R$: