thin-lens equation magnification and optical instruments

A thin lens is described by the equation $1/f = 1/d_o + 1/d_i,$ where $f$ is the focal length, $d_o$ is the object distance from the lens, and $d_i$ is the image distance from the lens. In the real-is-positive convention, a converging lens has $f>0$, a diverging lens has $f<0$, while real objects and images are taken with a $+$ sign and virtual ones with a $-$ sign. The linear magnification $m = h_i/h_o = -d_i/d_o$: the sign of $m$ gives the orientation (a minus means an inverted image), and $|m|$ tells how many times larger the image is than the object. Optical instruments — magnifier, microscope, telescope, camera — reduce to one or several lenses, so their analysis begins with the thin-lens equation and magnification. Limits of validity: paraxial rays near the principal optical axis and a thin lens without aberrations. For two thin lenses in contact, the optical powers add: $1/F = 1/f_1 + 1/f_2$.

A converging lens has $f = 10$ cm. An object is placed at a distance $d_o = 15$ cm from the lens. Find $d_i$ and $m$, and state whether the image is real and how it is oriented.

In a projector, a slide is illuminated through a converging lens with $f = 10.0$ cm, and a sharp image forms on a screen at $d_i = 2.00$ m from the lens. Find the slide-to-lens distance and the magnification; explain how the slide must be inserted.

Two thin lenses are placed in contact: a converging lens with $f_1 = +20$ cm and a diverging lens with $f_2 = -30$ cm. An object is placed at $d_o = 30$ cm in front of this system. Find the equivalent focal length $F$, the image distance $d_i$, the magnification $m$, and describe the image; state the role of the diverging lens.

Safe self-study activity: take a converging lens (a magnifying glass) and move it away from a window toward a white screen (a sheet of paper) until a sharp image of a distant object (a tree or a lamp outside the window) forms on the screen. Measure the lens-to-screen distance — this is approximately $f$. Then place a small object (e.g., a coin) at distances $d_o = f/2$, $d_o = 3f/2$, and $d_o = 2f$, and observe the image; for a quantitative check, plot $1/d_i$ against $1/d_o$ and verify that the points lie on a straight line with slope $-1$ and intercept $1/f$. Never point the lens at the Sun: focused light can cause burns and eye damage, so perform solar observations only in a simulation (e.g., PhET "Geometric Optics").

Show how the thin-lens equation links object distance, image distance, and focal length, and how the sign of $m$ encodes image orientation.

A ray diagram with three principal rays through a converging lens, marking $f$, $d_o$, $d_i$ and heights $h_o$, $h_i$, together with a graph of $1/d_i$ versus $1/d_o$: a straight line with slope $-1$ and intercept $1/f$.

You photograph a friend who stands $1.5\ \text{m}$ away. The phone's main lens has a focal length $f = 4.3\ \text{mm} = 4.3\times10^{-3}\ \text{m}$. With the Cartesian convention $u = -1.5\ \text{m}$, the thin-lens equation gives $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{4.3\times10^{-3}} + \frac{1}{-1.5} = 232.6 - 0.67 \approx 231.9\ \text{m}^{-1}$, so $v = 4.31\times10^{-3}\ \text{m} = 4.31\ \text{mm}$. The sensor therefore sits $4.31\ \text{mm}$ behind the lens. The linear magnification is $m = \frac{v}{u} = \frac{4.31\times10^{-3}}{-1.5} = -2.9\times10^{-3}$, so your $1.7\ \text{m}$ tall friend forms an image $1.7 \times 2.9\times10^{-3} = 4.9\ \text{mm}$ tall on the sensor.

This exercises the thin-lens equation solved for $v$ when $u$ is much larger than $f$, and the definition of lateral magnification $m = v/u$. The negative sign of $m$ shows that the image on the sensor is real and inverted.

A pair of reading glasses is marked $+2.0\ \text{D}$, so its focal length is $f = 1/2.0\ \text{D} = 0.50\ \text{m} = 50\ \text{cm}$. You hold a book $25\ \text{cm}$ in front of the lens: $u = -25\ \text{cm}$. Then $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{50} - \frac{1}{25} = -\frac{1}{50}$, giving $v = -50\ \text{cm}$ and $m = v/u = (-50)/(-25) = +2.0$. The image is virtual, upright and twice as large, located $50\ \text{cm}$ in front of the lens.

It exercises the sign convention: a negative image distance means a virtual image on the same side as the object, and a positive magnification means it is upright. This is exactly how a converging lens acts as a magnifier when the object lies inside the focal length ($25\ \text{cm} < 50\ \text{cm}$).

A classroom projector has $f = 10\ \text{cm} = 0.10\ \text{m}$ and throws an image onto a screen $2.0\ \text{m}$ away, so $v = +2.0\ \text{m}$. From $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{2.0} - \frac{1}{0.10} = 0.50 - 10 = -9.5\ \text{m}^{-1}$, the slide must sit at $u = -0.105\ \text{m} = -10.5\ \text{cm}$, just outside the focal point. The magnification is $m = v/u = 2.0/(-0.105) = -19$, so a $36\ \text{mm}$ wide slide gives an image $36\ \text{mm} \times 19 = 684\ \text{mm} \approx 68\ \text{cm}$ wide, inverted on the screen.

It uses the same thin-lens equation arranged for a real image ($v > 0$) with $|u|$ slightly larger than $f$, and the magnification $m = v/u$ to predict image size. The negative $m$ is why slides are loaded upside down.