Worked examples · Sheet 3

Images formed by thin lenses

Theory used: Thin lens equation Sign convention and magnification Thin lens Principal rays of a lens Lensmaker's equation Systems of two lenses Object and image

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

An object of height $y = 2.0 cm$ is at $p = 30 cm$ from a thin converging lens with focal length $f = 12 cm$ . Determine the position, height and characteristics of the image.

Construction of the image with the three principal rays for

f = 12 cm

and

p = 30 cm

; the heights are drawn on a larger scale than the distances

\frac{1}{q} = \frac{1}{f} - \frac{1}{p}, m = - \frac{q}{p}, y^{'} = m y

\frac{1}{q} m y^{'} = \frac{1}{12 cm} - \frac{1}{30 cm} = \frac{5 - 2}{60 cm} = \frac{1}{20 cm} ⟹ q = 20 cm = - \frac{20}{30} \approx - 0.67 = - 0.67 \cdot 2.0 cm \approx - 1.3 cm

The lens is thin and the object is small compared with the distances, so the thin lens equation applies with $f > 0$ , because the lens is converging, and $p > 0$ , because the object is real. One obtains $q = 20 cm$ , positive, hence the image is real and forms on the other side of the lens, where a screen captures it.

The magnification is negative and less than $1$ in absolute value, so the image is inverted and reduced, $1.3 cm$ tall. The result is consistent with the position of the object: $p = 30 cm$ exceeds $2 f = 24 cm$ , and for an object beyond twice the focal length the image falls between $f$ and $2 f$ ; indeed $12 cm < q < 24 cm$ . The ray diagram confirms the computation: the ray parallel to the axis, the ray through the center and the ray through the focus $F$ meet at the tip of the image.

Exercise 2

A converging lens with $f = 8.0 cm$ is used as a magnifying glass. At what distance from the lens must an object be placed to see an upright image magnified three times? Where does the image form?

m = - \frac{q}{p} = 3 ⟹ q = - 3 p

\frac{1}{p} - \frac{1}{3 p} = \frac{1}{f} ⟹ \frac{2}{3 p} = \frac{1}{f} ⟹ p = \frac{2}{3} f

p = \frac{2}{3} \cdot 8.0 cm \approx 5.3 cm, q = - 3 \cdot \frac{16}{3} cm = - 16 cm

An upright image has positive magnification, so $m = 3$ , and from the magnification formula $q = - 3 p$ : since $p > 0$ , it follows that $q < 0$ , that is, the image is virtual. Substituting $q = - 3 p$ into the thin lens equation leaves an equation in the single unknown $p$ , which gives $p = \frac{2}{3} f = \frac{16}{3} cm \approx 5.3 cm$ .

The object is between the focus and the lens, as it must be for a magnifying glass. The image is at $16 cm$ from the lens, on the same side as the object; the eye, placed on the other side, sees it through the lens upright and three times larger.

Exercise 3

A biconcave glass lens with $n = 1.50$ has both faces with radius of curvature $30 cm$ . Compute the focal length and the optical power of the lens in air, and the image of an object placed at $20 cm$ from the lens.

R_{1} = - 30 cm, R_{2} = + 30 cm, \frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{f} P \frac{1}{q} m = 0.50 \cdot (- \frac{1}{30 cm} - \frac{1}{30 cm}) = - \frac{1}{30 cm} ⟹ f = - 30 cm = \frac{1}{- 0.30 m} \approx - 3.3 D = - \frac{1}{30 cm} - \frac{1}{20 cm} = - \frac{5}{60 cm} ⟹ q = - 12 cm = - \frac{- 12}{20} = 0.60

For the lensmaker’s equation the signs of the radii are needed. The first face is concave toward the incoming light, so its center of curvature is on the side from which the light arrives and $R_{1} < 0$ ; the second face has its center of curvature on the side from which the light exits, so $R_{2} > 0$ . The formula gives $f = - 30 cm$ : the lens is diverging, like every lens in air that is thinner at the center than at the edges, and its optical power, computed with $f$ in meters, is negative.

With the lens equation, $f$ negative and $p = 20 cm$ , one obtains $q = - 12 cm$ : the image is virtual, on the same side as the object, at $12 cm$ from the lens. The magnification $0.60$ is positive and less than $1$ , so the image is upright and reduced, as for every real object in front of a diverging lens.

Exercise 4

Two thin converging lenses, $L_{1}$ with $f_{1} = 10 cm$ and $L_{2}$ with $f_{2} = 15 cm$ , are placed on the same axis at distance $D = 40 cm$ . An object is $15 cm$ in front of $L_{1}$ . Determine the position of the final image, the overall magnification and the characteristics of the image.

\frac{1}{q _{1}} = \frac{1}{f _{1}} - \frac{1}{p _{1}}, p_{2} = D - q_{1}, \frac{1}{q _{2}} = \frac{1}{f _{2}} - \frac{1}{p _{2}}, m = m_{1} m_{2} = \frac{q _{1}}{p _{1}} \cdot \frac{q _{2}}{p _{2}}

\frac{1}{q _{1}} p_{2} \frac{1}{q _{2}} m = \frac{1}{10 cm} - \frac{1}{15 cm} = \frac{1}{30 cm} ⟹ q_{1} = 30 cm, m_{1} = - \frac{30}{15} = - 2 = 40 cm - 30 cm = 10 cm = \frac{1}{15 cm} - \frac{1}{10 cm} = - \frac{1}{30 cm} ⟹ q_{2} = - 30 cm, m_{2} = - \frac{- 30}{10} = 3 = (- 2) \cdot 3 = - 6

The system is solved one lens at a time, because the image formed by the first lens acts as the object for the second. The first lens forms a real image $30 cm$ beyond $L_{1}$ , inverted and magnified twice. This image is $10 cm$ before $L_{2}$ , so it is a real object for the second lens, with $p_{2} = 10 cm$ .

The object is closer to $L_{2}$ than its focal length, hence the second lens acts as a magnifying glass: $q_{2} = - 30 cm$ , that is, the final image is virtual, $30 cm$ before $L_{2}$ , and therefore $10 cm$ beyond $L_{1}$ , between the two lenses. The two successive images are magnified by $- 2$ and by $3$ , and the overall magnification is the product, $- 6$ : with respect to the object the final image is virtual, inverted and six times larger.