2.4 Thin lenses  (Page 7/13)

 Page 7 / 13

Using the lens maker’s equation

Find the radius of curvature of a biconcave lens symmetrically ground from a glass with index of refractive 1.55 so that its focal length in air is 20 cm (for a biconcave lens, both surfaces have the same radius of curvature).

Strategy

Use the thin-lens form of the lens maker’s equation:

$\frac{1}{f}=\left(\frac{{n}_{2}}{{n}_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$

where ${R}_{1}<0$ and ${R}_{2}>0$ . Since we are making a symmetric biconcave lens, we have $|{R}_{1}|=|{R}_{2}|$ .

Solution

We can determine the radius R of curvature from

$\frac{1}{f}=\left(\frac{{n}_{2}}{{n}_{1}}-1\right)\left(-\frac{2}{R}\right).$

Solving for R and inserting $f=-20\phantom{\rule{0.2em}{0ex}}\text{cm},{n}_{2}=1.55,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{n}_{1}=1.00$ gives

$R=-2f\left(\frac{{n}_{2}}{{n}_{1}}-1\right)=-2\left(-20\phantom{\rule{0.2em}{0ex}}\text{cm}\right)\left(\frac{1.55}{1.00}-1\right)=22\phantom{\rule{0.2em}{0ex}}\text{cm}.$

Converging lens and different object distances

Find the location, orientation, and magnification of the image for an 3.0 cm high object at each of the following positions in front of a convex lens of focal length 10.0 cm. $\text{(a)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=50.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ , $\text{(b)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=5.00\phantom{\rule{0.2em}{0ex}}\text{cm}$ , and $\text{(c)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=20.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ .

Strategy

We start with the thin-lens equation $\frac{1}{{d}_{\text{i}}}+\frac{1}{{d}_{\text{o}}}=\frac{1}{f}$ . Solve this for the image distance ${d}_{\text{i}}$ and insert the given object distance and focal length.

Solution

1. For ${d}_{\text{o}}=50\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10\phantom{\rule{0.2em}{0ex}}\text{cm}$ , this gives
$\begin{array}{cc}\hfill {d}_{\text{i}}& ={\left(\frac{1}{f}-\frac{1}{{d}_{\text{o}}}\right)}^{-1}\hfill \\ & ={\left(\frac{1}{10.0\phantom{\rule{0.2em}{0ex}}\text{cm}}-\frac{1}{50.0\phantom{\rule{0.2em}{0ex}}\text{cm}}\right)}^{-1}\hfill \\ & =12.5\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$

The image is positive, so the image, is real, is on the opposite side of the lens from the object, and is 12.6 cm from the lens. To find the magnification and orientation of the image, use
$m=\text{−}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=\text{−}\frac{12.5\phantom{\rule{0.2em}{0ex}}\text{cm}}{50.0\phantom{\rule{0.2em}{0ex}}\text{cm}}=-0.250.$

The negative magnification means that the image is inverted. Since $|m|<1$ , the image is smaller than the object. The size of the image is given by
$|{h}_{\text{i}}|=|m|{h}_{\text{o}}=\left(0.250\right)\left(3.0\phantom{\rule{0.2em}{0ex}}\text{cm}\right)=0.75\phantom{\rule{0.2em}{0ex}}\text{cm}$
2. For ${d}_{\text{o}}=5.00\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10.0\phantom{\rule{0.2em}{0ex}}\text{cm}$
$\begin{array}{cc}\hfill {d}_{\text{i}}& ={\left(\frac{1}{f}-\frac{1}{{d}_{\text{o}}}\right)}^{-1}\hfill \\ & ={\left(\frac{1}{10.0\phantom{\rule{0.2em}{0ex}}\text{cm}}-\frac{1}{5.00\phantom{\rule{0.2em}{0ex}}\text{cm}}\right)}^{-1}\hfill \\ & =-10.0\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$

The image distance is negative, so the image is virtual, is on the same side of the lens as the object, and is 10 cm from the lens. The magnification and orientation of the image are found from
$m=\text{−}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=\text{−}\frac{-10.0\phantom{\rule{0.2em}{0ex}}\text{cm}}{5.00\phantom{\rule{0.2em}{0ex}}\text{cm}}=+2.00.$

The positive magnification means that the image is upright (i.e., it has the same orientation as the object). Since $|m|>0$ , the image is larger than the object. The size of the image is
$|{h}_{\text{i}}|=|m|{h}_{\text{o}}=\left(2.00\right)\left(3.0\phantom{\rule{0.2em}{0ex}}\text{cm}\right)=\phantom{\rule{0.2em}{0ex}}6.0\phantom{\rule{0.2em}{0ex}}\text{cm}.$
3. For ${d}_{\text{o}}=20\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10\phantom{\rule{0.2em}{0ex}}\text{cm}$
$\begin{array}{cc}\hfill {d}_{\text{i}}& ={\left(\frac{1}{f}-\frac{1}{{d}_{\text{o}}}\right)}^{-1}\hfill \\ & ={\left(\frac{1}{10.0\phantom{\rule{0.2em}{0ex}}\text{cm}}-\frac{1}{20.0\phantom{\rule{0.2em}{0ex}}\text{cm}}\right)}^{-1}\hfill \\ & =20.0\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$

The image distance is positive, so the image is real, is on the opposite side of the lens from the object, and is 20.0 cm from the lens. The magnification is
$m=\text{−}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=\text{−}\frac{20.0\phantom{\rule{0.2em}{0ex}}\text{cm}}{20.0\phantom{\rule{0.2em}{0ex}}\text{cm}}=-1.00.$

The negative magnification means that the image is inverted. Since $|m|=1$ , the image is the same size as the object.

When solving problems in geometric optics, we often need to combine ray tracing and the lens equations. The following example demonstrates this approach.

Choosing the focal length and type of lens

To project an image of a light bulb on a screen 1.50 m away, you need to choose what type of lens to use (converging or diverging) and its focal length ( [link] ). The distance between the lens and the lightbulb is fixed at 0.75 m. Also, what is the magnification and orientation of the image?

Strategy

The image must be real, so you choose to use a converging lens. The focal length can be found by using the thin-lens equation and solving for the focal length. The object distance is ${d}_{\text{o}}=0.75\phantom{\rule{0.2em}{0ex}}\text{m}$ and the image distance is ${d}_{\text{i}}=1.5\phantom{\rule{0.2em}{0ex}}\text{m}$ .

Solution

Solve the thin lens for the focal length and insert the desired object and image distances:

$\begin{array}{cc}\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}\hfill & =\frac{1}{f}\hfill \\ \hfill f& ={\left(\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}\right)}^{-1}\hfill \\ & ={\left(\frac{1}{0.75\phantom{\rule{0.2em}{0ex}}\text{m}}+\frac{1}{1.5\phantom{\rule{0.2em}{0ex}}\text{m}}\right)}^{-1}\hfill \\ & =0.50\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \end{array}$

The magnification is

$m=\text{−}\frac{{d}_{i}}{{d}_{\text{o}}}=\text{−}\frac{1.5\phantom{\rule{0.2em}{0ex}}\text{m}}{0.75\phantom{\rule{0.2em}{0ex}}\text{m}}=-2.0.$

Significance

The minus sign for the magnification means that the image is inverted. The focal length is positive, as expected for a converging lens. Ray tracing can be used to check the calculation (see [link] ). As expected, the image is inverted, is real, and is larger than the object. A light bulb placed 0.75 m from a lens having a 0.50-m focal length produces a real image on a screen, as discussed in the example. Ray tracing predicts the image location and size.

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