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Q102P

Expert-verified
Found in: Page 1157

University Physics with Modern Physics

Book edition 14th edition
Author(s) Hugh D. Young, Roger A. Freedman
Pages 1596 pages
ISBN 9780321973610

In setting up an experiment for a high school biology lab, you use a concave spherical mirror to produce real images of a 4.00 - mm -tall firefly. The firefly is to the right of the mirror, on the mirror’s optic axis, and serves as a real object for the mirror. You want to determine how far the object must be from the mirror’s vertex (that is, object distance ) to produce an image of a specified height. First, you place a square of white cardboard to the right of the object and find what its distance from the vertex needs to be so that the image is sharply focused on it. Next you measure the height of the sharply focused images for five values of . For each value, you then calculate the lateral magnification . You find that if you graph your data with on the vertical axis and on the horizontal axis, then your measured points fall close to a straight line. (a) Explain why the data plotted this way should fall close to a straight line. (b) Use the graph in Fig. P34.102 to calculate the focal length of the mirror. (c) How far from the mirror’s vertex should you place the object in order for the image to be real 8.00 mm, tall, and inverted? (d) According to Fig. P34.102, starting from the position that you calculated in part (c), should you move the object closer to the mirror or farther from it to increase the height of the inverted, real image? What distance should you move the object in order to increase the image height from to 12.00 mm? (e) Explain why approaches zero as approaches . Can you produce a sharp image on the cardboard when s = 25 cm ? (f) Explain why you can’t see sharp images on the cardboard when s < 25 cm (and is positive).

1. As the equation of curve $s=-f\frac{1}{m}+f$ is linear, therefore the data plotted should be close to a straight line.
2. The focal length of the mirror is 24 cm.
3. If mirror’s vertex is placed 36 cm from the object, the image will be real, 8.00 mm tall, and inverted.
4. The object should be closer to the mirror, to increase the height of the inverted, real image.
5. As the distance of object is approximately equal to the focal length, therefore, approaches zero as approaches 25 cm .
6. As the virtual image is formed, therefore, the image cannot be sharp.
See the step by step solution

Step 1: Define the focal length and magnification.

The focal length is the distance between the convex or concave mirror and the focal point of the mirror.

The relation between the distance of object s, the distance of the image s' and the focal length is

$\frac{1}{f}=\frac{1}{s}+\frac{1}{{s}^{\mathrm{\prime }}}$

The ratio of the distance of the image s' to the ratio of the object s is known as magnification .

$m=-\frac{s\text{'}}{{s}^{\mathrm{\prime }}}$

The ratio of real image y' and focal length ${f}_{2}$ is the distance of the image u' .

${s}^{\mathrm{\prime }}=\frac{{y}^{\mathrm{\prime }}}{{f}_{2}}$

The ratio of real image y' and focal length ${f}_{1}$ is the distance of the object u.

${s}^{\mathrm{\prime }}=\frac{{y}^{\mathrm{\prime }}}{{f}_{1}}$

Step 2: Determine the whether the curve is from straight line or not also find the focal length.

From the graph the distance of the image is and $\frac{1}{m}$ is lateral magnification.

So, by the magnification formula

s' = - ms

By the relation between the distance of object , the distance of the image and the focal length f , the distance of image s is:

$\begin{array}{r}\frac{1}{s}+\frac{1}{{s}^{\mathrm{\prime }}}=\frac{1}{f}\\ \frac{1}{s}+\left(-\frac{1}{ms}\right)=\frac{1}{f}\\ s=-f\frac{1}{m}+f\end{array}$

So, the intercept the equation is and the slope is -f and graph given is linear.

From the equation $s=-f\frac{1}{m}+f$

$\begin{array}{r}f=-\text{slope}\\ =-\left(\frac{50\mathrm{cm}-38\mathrm{cm}}{-1.0-\left(-0.5\right)}\right)\\ =24\mathrm{cm}\end{array}$

Hence, as the equation of curve $s=-f\frac{1}{m}+f$ is linear, therefore the data plotted should be close to a straight line and the focal length of the mirror is 24cm.

Step 3: Determine the vertex place and the object location.

Given that, y' = 8 mm

The lateral magnification is:

$\begin{array}{r}\frac{1}{m}=-\frac{y}{{y}^{\mathrm{\prime }}}\\ =-\frac{4}{8}\\ =-0.5\end{array}$

Now, the distance of object is:

$\begin{array}{r}s=-f\frac{1}{m}+f\\ =-\left(24\mathrm{cm}\right)\left(-0.5\right)+24\mathrm{cm}\\ =36\mathrm{cm}\end{array}$

Since the distance is inversely proportional to the magnification. Therefore, with increase in magnification, there is increase in height of the image the object will be closer to the lens.

$s\propto \frac{1}{m}$

Hence, if mirror’s vertex is placed 36 cm from the object, the image will be real 8.00 mm , tall, and inverted and the object will be closer to the mirror to increase the height of the inverted, real image.

Step 4: Explain $$1/m$$ approaches to zero as approaches to $$25{\rm{ cm}}$$.

As the distance of object approaches 25 cm the magnification increases until it reaches $\infty$ . Thus 1/m will approach to zero. The magnification will be because is very close to the focal length.

Since the virtual image is formed on carboard when s < 25 cm . Therefore, image could not be sharp.

Hence, when the distance is approximately equal to the focal length therefore, 1 / m approaches zero as approaches 25 cm and since the virtual image is formed, therefore, the image cannot be sharp.