Question

A fruit fly of height H sits in front of lens 1 on the central axis through the lens.The lens forms an image of the fly at a distance d = 30 cm from the fly; the image has the fly's orientation and height HI = 2.5 H. What are (a) the focal length f1 of the lens and (b) the object distance p1 of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at d = 30 cm that has the same orientation as the fly, but now HI = 0.86 H. What are (c) f2 and (d) p2?

Answer 1

Answer on Question #55829, Physics / Optics

A fruit fly of height H sits in front of lens 1 on the central axis through the lens. The lens forms an image of the fly at a distance $d = 30 \, \text{cm}$ from the fly; the image has the fly's orientation and height $H_{1} = 2.5 \, \text{H}$. What are (a) the focal length $f_{1}$ of the lens and (b) the object distance $p_{1}$ of the fly?

The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at $d = 30$ cm that has the same orientation as the fly, but now $H_{1} = 0.86$ H. What are (c) $f_{2}$ and (d) $p_{2}$?

Solution:

Definitions of the terms :

- p = object distance

- i = image distance

- d = distance between image and object

- f = focal length

- m = magnification

In this case $m > +1$, and we know that lens 1 is converging (producing a virtual image), so that our result for focal length should be positive.

Since

$$\left| i_{1} + p_{1} \right| = d$$

and

$$m = -\frac{i_{1}}{p_{1}} = 2.5$$

we find

$$i_{1} = -2.5 p_{1}$$ $$\left| -2.5 p_{1} + p_{1} \right| = d$$ $$p_{1} = \frac{d}{1.5} = \frac{30}{1.5} = 20 \, \text{cm}$$

and

$$i_{1} = -2.5 * 20 = -50 \, \text{cm}$$

Thin lens equation

$$\frac{1}{p_{1}} + \frac{1}{i_{1}} = \frac{1}{f_{1}}$$

sign rules

- p is positive = object is in front of the lens

- i is negative = image is on the same side of the object (virtual image !)

- f is positive = converging lens

Substitute the values, you get,

$$\frac{1}{20} - \frac{1}{50} = \frac{1}{f_{1}},$$ $$\frac{3}{100} = \frac{1}{f_{1}}$$ $$f _ {1} = \frac {100}{3} = +33.3 \mathrm{cm}$$

(c) In this case $0 < m < 1$ and we know that lens 2 is diverging (producing a virtual image), so that our result for focal length should be negative. Since $|p_2 + i_2| = 30$ cm and

$$i _ {2} = -0.86 p _ {2},$$

we find

$$\left| p _ {2} - 0.86 p _ {2} \right| = 30 \mathrm{cm}$$ $$p _ {2} = \frac {30}{0.14} = 214.29 \mathrm{cm}$$

and

$$i _ {2} = -0.86 * \frac {30}{0.14} = -184.29 \mathrm{cm}$$

Substituting these into Thin lens equation leads to

$$\frac {0.14}{30} - \frac {0.14}{30 * 0.86} = \frac {1}{f _ {2}},$$ $$- \frac {49}{64500} = \frac {1}{f _ {2}}$$ $$f _ {2} = - \frac {64500}{49} = -1316.33 \mathrm{cm}$$

Answer: (a) $f_{1} = \frac{100}{3} = +33.3 \mathrm{cm}$; (b) $p_{1} = 20 \mathrm{cm}$

(c) $f_{2} = -1316.33 \mathrm{cm}$; (d) $p_{2} = 214.29 \mathrm{cm}$

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