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Q107P

Expert-verifiedFound in: Page 1044

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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** ${\mathit{d}}{\mathbf{=}}{\mathbf{20}}{\mathbf{\text{cm}}}$** from the fly; the image has the fly’s orientation and height** ${{\mathit{H}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathit{H}}$**. What are (a) the focal length ${{\mathit{f}}}_{{\mathbf{1}}}$** **of the lens and (b) the object distance ** ${{\mathit{p}}}_{{\mathbf{1}}}$** of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at** ${\mathit{d}}{\mathbf{=}}{\mathbf{20}}{\mathbf{\text{cm}}}$** that has the same orientation as the fly, but now** ${{\mathit{H}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{50}}{\mathbf{}}{\mathit{H}}$**. What are (c) ** ${{\mathit{f}}}_{{\mathbf{2}}}$** and (d) ** ${{\mathit{p}}}_{{\mathbf{2}}}$**?**

- The focal length ${f}_{1}$ of the lens is $+40\text{cm}$.
- The object distance ${p}_{1}$ of the fly is $20\text{cm}$.
- The focal length ${f}_{2}$ is $-40\text{cm}$.
- The object distance ${p}_{2}$ is $40\text{cm}$.

- Image distance from the fly for lens 1, $\left|{i}_{1}+{p}_{1}\right|=20\text{cm}$
- Image distance from the fly for lens 2, $\left|{i}_{2}+{p}_{2}\right|=20\text{cm}$
- The height of the fly’s image, ${H}_{1}=2.0H$

**Find the object distance from the lens 1. After that, you find the focal length of the lens 1. Again the object is situated in front of lens 2; then by using the concept of magnification, you find the image distance due to lens 2. After that, you can find the focal length from the lens 2.**

Formulae:

The lens formula is,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ …(i)

The magnification formula of the lens is,

$m=\frac{H\text{'}}{H}=-\frac{i}{p}$ …(ii)

If magnification is greater than one $\left({m}_{1}>1\right)$, then the lens is converging, and for a converging lens, the focal length is positive.

From the given information and equation (ii), the relation of the image distance to the object distance can be given as follows:

localid="1663062400373" $\frac{2H}{H}=-\frac{{i}_{1}}{{p}_{1}}\phantom{\rule{0ex}{0ex}}$

${i}_{1}=-2{p}_{1}$ ...(iii)

Now, using the given data, you get the image and object distance for lens 1 as follows:

$\left|{p}_{1}+{i}_{1}\right|=20\text{cm}\phantom{\rule{0ex}{0ex}}\left|{p}_{1}+(-2{p}_{1})\right|=20\text{cm}\phantom{\rule{0ex}{0ex}}{p}_{1}=20\text{cm}$

Now, using this value in equation (iii), you get the image distance as:

${i}_{1}=-2\times 20\text{cm}\phantom{\rule{0ex}{0ex}}=-40\text{cm}$

Substituting the above values into equation (I), you get the focal length of the lens 1 as follows:

${f}_{1}=\frac{{p}_{1}{i}_{1}}{{p}_{1}+{i}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{\left(20\text{cm}\right)\left(-40\text{cm}\right)}{\left(20-40\right)\text{cm}}\phantom{\rule{0ex}{0ex}}=+40\text{cm}$

Here, the focal length is positive for a converging lens.

Hence, the focal length value is $+40\text{cm}$.

From result of part (a), the object distance ${p}_{1}$ of the fly is $20\text{cm}$.

In this case, magnification is less than 1 and greater than zero and thus, the lens 2 is a diverging lens. So, the image produced by the lens 2 is virtual, and our result for the focal length is negative.

From the given information and equation (ii), the relation of the image distance to the object distance can be given as follows:

$\frac{H}{2H}=-\frac{{i}_{2}}{{p}_{2}}$

${i}_{2}=-\frac{{p}_{2}}{2}$ …(iv)

Now, using the given data, you get the image and object distance for lens 2 as follows:

$\left|{p}_{2}+-\frac{{p}_{2}}{2}\right|=20\text{cm}\phantom{\rule{0ex}{0ex}}\frac{{p}_{2}}{2}=20\text{cm}\phantom{\rule{0ex}{0ex}}{p}_{2}=40\text{cm}$

Now, using this value in equation (iv), you get the image distance as:${i}_{2}=-\frac{40\text{cm}}{2}\phantom{\rule{0ex}{0ex}}=-20\text{cm}$

Substituting the above values into equation (i), you get the focal length of the lens 1 as follows:

${f}_{2}=\frac{{i}_{2}{p}_{2}}{{p}_{2}+{i}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(40\text{cm}\right)\left(-20\text{cm}\right)}{\left(40-20\right)\text{cm}}\phantom{\rule{0ex}{0ex}}=-\text{40 cm}$

Because focal length is negative, the lens is diverging.

Hence, the focal length of lens 2 is $-\text{40 cm}$.

From result of part (c), the object distance ${p}_{2}$ of the fly is $40\text{cm}$

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