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Q105P

Expert-verifiedFound in: Page 1043

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 34-51, a box is somewhere at the left, on the central axis of the thin converging lens. The image ${{\mathit{I}}}_{{\mathbf{m}}}$** ** of the box produced by the plane mirror is** ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{cm}}}$ ** “inside” the mirror. The lens–mirror separation is** ${\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}$**, and the focal length of the lens is** ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{cm}}}$**. (a) What is the distance between the box and the lens? Light reflected by the mirror travels back through the lens, which produces a final image of the box. (b) What is the distance between the lens and that final image?**

- The distance between box and lens is $3\text{cm}$.
- The distance between the lens and final image is $2.33\text{cm}$.

- Lens-mirror separation, $10\text{cm}$
- Distance of ${I}_{m}$ produced by the plane mirror, $4\text{cm}$
- Focal length, $f=2\text{cm}$

**Consider a thin converging lens in which the thickest part is very small in width relative to the object distance, the image distance, and the focal length. In this case, the object for the mirror that results in that box image is equally in front of the mirror about 4 cm. This object is the first image formed by the system produced by the first transmission through the lens, i.e., 10 - 4 = 6cm, and focal length is f = 2 cm. For the second transmission through the lens, to find the distance between the lens and the final image, we can use distance and focal length from the first one.**

**Formulae:**

**The lens formula, **

**$\frac{\mathbf{1}}{\mathit{f}}{\mathbf{=}}\frac{\mathbf{1}}{\mathit{p}}{\mathbf{+}}\frac{\mathbf{1}}{\mathit{i}}$ ...**(i)

**Here f is the focal length, i is the image distance, p is the object distance, m is the magnification.**

The object for the mirror that results in that box image is equally in front of the mirror. Thus, for the first transmission the image distance is found to be:

${i}_{1}=10\text{cm}-4\text{cm}\phantom{\rule{0ex}{0ex}}=6\text{cm}\phantom{\rule{0ex}{0ex}}$

And the focal length to be: $f=2\text{cm}$ .

Now, using the above values in equation (i), the object distance (box) from the lens as follows:

localid="1663025945731" $\frac{1}{2\text{cm}}=\frac{1}{p}+\frac{1}{6\text{cm}}\phantom{\rule{0ex}{0ex}}\frac{1}{p}=\frac{1}{2\text{cm}}-\frac{1}{6\text{cm}}\phantom{\rule{0ex}{0ex}}=\frac{3-1}{6\text{cm}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3\text{cm}}\phantom{\rule{0ex}{0ex}}p=3\text{cm}$

Hence, the distance between box and the lens is $3\text{cm}$.

The box image serves as an object for the light returning through the lens $f=2\text{cm}$.

At this point, the object distance is found to be:

role="math" localid="1663025875085" $p=10\text{cm}+4\text{cm}\phantom{\rule{0ex}{0ex}}=14\text{cm}\phantom{\rule{0ex}{0ex}}$ .

Thus, using the above values in equation (i), the distance between the lens and the final image can be calculated as follows:

role="math" localid="1663025833390" $\frac{1}{2\text{cm}}=\frac{1}{14\text{cm}}+\frac{1}{i}\phantom{\rule{0ex}{0ex}}\frac{1}{i}=\frac{1}{2\text{cm}}-\frac{1}{14\text{cm}}\phantom{\rule{0ex}{0ex}}=\frac{7-1}{14\text{cm}}\phantom{\rule{0ex}{0ex}}=\frac{6}{14\text{cm}}\phantom{\rule{0ex}{0ex}}i=2.33\text{cm}\phantom{\rule{0ex}{0ex}}$

Hence, the distance between the image and the lens is $2.33\text{cm}$.

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