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Q104P

Expert-verifiedFound in: Page 1158

Book edition
14th edition

Author(s)
Hugh D. Young, Roger A. Freedman

Pages
1596 pages

ISBN
9780321973610

**The science museum where you work is constructing a new display. You are given a glass rod that is surrounded by air and was ground on its left-hand end to form a hemispherical surface there. You must determine the radius of curvature of that surface and the index of refraction of the glass. Remembering the optics portion of your physics course, you place a small object to the left of the rod, on the rod’s optic axis, at a distance s from the vertex of the hemispherical surface. You measure the distance of the image from the vertex of the surface, with the image being to the right of the vertex. Your measurements are as follows:**

**Recalling that the object–image relationships for thin lenses and spherical mirrors include reciprocals of distances, you plot your data as 1/s' versus 1/s . (a) Explain why your data points plotted this way lie close to a straight line. (b) Use the slope and y - intercept of the best-fit straight line to your data to calculate the index of refraction of the glass and the radius of curvature of the hemispherical surface of the rod. (c) Where is the image if the object distance is ? **

- The slope of curve is - 1 / n therefore, point plotted lie close to a straight line.
- The index of refraction is 1.45 .
- If the object distance is 15 cm , the distance of the image is - 66.9 cm.

**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 , the distance of the image and the focal length is

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

The Lensmaker’s formula is

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

Where, is the refraction index of lens, ${R}_{1}$ and ${R}_{2}$ are the first and second curvature of the lens respectively.

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

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

The Lensmaker’s formula is

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

Using two equations to write the equation of curve

$\frac{1}{{s}^{\mathrm{\prime}}}=-\frac{1}{ns}+\frac{n-1}{nR}$

The above equation written in the form y = m + c . Where slope is $-\frac{1}{n}$ and intercept of curve is $\frac{n-2}{nR}$.

Hence, the slope of curve is therefore, point plotted lie close to a straight line.

The slope of curve is - 1 / n .

$\begin{array}{r}-\frac{1}{n}=\frac{(1/271.6\mathrm{cm})-(1/148.3\mathrm{cm})}{(1/22.5\mathrm{cm})-(1/25\mathrm{cm})}\\ -\frac{1}{n}=-0.688\\ n=1.45\end{array}$

Hence, the refraction index is 1.45.

Given that, R = 10 cm

For s = 15 cm , the distance of the image is:

$\begin{array}{r}\frac{1}{{s}^{\mathrm{\prime}}}=-\frac{1}{ns}+\frac{n-1}{nR}\\ \frac{1}{{s}^{\mathrm{\prime}}}=\frac{-1}{\left(1.45\right)\left(15\mathrm{cm}\right)}+\frac{1.45-1}{\left(1.45\right)\left(10\right)}\\ \frac{1}{{s}^{\mathrm{\prime}}}=\frac{1}{-66.9\mathrm{cm}}\\ {s}^{\mathrm{\prime}}=-66.9\mathrm{cm}\end{array}$

Hence, if the object distance is 15 cm, the distance of the image is - 66.9 cm .

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