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Chapter 2: Problem 46

An object is located in water \(15 \mathrm{cm}\) from the vertex of a concave surface made of glass with a radius of curvature 10 \(\mathrm{cm} .\) Where does the image by refraction form and what is its magnification? Use \(n_{\text {water }}=4 / 3\) and \(n_{\text {glaxs }}=1.5\)

Short Answer

Expert verified
The image forms approximately \(6.92 \mathrm{cm}\) from the vertex of the concave surface in the glass, and the magnification of the image is \(0.4\).

Step by step solution

01

Write down the given values

We are given the following information: - Object distance from glass surface, \(s_i = 15 \mathrm{cm}\). - Radius of curvature of glass surface, \(R = 10 \mathrm{cm}\) - Refractive index of water, \(n_{\text {water}} = \frac{4}{3}\) - Refractive index of glass, \(n_{\text {glass}} = 1.5\)
02

Lensmaker's equation for refraction

We can now use the lensmaker's equation for refraction. The lensmaker's equation for a curved surface is given by: \( \frac{n_2}{s_2} - \frac{n_1}{s_1} = \frac{n_2 - n_1}{R} \) where \(n_1\) and \(n_2\) are the refractive indices of the object and image side media, respectively (here, water and glass), \(s_1\) is the object distance, \(s_2\) is the image distance, and \(R\) is the radius of curvature. Plugging in the given values: \( \frac{1.5}{s_2} - \frac{4/3}{15} = \frac{1.5 - 4/3}{10} \)
03

Solve for the image distance (\(s_2\))

Now we need to solve for \(s_2\), the image distance: \( \frac{1.5}{s_2} - \frac{1}{5} = \frac{1/6}{10} \) First, let's simplify the equation: \( \frac{1.5}{s_2} = \frac{1}{5} + \frac{1}{60} \) Now, find a common denominator and combine fractions: \( \frac{1.5}{s_2} = \frac{13}{60} \) Next, we can find \(s_2\) by taking the reciprocal of both sides: \( \frac{s_2}{1.5} = \frac{60}{13} \) Finally, we find the value of \(s_2\): \( s_2 = \frac{60}{13} \times 1.5 = \frac{90}{13} \approx 6.92 \mathrm{cm} \) So the image distance is approximately \(6.92 \mathrm{cm}\) from the vertex of the concave surface in the glass.
04

Calculate the magnification

Now that we have the image distance, we can calculate the magnification: Magnification, \(M =\frac{s_2}{s_1} \) Substituting the values we found: \( M = \frac{6.92 \,\text{cm}}{15 \,\text{cm}} =\frac{90/13}{15}= \frac{90/13}{195/13} = \frac{90}{195} = \frac{2}{5}\) The magnification is, therefore, \(0.4\). In conclusion, the image forms approximately \(6.92 \mathrm{cm}\) from the vertex of the concave surface in the glass, and the magnification of the image is \(0.4\).

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