An object is located in air \(30 \mathrm{cm}\) from the vertex of a concave surface made of glass with a radius of curvature 10 cm. Where does the image by refraction form and what is its magnification? Use \(n_{\text {air }}=1\) and \(n_{\text {glass }}=1.5\)

We are given: - Distance of the object from the vertex (concave surface) = \(d_o\) = 30 cm - Radius of curvature of the concave surface = \(R\) = 10 cm - Refractive index of air = \(n_{\text{air}}\) = 1 - Refractive index of glass = \(n_{\text{glass}}\) = 1.5 Our goal is to find the distance of the image formed by refraction (denoted as \(d_i\)) and the magnification (denoted as \(M\)).

The lensmaker's formula relates the focal length (\(f\)) of the lens, the refractive index of the lens (\(n_{\text{lens}}\)), the refractive index of the surrounding medium (\(n_{\text{medium}}\)), and the radius of curvature of the lens surface (\(R\)). The lensmaker's formula is given by: \[\frac{1}{f} = (n_{\text{lens}} - n_{\text{medium}}) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\] For a single refracting surface like a concave surface, we consider \(R_2\) as infinity. Therefore, the formula becomes: \[\frac{1}{f} = (n_{\text{lens}} - n_{\text{medium}}) \left(\frac{1}{R_1}\right)\] Now, substituting the given values, we get: \[\frac{1}{f} = (1.5 - 1) \left(\frac{1}{10}\right)\] Solve for the focal length \(f\): \[f = \frac{1}{0.5 \cdot 0.1} = 20\,\text{cm}\]

The thin lens formula relates the distance of the object, the distance of the image, and the focal length: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] We can substitute the known values and solve for \(d_i\): \[\frac{1}{20} = \frac{1}{30} + \frac{1}{d_i}\] Rearrange the equation and solve for \(d_i\): \[\frac{1}{d_i} = \frac{1}{20} - \frac{1}{30}\] \[d_i = \frac{1}{\frac{1}{20} - \frac{1}{30}} = 60\,\text{cm}\] The image is formed at a distance of 60 cm from the concave surface.

The magnification (\(M\)) of an image can be determined by the ratio of the image distance (\(d_i\)) to the object distance (\(d_o\)): \[M = \frac{d_i}{d_o}\] Substitute the known values and solve for \(M\): \[M = \frac{60\,\text{cm}}{30\,\text{cm}} = 2\] The magnification of the image is 2. To conclude, the image is formed at a distance of 60 cm from the concave surface and the image magnification is 2.