A hemispherical glass lens with a refractive | vedclass

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(B) The refraction at a spherical surface is given by the formula:$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$Here, the light travels from th

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$65$

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(B) The refraction at a spherical surface is given by the formula:$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$Here, the light travels from the liquid $(n_1 = 1.3)$ into the glass $(n_2 = 1.5)$.For a parallel beam of light, the object distance $u = -\infty$.The radius of curvature $R$ for the convex surface is $+10 \,cm$.Substituting these values into the formula:$\frac{1.5}{v} - \frac{1.3}{-\infty} = \frac{1.5 - 1.3}{10}$$\frac{1.5}{v} - 0 = \frac{0.2}{10}$$\frac{1.5}{v} = \frac{0.2}{10}$$v = \frac{1.5 	imes 10}{0.2} = \frac{15}{0.2} = 75 \,cm$Wait, re-evaluating the refraction at the second surface (plane surface):The light enters the glass and then exits into the liquid. The first refraction at the curved surface forms an image at $v_1 = 75 \,cm$ from the pole.Since the lens is hemispherical, the distance from the pole to the plane surface is $R = 10 \,cm$.The object distance for the second surface is $u_2 = v_1 - R = 75 - 10 = 65 \,cm$.For the second surface (plane), $n_1 = 1.5$ and $n_2 = 1.3$. The radius $R_2 = \infty$.Using $\frac{n_2}{v_2} - \frac{n_1}{u_2} = \frac{n_2 - n_1}{R_2}$:$\frac{1.3}{v_2} - \frac{1.5}{65} = 0$$v_2 = \frac{1.3 	imes 65}{1.5} = \frac{84.5}{1.5} = 56.33 \,cm$ from the plane surface.The distance from the center of the lens (the plane surface) is $56.33 \,cm$.

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