Question

At what distance d1 (in cm) from the center of the flask is the center of this image located? In the window of the physics laboratory, an inverted image of the neighboring school was photographed, created by a spherical glass flask with a radius of R = 60 mm filled with water. Take µwater = 4/3

Answer 1

12

Answer 2

The problem involves refraction through a spherical flask filled with water, acting as a lens. The key is to trace the path of light as it refracts at each surface of the sphere.

1. Refraction at the First Surface (Air to Water):

   • Object distance, $u_1 = -\infty$ (distant school)
   • Radius of curvature, $R_1 = +6$ cm (convex surface)
   • Refractive index of air, $\mu_1 = 1$
   • Refractive index of water, $\mu_2 = \frac{4}{3}$

   Using the spherical refracting surface formula:

   $$\frac{\mu_2}{v_1} - \frac{\mu_1}{u_1} = \frac{\mu_2 - \mu_1}{R_1}$$ $$\frac{4/3}{v_1} - \frac{1}{-\infty} = \frac{4/3 - 1}{6}$$ $$\frac{4}{3v_1} = \frac{1}{18}$$ $$v_1 = 24 \text{ cm}$$

   This intermediate image is formed 24 cm from the first surface.

2. Refraction at the Second Surface (Water to Air):

   • The intermediate image acts as the object for the second surface.
   • The distance between the two surfaces is the diameter of the sphere, $2R = 12$ cm.
   • The object distance from the second surface, $u_2 = 24 - 12 = 12$ cm.
   • Radius of curvature, $R_2 = -6$ cm (concave surface)
   • $\mu_1 = \frac{4}{3}$
   • $\mu_2 = 1$

   Using the spherical refracting surface formula:

   $$\frac{\mu_2}{v_2} - \frac{\mu_1}{u_2} = \frac{\mu_2 - \mu_1}{R_2}$$ $$\frac{1}{v_2} - \frac{4/3}{12} = \frac{1 - 4/3}{-6}$$ $$\frac{1}{v_2} - \frac{1}{9} = \frac{1}{18}$$ $$\frac{1}{v_2} = \frac{1}{18} + \frac{1}{9} = \frac{1}{6}$$ $$v_2 = 6 \text{ cm}$$

   The final image is formed 6 cm from the second surface.

3. Distance from the Center:

   The second surface is located at a distance of $R = 6$ cm from the center of the sphere. The final image is formed 6 cm to the right of the second surface. Therefore, the distance of the final image from the center of the sphere is:

   $$d_1 = 6 \text{ cm} + 6 \text{ cm} = 12 \text{ cm}$$

Therefore, the distance $d_1$ from the center of the flask to the center of the image is 12 cm.