Question

Water (with refractive index $= \frac{4}{3}$) in a tank is 72 cm deep. Oil of refractive index 7/4 lies on water making a thin convex surface of radius of curvature R = 12 cm as shown. An object 'S' is placed 24 cm above the surface. The location of its image is at 'x' cm above the bottom of the tank. Then 'x' is ______ cm.

Answer 1

8

Answer 2

The problem involves refraction of light through multiple interfaces. We use the spherical refracting surface formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.

Step 1: Refraction at the oil-air interface. Object S is in air ($n_1 = 1.0$), light refracts into oil ($n_2 = 7/4$). Object distance $u_1 = -24$ cm (light travels from object to surface). Radius of curvature $R_1 = +12$ cm (convex surface). Applying the formula: $\frac{7/4}{v_1} - \frac{1.0}{-24} = \frac{7/4 - 1.0}{12}$ $\frac{7}{4v_1} + \frac{1}{24} = \frac{3/4}{12} = \frac{1}{16}$ $\frac{7}{4v_1} = \frac{1}{16} - \frac{1}{24} = \frac{3-2}{48} = \frac{1}{48}$ $v_1 = \frac{7 \times 48}{4} = 84 \text{ cm}$ The image is formed 84 cm below the oil-air interface.

Step 2: Refraction at the oil-water interface. The oil-water interface is flat, so $R_2 = \infty$. Light travels from oil ($n_1 = 7/4$) to water ($n_2 = 4/3$). The image from Step 1 acts as the object. Since the oil layer is thin, we assume the oil-water interface is at the same level as the vertex of the oil-air interface. Object distance $u_2 = -84$ cm (image is 84 cm below the interface, in the medium of incidence). Applying the formula: $\frac{4/3}{v_2} - \frac{7/4}{-84} = \frac{4/3 - 7/4}{\infty}$ $\frac{4}{3v_2} + \frac{7}{4 \times 84} = 0$ $\frac{4}{3v_2} = - \frac{7}{336} = - \frac{1}{48}$ $v_2 = - \frac{4 \times 48}{3} = -64 \text{ cm}$ This image is formed 64 cm above the oil-water interface.

Step 3: Location of the image relative to the bottom of the tank. The water depth is 72 cm. The image is 64 cm above the oil-water interface. Distance from the bottom = Total water depth - Distance of image above oil-water interface Distance from bottom = $72 \text{ cm} - 64 \text{ cm} = 8 \text{ cm}$. Therefore, $x = 8$ cm.