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A solid glass sphere (n = 1.50) of radius 6.0 cm has a tiny air bubble inside…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solid glass sphere ( $n = 1.50$ ) of radius $6.0\ \text{cm}$ has a tiny air bubble inside. The bubble is on a diameter, $3.0\ \text{cm}$ below the nearest point of the surface. First you look at the bubble from the near side along the diameter. Find where the bubble seems to be, measured from that surface, and the magnification. Then you look from the opposite side along the same diameter. Find where it seems to be, measured from the surface that faces you.

Need a reminder? The key ideaShow

For a spherical surface, $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$ and $m = \dfrac{n_1 v}{n_2 u}$ . The light comes from the left, $u < 0$ for a real object, and $R > 0$ when the centre of curvature is on the right, on the side of the outgoing light.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $n_1 = 1.50$ and $n_2 = 1.00$ . Looking from either side, the centre of the sphere is on the same side as the bubble, so $R = -6.0\ \text{cm}$ .

Step 1. From the near side. $u = -3.0\ \text{cm}$ . $\dfrac{n_2}{v} = \dfrac{n_2 - n_1}{R} + \dfrac{n_1}{u} = \dfrac{-0.50}{-6.0\ \text{cm}} + \dfrac{1.50}{-3.0\ \text{cm}} = 0.0833\ \text{cm}^{-1} - 0.500\ \text{cm}^{-1} = -0.417\ \text{cm}^{-1}$ . So $v = -2.4\ \text{cm}$ .

Step 2. The magnification. $m = \dfrac{1.50 \times (-2.4\ \text{cm})}{1.00 \times (-3.0\ \text{cm})} = 1.2$ .

Step 3. From the far side. $u = -9.0\ \text{cm}$ , and $R$ is still $-6.0\ \text{cm}$ . $\dfrac{1.00}{v} = 0.0833\ \text{cm}^{-1} + \dfrac{1.50}{-9.0\ \text{cm}} = 0.0833\ \text{cm}^{-1} - 0.167\ \text{cm}^{-1} = -0.0833\ \text{cm}^{-1}$ . So $v = -12\ \text{cm}$ .

Step 4. The magnification. $m = \dfrac{1.50 \times (-12\ \text{cm})}{1.00 \times (-9.0\ \text{cm})} = 2.0$ .

Answer: $2.4\ \text{cm}$ with $m = 1.2$ , and $12\ \text{cm}$ with $m = 2.0$ .

Check. The bubble is $3.0\ \text{cm}$ from the near surface, so it is $9.0\ \text{cm}$ from the far one. From the far side it seems to be $12\ \text{cm}$ from that surface, which is exactly on the near surface. A bubble in a glass ball looks nearer than it is from one side and farther from the other.

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