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A small tank is closed at one end by a thin biconvex lens of glass with refractive index 1.50…

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

A small tank is closed at one end by a thin biconvex lens of glass with refractive index $1.50$ . Both faces have a radius of curvature of $25\ \text{cm}$ . The tank is full of water of refractive index $1.33$ , so the lens has air on its outer side and water on its inner side. A bright point source is $60\ \text{cm}$ from the lens, in the air, on the axis. Find the power of the lens in these two media, its second focal length (on the water side), the distance from the lens to the image in the water, and the magnification. Use the New Cartesian convention and give the sign of the magnification.

Need a reminder? The key ideaShow

With the object and the screen $D$ apart, a lens that gives a sharp image at two positions $d$ apart has $f = \dfrac{D^2 - d^2}{4D}$ . The two magnifications are $m$ and $\dfrac{1}{m}$ , so $h = \sqrt{h_1h_2}$ . In a liquid, $\dfrac{1}{f_m} = \left(\dfrac{n_l}{n_m} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ . With two different media, $\dfrac{n_3}{v} - \dfrac{n_1}{u} = P$ and $m = \dfrac{n_1v}{n_3u}$ .

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. $n_1 = 1.00$ , $n_g = 1.50$ , $n_3 = 1.33$ , $R_1 = +25\ \text{cm}$ , $R_2 = -25\ \text{cm}$ and $u = -60\ \text{cm}$ .

Plan. We use the surface formula for each face and add them. All distances follow the New Cartesian convention.

Step 1. The power. $P = \dfrac{1.50 - 1.00}{25\ \text{cm}} + \dfrac{1.33 - 1.50}{-25\ \text{cm}} = 0.0200\ \text{cm}^{-1} + 0.0068\ \text{cm}^{-1} = 0.0268\ \text{cm}^{-1} = 2.68\ \text{D}$ .

Step 2. The second focal length. $f_2 = \dfrac{n_3}{P} = \dfrac{1.33}{0.0268\ \text{cm}^{-1}} = 49.6\ \text{cm}$ . The first focal length is $f_1 = -\dfrac{1.00}{0.0268\ \text{cm}^{-1}} = -37.3\ \text{cm}$ .

Step 3. The image. We want $v$ , so $\dfrac{n_3}{v} = \dfrac{n_1}{u} + P = -\dfrac{1}{60\ \text{cm}} + 0.0268\ \text{cm}^{-1} = 0.010133\ \text{cm}^{-1}$ . Then $v = \dfrac{1.33}{0.010133\ \text{cm}^{-1}} = 131\ \text{cm}$ , in the water.

Step 4. The magnification. $m = \dfrac{n_1v}{n_3u} = \dfrac{1.00 \times 131.25\ \text{cm}}{1.33 \times (-60\ \text{cm})} = -1.64$ . The image is real and inverted.

Answer: $2.68\ \text{D}$ , $49.6\ \text{cm}$ , $131\ \text{cm}$ and $-1.64$ .

Check. The ratio $\dfrac{f_2}{|f_1|} = \dfrac{49.6\ \text{cm}}{37.3\ \text{cm}} = 1.33$ , which is the ratio of the indices of the two media.

Watch out. Leaving out the $n_3$ in $\dfrac{n_3}{v}$ would give $98.7\ \text{cm}$ , which is wrong.

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