Lens Focal Length in Water
A student has a biconvex lens made of crown glass of refractive index 1.52; both surfaces have radius of curvature 20 cm. In air, using a distant window as the object, he measures a focal length of 19.5 cm. He then immerses the lens in water of refractive index 1.33 and sends a parallel laser beam through it, finding the focus at 55 cm from the lens. Using the lens maker’s formula with the same radii, he predicts about 72 cm and concludes that either the water is contaminated or the lens is not crown glass. The teacher asks him to check the sign convention, to calculate what refractive index would explain the 55 cm result, and to comment on the statement that a lens must have a shorter focal length in water because water is denser than air.
1. Calculate the predicted focal lengths in air and water, compare them with the measurements, and judge the contamination claim.
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The lens maker’s formula is 1/f = (n − 1)(1/R1 − 1/R2). For a biconvex lens, R1 = +20 cm and R2 = −20 cm, so (1/R1 − 1/R2) = 1/20 + 1/20 = 0.10 cm−1. In air, 1/f = (1.52 − 1)(0.10) = 0.052 cm−1, giving f = 19.2 cm, consistent with the measured 19.5 cm. In water the relative refractive index is 1.52/1.33 = 1.143, so 1/f = (0.143)(0.10) = 0.0143 cm−1 and f = 70 cm. The measured 55 cm is 21% shorter than predicted. To explain 55 cm, the required relative index is nrel = 1 + 1/(55 × 0.10) = 1.182, so nglass = 1.182 × 1.33 = 1.57, closer to flint glass than crown glass. Contamination of water with salt raises its refractive index, which would make the relative index smaller and the focal length longer, not shorter. So contamination cannot explain the 55 cm result; a different glass or a measurement error is more likely.
2. Explain the sign convention used for the two surfaces and what would happen if both radii were taken as positive.
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For a biconvex lens the two centres of curvature lie on opposite sides of the lens. Using the Cartesian convention, if light travels left to right, the first surface encountered is convex toward the incident light and has R1 = +20 cm, while the second surface has its centre on the incident side and R2 = −20 cm. The bracket 1/R1 − 1/R2 then becomes 0.10 cm−1. If both radii were taken as positive, the bracket would be 1/20 − 1/20 = 0, predicting infinite focal length, which contradicts the measured 19.5 cm in air. The air measurement itself confirms that the signs must be opposite.
3. Evaluate the statement: a lens must have a shorter focal length in water because water is denser than air.
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The statement is false. Refraction at a spherical surface depends on the relative refractive index of the lens with respect to the surrounding medium, not on density alone. In air the relative index is about 1.52, while in water it is about 1.14. The lens is therefore much weaker in water, and its focal length increases from about 19 cm to about 70 cm. A higher-index surrounding medium reduces the bending at each surface. If the surrounding medium had a refractive index greater than that of the lens, the lens could even become diverging, with a negative focal length. Density is not the controlling quantity; the refractive-index contrast is.