A Lens That 'Stops Working' in Water
A student sets up a lens experiment on an optical bench. The thin biconvex lens is made of glass of refractive index 1.50 and both of its surfaces have a radius of curvature of 20 cm. A pin fixed at the 30 cm mark serves as the object, and the student obtains a sharp image on a screen. The student then repeats the experiment with the same lens and the same object pin, but with the lens and pin immersed in a transparent liquid of refractive index 1.33 held in a shallow glass tank. No position of the screen gives a sharp image, and the student concludes that the lens has been damaged or that the liquid has spoiled its surfaces. A second student proposes using a liquid of refractive index 1.60 instead, predicting that the image will then be formed closer to the lens than it was in air. You are asked to compute what happens in each case and to judge both statements.
1. Use the lens maker's formula to find the focal length of the lens in air, and locate the image of the pin at u = 30 cm. Describe the image.
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Lens maker's formula: 1/f = (n - 1)(1/R1 - 1/R2). For a biconvex lens with R1 = +20 cm and R2 = -20 cm:
1/f = (1.50 - 1)(1/20 + 1/20) = 0.5 x 0.10 = 0.05 cm^-1, so f = 20 cm.
Thin lens equation, with u = -30 cm (object on the left):
1/v = 1/f + 1/u = 1/20 - 1/30 = (3 - 2)/60 = 1/60.
v = +60 cm: a real image 60 cm on the far side of the lens, so the screen sits at 60 cm. Magnification m = v/u = 60/(-30) = -2: inverted and twice the size of the pin. This is the standard real, inverted, magnified image of an object lying between f and 2f.
2. Find the focal length of the same lens in the liquid of refractive index 1.33, locate the image, and explain why no screen position works.
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What matters is the relative refractive index of glass with respect to the surrounding medium, n_rel = 1.50/1.33 = 1.125 (using 4/3 gives exactly 1.125).
1/f' = (1.125 - 1)(1/20 + 1/20) = 0.125 x 0.10 = 0.0125 cm^-1, so f' = 80 cm, four times the focal length in air. The liquid has 'robbed' the lens of most of its converging power, because the bending of light at each surface depends on the difference of refractive indices across it.
Now the object at 30 cm lies well inside the focal length (30 cm < 80 cm):
1/v = 1/f' + 1/u = 1/80 - 1/30 = (3 - 8)/240 = -5/240, giving v = -48 cm.
The image is virtual, 48 cm on the same side as the object, with m = v/u = (-48)/(-30) = +1.6, erect and magnified. A virtual image is formed by diverging rays and can never be caught on a screen, which is exactly what the student observed. The lens is undamaged; the student has simply placed the object inside the (new) focal length. If a screen were still wanted, the pin would have to be moved beyond 80 cm from the lens.
3. Judge the second student's prediction for a liquid of refractive index 1.60. Support your judgement with a calculation and a physical explanation.
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n_rel = 1.50/1.60 = 0.9375, which is less than one: the glass is now optically rarer than its surroundings.
1/f'' = (0.9375 - 1)(0.10) = -0.00625 cm^-1, so f'' = -160 cm.
The lens has become diverging, with a virtual focal point 160 cm on the object side. For an object at 30 cm:
1/v = 1/f'' + 1/u = -1/160 - 1/30 = -(3 + 16)/480, giving v = -25.3 cm; m = (-25.3)/(-30) = +0.84. The image is virtual, erect and diminished, and it is closer to the lens than the object.
So the second student is half right and half dangerously wrong. The image does come closer to the lens, but it is not a real image and can never be caught on a screen; no screen position will work in this liquid either. The physical rule is simple and worth memorising: a converging lens stays converging only while the refractive index of the lens exceeds that of the medium. When n_medium > n_lens the sign of the focal length reverses, and a biconvex lens behaves like a diverging lens. The correct conclusion is that the observed failure in the first liquid is a predictable optics effect, not damage, and that the second liquid would make matters worse, not better.