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What happens to the index of refraction and focal length of a lens that is initially in air that is then placed in water?

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As Bill Nye used to say, "Try it!". A magnifying glass, a kitchen sink, and sunlight should suffice. – DumpsterDoofus Apr 29 '14 at 12:46
up vote 4 down vote accepted

The index of refraction of the lens will remain the same (the medium remains the same, whether in a vacuum, in air, immersed in water, etc.), but the focal length will increase.

The index of refraction of water is greater than that of air, so light entering through the lens will be refracted at a smaller angle with respect to the horizontal of the lens than it would in air.

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ah got it! thanks. so the index of refraction of a lens stays constant b/c the ratio of c to velocity of light in that lens' glass does not change. the focal length would increase b/c power decreases due to the lensmaker equation. and refracted light rays would refract at a smaller angle in water as compared to air. Do all 3 of these conclusions hold for diverging and converging lenses? – stackseverywhere Apr 30 '14 at 14:17
Yes, your statements will hold true for both converging and diverging lenses (given that we focus on the virtual image of the diverging lens). Pictorially, this is easily conceptualised based on Snell's Law; you can draw out a diagram of a simple lens and compare visually where the focal length between the angles. – Daniel May 1 '14 at 10:33

It depends, suppose the refractive index of the lens is same as that of water. Then, no refraction takes place, it means the focal length of the lens becomes infinite. For example, glycerin and Pyrex rod has approximately same refractive index. If you immerse Pyrex rod in glycerin, Pyrex becomes invisible.

Index of refraction doesn't change, but focal length changes according to the medium in which you place.

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The effect of Snell's law is that the equalivent defraction is set by the ratios of the individual refactravive indexes. So if your lens has a refractive index of 4 in air, and water having $1 \frac 13$, then the effect of putting the lens in water is $4 \div 1\frac 13$ or the same solution as the lense of refractive index $3$ in air.

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