Single Choice

A thin lens is made with a material having refractive index $$\displaystyle \mu =1.5$$. Both the sides are convex. It is dipped in water ($$\displaystyle \mu =1.33$$), it will behave like:

AA convergent lens
Correct Answer
BA divergent lens
CA rectangular slab
DA prism

Solution

Answer is A.

The focal length of a lens in air can be calculated from the lens maker's equation:
$$1/f = (n - 1) [1/R1 - 1/R2 + (n - 1) d / (n R1 R2)] $$
where ,
f is the focal length of the lens,
n is the refractive index of the lens material,
R1 is the radius of curvature of the lens surface closest to the light source,
R2 is the radius of curvature of the lens surface farthest from the light source, and
d is the thickness of the lens (the distance along the lens axis between the two surface vertices).
For thin lenses, "d" can be ignored so:
$$1/f = (n - 1) [1/R1 - 1/R2] $$
Therefore,
Let f1 be the focal length of lens in air. That is,
$$1/f1 = (n - 1) [1/R1 - 1/R2] $$
$$[1/R1 - 1/R2] = 1/f1(n-1) = 2/f1 $$

Now, let us consider the lens when lens is dipped in a medium of refractive index = 1.33.

Let f2 be the focal length of lens when dipped in medium, say water.
Then, $$1/f2 = (n - 1) [1/R1 - 1/R2] = (n/n' - 1) [1/R1 - 1/R2] $$ (n' - refractive index of the medium)

$$\implies 1/f2 = (1.5/1.33-1) \times 2/f1 = 0.34/1.33 (f1)$$

That is, f2 = 3.91(f1).

As the sign of f2 is same as that of f1 the lens will behave as a converging lens.

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