A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33 . Will the lens behave as a converging or a diverging lens? Give reasons.

When a biconvex lens is immersed in a medium with a higher refractive index (in this case, water with a refractive index of 1.33) than the refractive index of the lens material (1.25), it will behave as a converging lens. Here's the reason why:

The behavior of a lens depends on the relative refractive indices of the lens material and the surrounding medium. In this case, the lens material has a lower refractive index (1.25) compared to the refractive index of water (1.33). When light passes from a medium with a lower refractive index to a medium with a higher refractive index (as is the case when light enters the lens from the surrounding water), the light rays will bend toward the normal (the imaginary line perpendicular to the surface).

This means that as light enters the lens from the water, the light rays will bend toward the center of the lens. When these light rays exit the lens on the other side, they will bend away from the normal because they are going from a medium with a higher refractive index to a medium with a lower refractive index. This bending of light causes the rays to converge on the other side of the lens, and the lens behaves as a converging lens.

So, in summary, when a biconvex lens with a refractive index of 1.25 is immersed in water with a refractive index of 1.33, it behaves as a converging lens due to the difference in refractive indices between the lens material and the surrounding medium.