Gauss M.D. said:Homework Statement
A biconvex lens with radius of curvature 15 cm has a focal length of 15 cm in air. What is its focal length if it is submerged in water (n=1.33)?
Homework Equations
1/f = (n-1)(1/R1 - 1/R2)
The Attempt at a Solution
1/f = 1/0.15 = (n-1)(2/0.15)
Solving for n gives n = 1.5
So the new equation will be:
1/f = (1.5 - 1.33)(2/0.15) = 2.27
The focal distance of a lens submerged in water refers to the distance between the lens and the focal point when the lens is placed in water. This distance can vary depending on the refractive index of water and the type of lens being used.
When a lens is submerged in water, the focal distance changes due to the difference in the refractive index of water compared to air. This causes the light rays passing through the lens to bend, resulting in a shorter focal distance.
The focal distance of a lens submerged in water can be affected by the curvature and refractive index of the lens, as well as the refractive index of the water. Additionally, the distance between the lens and the object being focused on, known as the object distance, can also impact the focal distance.
No, the focal distance of a lens submerged in water is not the same as in air. The refractive index of air is lower than that of water, causing the light rays passing through the lens to bend differently and resulting in a shorter focal distance.
The focal distance of a lens submerged in water can be calculated using the lens maker's formula, which takes into account the refractive index of the lens and the medium (water) it is submerged in, as well as the curvature of the lens. This formula is: 1/f = (n - 1) * (1/R1 - 1/R2), where f is the focal length, n is the ratio of the refractive indices, and R1 and R2 are the radii of curvature of the two surfaces of the lens.