Anearsighted eye. a certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye.
consider the simplified model of the eye described in exercise. if the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described in exercise, what is the distance from the cornea vertex to the retina? what does this tell you about the shape of the nearsighted eye?
exercise:
curvature of the cornea. in a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40 and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. what should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea's vertex is focused on the retina?

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s