Complete the passage with the options that explain why converging lenses are used to correct farsightedness (hyperopia), whereas diverging lenses are used to correct nearsightedness (myopia).
In order to see an object, the cornea and lens of the eye form an image directly on the retina. The eyes of a nearsighted person cause the light rays of a object to converge too and therefore focus the image the retina. A diverging lens placed in front of a nearsighted eye compensates for the convergence so that an image forms properly on the retina. The eyes of a farsighted person cause the light rays of a object to converge too and therefore focus the image the retina. A " converging lens placed in front of a farsighted eye compensates for the convergence so that an image forms properly on the retina.

Explanation:

In order to see an object, the cornea and lens of the eye form an image directly on the retina. The eyes of a nearsighted person cause the light rays of a distant object to converge too much and therefore focus the image __before the retina. A diverging lens placed in front of a nearsighted eye compensates for the _greater convergence so that an image forms properly on the retina. The eyes of a farsighted person cause the light rays of a _nearby object to converge too _less and therefore focus the image __behind the retina. A " converging lens placed in front of a farsighted eye compensates for the __lesser convergence so that an image forms properly on the retina.

In case of person suffering from near nearsightedness , convergence of light is too much so light gets focused right before retina . It is corrected by diverging lens . Opposite is the case with person suffering from far farsightedness.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s