Darcy suffers from farsightedness equally severely in both eyes. The focal length of either of Darcy's eyes is 19.8 mm in its most accommodated state (i. e, when the eye is focusing on the closest object that it can clearly see) What lens strength (a. k.a., lens power) of contact lenses should be prescribed to correct the farsightedness in Darcy's eyes? (Assume the lens-to retina distance of Darcy's eyes is 2.00 cm, and the contact lenses are placed a negligibly small distance from the front of Darcy's eyes) i. 1.64D
ii. 2.98 D
iii. 2.19 D
iv. 3.49 D
v. 1.37 D

Explanation:

Power of defective eye = (1 / near point )+ (1 / lens-retina distance )

1 / .0198 m  =  (1 / near point )+ (1 / .02 m  )

= 50.50 = (1 / near point )+ 50

0.50 = 1 / near point

near point = 1 / .5 = 2m

= 200 cm .

for near point at 25 cm , convex lens will be required.

u = - 25 cm , v = - 200 cm ,

1 / v - 1 / u = 1 /f for lens

- 1 / 200 +  1 / 25 = 1 / f

= .04 - .005 = 1 /f

.035 = 1 / f

f = 28.57 cm = 0.2857 m

power = 1 /f

= 1 / 0.2857

= 3.5 D .

i v ) option is correct .

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles