A. a light wave travels trough glass (n = 1.5) at an angle of 28°. what angle does it have when it passes from the glass into water (n = 1.33)?

b. draw a ray diagram to locate the image of the object below. write one or two sentences to describe the type of image, its size, relative to the object and its position with respect to the lens. (image attached)

c, an object is located 94.0 cm from a concave mirror. the focal length is 34.0 cm. what is the image distance? is the image real or virtual?

d. an object is located 65.0 cm from a concave lens. the image distance is -11.8 cm. what is the magnification?

e. for a person who is nearsighted, seeing objects that are far away is difficult. the light from these objects does not get focused far enough back in the eye. write one or two sentences to identify the type of lens that can be used to correct nearsightedness and describe why it works.

a) sin(28)/sin(x)=1.33/1.5, so x=31.97 deg

b) The image is real, inverted, smaller than the original

c) u=94 cm, f=34 cm, v - ?

1/f=1/u+1/v, so

v=fu/(u-f)=94*34/(94-34)=51.27 cm

d) m=v/u=11.8/65=0.182

e) We need diverging (concave) lens, they are thinner at the center and thicker at the edge. This lens moves the focus of light from a point in front of the retina backwards so the light becomes focused exactly on retina.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .