Question

PLEASE HELP!!!!!!!!!!!!!

a. A light wave travels through 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, it’s size relative to the object, and it’s position with respect to the lens.

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.

PLEAAAASE HELP MEEE

Answer 1

A. The angle with which the light passes from glass into water is approximately 31.97°

B. The image formed is, real, inverted, and smaller than the object

C. The image distance is 53.2
`\overline 6` cm

The image is real

D. The magnification is approximately 0.182

E. The type of lens required is a concave lens which corrects the focus of a powerful eye lens or a long eyeball

Step-by-step explanation:

The given parameters for the light wave are;

The refractive index of the glass medium, n₁ = 1.5

The angle of incidence of the light, θ₁ = 28°

The refractive index of water, n₂ = 1.33

A. Let θ₂ represent the angle with which the light passes from glass into water

Snell's law is mathematically stated as follows;

n₁·sin(θ₁) = n₂·sin(θ₂)

According to Snell's law, we have;

1.5 × sin(28°) = 1.33 × sin(θ₂)

sin(θ) = (1.5 × sin(28°))/1.33 = 0.52947920615

∴ θ₂ = arcsin(0.52947920615) ≈ 31.97°

The angle with which the light passes from glass into water, θ₂ ≈ 31.97°

B. Please find attached the drawing of the image created with Microsoft Visio

From the drawing, we have that the image formed is real inverted smaller than the object and between the focus and radius of curvature

C. The location of the object from the mirror, u = 94.0 cm

The focal length of the mirror, f = +34.0 cm

The mirror formula for a concave mirror is given as follows;

`(1)/(f) = (1)/(u) + (1)/(v)`

Therefore, we have;

`(1)/(34.0) = (1)/(94.0) + (1)/(v)`

`(1)/(v) = (1)/(34.0) - (1)/(94.0) = (15)/(799)`

`\therefore (v)/(1) = v =(799)/(15) = 56.2\overline 6`

The image distance, v = 799/15 = 53.2
`\overline 6` cm

Therefore the image is in front of the mirror, which is a real image

The image is real

D. The given parameters of the object and lens are;

The distance of the object from the lens, u = 65.0 cm

The distance of the image formed from the lens, i = -18 cm

The magnification, M, is given as follows;

`M = (-v)/(u)`

Therefore;

`M = (-(-11.8))/(65.0) = (59)/(325) \approx 0.18153846153`

The magnification, M ≈ 0.182

E. For a person who is near sighted, the light from objects from distant location are almost parallel and are brought to a focus in front of the retina and therefore, they appear blurred

The use of a concave lens is required to correct nearsightedness as it causes the light to appear come from a point before the focus such that the divergent rays of the object refracted through the concave lens comes to a focus at the.