Question

a. A light wave moves through glass (n = 1.5) at an angle of 15°. What angle will it have when it moves from the glass into water (n = 1.33)? (4 points)


b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (8 points)




c. An object is located 65 cm from a concave mirror with a focal length of 45 cm. What is the image distance? Is the image real or virtual? (6 points)

Answer in detail with the correct units and steps to solve. Will mark brainliest.

Answer 1

A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:

a. The angle that the light wave would have is $7.5^{o}$.

b. The type of mirror that can be used is a plane mirror.

c. The image distance is 146.3 cm.

ii. The image formed by the mirror is a real image.

a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.

Thus from Snell's law, we have:

refractive index, n, = $\frac{Sin i}{Sin r}$

where: i is the angle of incidence, and r is the refracted angle.

Now given that n = 1.5, and i = 15.

Then;

n = $\frac{Sin i}{Sin r}$

1.5 = $\frac{Sin 15}{Sin r}$

Sin r = $\frac{0.2588}{1.5}$

        = 0.17253

r = $Sin^{-1}$ 0.17253

  = 9.936

r ≅ $10^{o}$

Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;

n = $\frac{Sin i}{Sin r}$

1.33 = $\frac{Sin 10}{Sin r}$

Sin r = $\frac{0.17365}{1.33}$

       = 0.1306

r = $Sin^{-1}$ 0.1306

 = 7.504

r = $7.5^{o}$

Therefore, the angle that the light wave would have is $7.5^{o}$.

b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.

The type of mirror that can be used is a plane mirror.

The ray diagram is attached to this answer.

c. From the mirror formula;

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.

Given; u = 65 cm, and f = 45 cm, then:

$\frac{1}{45}$ = $\frac{1}{65}$ + $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{1}{45}$ -  $\frac{1}{65}$

$\frac{1}{v}$ = $\frac{20}{2925}$

v = $\frac{2925}{20}$

v  = 146.25 cm

The image distance is 146.3 cm.

ii. The image formed by the mirror is a real image.

For more clarifications on the image formed by a concave mirror, visit: https://brainly.com/question/27877616

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