Practice Test on Optics

Name: KEY May 2, 2002

1. Finish the Ray Diagrams to show the location of the image. Show the image with correct orientation:

2. Draw the Ray Diagrams to show the location of the image. Show the image with correct orientation:

3. Red light has a wavelength of between 630-770 nM . The typical red light, then, would be around 700 nm or 700 X 10^-9 m . What is the typical frequency of red light? [Note the speed of light is 3.00 X 10 ⁸ m/s .

f = 4.29 X 10 ¹⁴ Hz

f*w=c

f = c/w = 3.00 X 10 ^{8 /} 700 X 10^-9 = 4.29 X 10 ¹⁴ Hz

4. The ideal radio antennae is ¼ the wavelength of the radio wave. What size antennae would be needed for the AM station 680 KHz (680 X 10³ Hz)?

d= 110m

f*w=c

w = c/f = 3.00 X 10 ^{8 /} 680 X 10³ = 441 m

d = ¼ w = ¼ * 441 = 110 m

5. What kind of instrument would have this lens arrangement? MICROSCOPE.

[NOTE: A telescope would have a convex lens and a concave lens!]

6. What do we know about the motion of a distant star if the top diagram shows the spectral lines of Helium and the bottom the spectral lines coming from that distant star?

a. _The star is moving away from us

b. What do we call this? RED SHIFT

7. An object 5cm tall is placed 60 cm in front of a 20 cm focal length converging (convex) lens. A. a. Based on the lens lab results, predict d_i and h_i where d_i is the distance of the image from the center of the lens and h_i is the height of the image.

Since do = 3f, di will be between 20-40 cm (pos) and hi will be less than 5 cm tall and inverted (neg) [i.e. d0 > 2F === 1F < di < 2F ]

b. Where is the image located? d_i = 30cm (real, note f < di < 2F )

1/f = 1/do + 1/di == solved == di = d_of / (d_o – f)

plug in: di = (60cm)(20cm)/ 60cm -20cm = 1200 cm² / 40 cm = 30 cm

c. What is the size of the image? h_i = -2.5 cm

hi/ho = -di/do solved: hi = -ho (di/do)

Plug in: hi = - (5cm) (30cm/60 cm) = -2.5 cm (inverted)

8. Now place the same object (from problem #7) only 12 cm from the same lens. Describe the new image. Given f = 20cm , do = 12 cm ho = 5 cm
1. d_i = -30cm
2. h_i = +12.5
3. Real or Virtual Image? VIRTUAL
4. Upright or Upside down? UPRIGHT

1/f = 1/do + 1/di == solved == di = d_of / (d_o – f)

plug in: di = (12cm)(20cm)/ (12cm -20cm) = - 30 cm

Note: Since di < f , we get a virtual image

For hi hi/ho = -di/do solved: hi = -ho (di/do)

Plug in: hi = - (5cm) (-30cm/12 cm) = +12.5 cm (This means the image is upright and larger!)

9. Consider the following Mirror:

Please note that “F” is the focus of the mirror and “C” is the center of curvature.

Will the image be real or virtual?

VIRTUAL

Because di < f

10.. A diverging (concave) lens has a focal length of 15.0 cm. A 5.0 cm tall object is placed 25.0 cm from the lens. Find the following properties of the image:

a. Magnification= 1.50 X

b. Size of image = 7.5 cm

c. Orientation of image = (same as object or inverse?) UPRIGHT (same as object)

d. Type of image = (virtual or real) VIRTUAL

Given: f= 15.0 cm h_o = 5.0 cm d_o = 25.0 cm

We must first find the location of the image, d_i.

1/f = 1/do + 1/di

solved: di = d_of / (d_o – f)

plugin: : di = (25.0 cm)(15.0 cm)/ (25.0 cm -15.0 cm ) = +37.5 cm

Magnification : M = For hi hi/ho = -di/do = (37.5cm) / (25.0) = 1.50 X (Magnification of Virtual Image)

solved: hi = -ho (di/do) = - (5.0 cm) * (37.5 cm) / (25.0 cm) = 7.5 cm

11.. An object is placed 20 cm in front of a diverging lens (concave lens). The image is ¾ the size of the object. What is the focal length of the lens?

___________________

Variables defined:

F = focal length

d_o = distance from object to center of lens

d_i = distance from image to center of lens

m = magnification of image

Equations Used:

Equation 1: 1/f = 1/d_o + 1/d_i

Equation 2: d_i = -md_o (How to obtain Equations 2: m = - d_i/d_o (or) d_i = -md_o )

Solving for focal length:

Substitute eqation 2 into equation 1 for di :

1/f = 1/d_o – 1/md_o = m/md_o – 1/md_o = (m-1)/md_o

Thus:

1/f = (m-1)/md_o

Invert:

f = mdo / (m – 1)

Plugin:

(+0.75) (20cm) / (0.75-1) = - 60 cm (NOTE: -f => diverging)