Optics Review #1

LCHSDr.E

In a vacuum, all electromagnetic waves have the same

(A) speed (B) phase (C) frequency(D) wavelength

When a light wave enters a new medium and is refracted, there must be a change in the light wave’s

(A) color (B) frequency (C) period (D) speed

The diagram represents a light ray reflecting from a plane mirror. The angle

of reflection for the light ray is

(A) 25° (B) 35° (C) 50° (D) 65°

What happens to the frequency and the speed of an electromagnetic wave as it passes from air into glass?

(A) The frequency decreases and the speed increases.(B) The frequency increases and the speed decreases. (C) The frequency remains the same and the speed increases.(D) The frequency remains the same and the speed decreases.

Which ray diagram best represents the phenomenon of refraction?

Mirror

The +/- Sign Conventions

f is + if the mirror is a concave mirrorf is - if the mirror is a convex mirrordi is + if the image is a real image and located on the object's

side of the mirror.di is - if the image is a virtual image and located behind the

mirror.hi is + if the image is an upright image (and therefore, also

virtual)hi is - if the image an inverted image (and therefore, also

real)

A A 4.0-cm tall light bulb is placed a distance 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a of 8.3 cm from a concave mirror having a

focal length of 15.2 cm. Determine the focal length of 15.2 cm. Determine the image distance.image distance.

1/f = 1/do + 1/d1/f = 1/do + 1/dii

1/(15.2 cm) = 1/(8.3 cm) + 1/d1/(15.2 cm) = 1/(8.3 cm) + 1/dii

0.0658 cm0.0658 cm-1-1 = 0.120 cm = 0.120 cm-1 -1 + 1/d+ 1/dii

-0.0547 cm-0.0547 cm-1-1 = 1/d = 1/dii

18.3 18.3 = d= dii

A 4.0-cm tall light bulb is placed a A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave distance of 8.3 cm from a concave

mirror having a focal length of 15.2 mirror having a focal length of 15.2 cm. Determine the image size. cm. Determine the image size.

((18.3 = 18.3 = di)di)

hhii/h/hoo = - d = - dii/d/doo

hhi i /(4.0 cm) = - (-18.2 cm)/(8.3 cm)/(4.0 cm) = - (-18.2 cm)/(8.3 cm)

hhi i = - (4.0 cm) • (-18.2 cm)/(8.3 cm)= - (4.0 cm) • (-18.2 cm)/(8.3 cm)

hhi i = 8.8 cm= 8.8 cm

Determine the image distance for a Determine the image distance for a 5.00-cm tall object placed 45.0 cm 5.00-cm tall object placed 45.0 cm

from a concave mirror having a focal from a concave mirror having a focal length of 15.0 cm. length of 15.0 cm.

ddi i = 22.5 cm= 22.5 cm

Determine the image height for a 5.00-Determine the image height for a 5.00-cm tall object placed 45.0 cm from a cm tall object placed 45.0 cm from a

concave mirror having a focal length of concave mirror having a focal length of 15.0 cm. 15.0 cm. (d(di i = 22.5 cm)= 22.5 cm)

hhi i = -2.5 cm= -2.5 cm

Determine the image distance for a 5.00-Determine the image distance for a 5.00-cm tall object placed 30.0 cm from a cm tall object placed 30.0 cm from a

concave mirror having a focal length of concave mirror having a focal length of 15.0 cm.15.0 cm.

ddii = 30.0 cm = 30.0 cm

Determine the image height for a 5.00-Determine the image height for a 5.00-cm tall object placed 30.0 cm from a cm tall object placed 30.0 cm from a

concave mirror having a focal length of concave mirror having a focal length of 15.0 cm.15.0 cm. (d(dii = 30.0 cm) = 30.0 cm)

hhii = -5.0 cm = -5.0 cm

Determine the image distance for a 5.00-Determine the image distance for a 5.00-cm tall object placed 20.0 cm from a cm tall object placed 20.0 cm from a

concave mirror having a focal length of concave mirror having a focal length of 15.0 cm. 15.0 cm.

ddii = 60.0 cm = 60.0 cm

Determine the image height for a 5.00-Determine the image height for a 5.00-cm tall object placed 20.0 cm from a cm tall object placed 20.0 cm from a

concave mirror having a concave mirror having a focal length of focal length of 15.0 cm. (d15.0 cm. (dii = 60.0 cm ) = 60.0 cm )

hhii = -15.0 cm = -15.0 cm

Snell’s Law

Determine the angle of refraction if the angle of incidence is 45 degrees.

Substitute into Snell's law equation and perform the necessary algebraic operations to

solve:1.52 • sin(45 degrees) = 1.33 • sin (θr)

1.075 = 1.33* sin (θr) 0.8081 = sin (θr) 53.9 degrees = θr

Determine the angle of refraction if the angle of incidence is 60 degrees.

1.33 • sin (60 degrees) = 2.42 • sin θr

1.152 = 2.42 • sin θr

0.4760 = sin θr

28.4 degrees = θr

The angle of incidence at first The angle of incidence at first boundary is 30 degrees. Use the given boundary is 30 degrees. Use the given n values and Snell's Law to calculate n values and Snell's Law to calculate the the θθr values at each boundary. The r values at each boundary. The angle of refraction at one boundary angle of refraction at one boundary

becomes the angle of incidence at the becomes the angle of incidence at the next boundary; e.g., the next boundary; e.g., the θθrr at the air- at the air-

flint glass boundary is the flint glass boundary is the θθii at the flint at the flint

glass-water boundary. glass-water boundary.

air - flint glass: 18 degrees air - flint glass: 18 degrees flint glass - water: 22 degreesflint glass - water: 22 degreeswater - diamond: 12 degreeswater - diamond: 12 degrees

diamond - zirconium: 13 degreesdiamond - zirconium: 13 degreescubic zirconium - air: 30 degreescubic zirconium - air: 30 degrees

Critical Angle

Critical Angle = sin-1 (nout/nin)

Calculate the critical angle for an Calculate the critical angle for an ethanol (1.36) – air (1) boundary. ethanol (1.36) – air (1) boundary.

Critical Angle = sin-1 (nout/nin)Critical Angle = sin-1 (1.0 / 1.36)

Critical Angle = 47.3 degrees

Calculate the critical angle for a flint glass Calculate the critical angle for a flint glass (1.58) – air boundary. (1.58) – air boundary.

Critical Angle = sinCritical Angle = sin-1-1 (n (noutout/n/ninin))))CA = sinCA = sin-1 -1 (1.0 / 1.58)(1.0 / 1.58)

CA = CA = 39.3 degrees39.3 degrees

Lens

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex distance of 45.7 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the image distance. Determine the image distance.

1/f = 1/do + 1/d1/f = 1/do + 1/dii

1/(15.2 cm) = 1/(45.7 cm) + 1/d1/(15.2 cm) = 1/(45.7 cm) + 1/dii

0.0658 cm0.0658 cm-1-1 = 0.0219 cm = 0.0219 cm-1 -1 + 1/d+ 1/dii

0.0439 cm0.0439 cm-1-1 = 1/d = 1/dii

22.8 cm = d22.8 cm = dii

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex distance of 45.7 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the image size. (Determine the image size. (22.8 cm = di)22.8 cm = di)

hhii/h/hoo = - d = - dii/d/doo

hhi i /(4.00 cm) = - (22.8 cm)/(45.7 cm)/(4.00 cm) = - (22.8 cm)/(45.7 cm)

hhi i = - (4.00 cm) • (22.8 cm)/(45.7 cm)= - (4.00 cm) • (22.8 cm)/(45.7 cm)

hhii = -1.99 cm= -1.99 cm

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex distance of 8.30 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the image distance.Determine the image distance.

1/f = 1/do + 1/d1/f = 1/do + 1/dii

1/(15.2 cm) = 1/(8.30 cm) + 1/d1/(15.2 cm) = 1/(8.30 cm) + 1/dii

0.0658 cm0.0658 cm-1-1 = 0.120 cm = 0.120 cm-1 -1 + 1/d+ 1/dii

-0.0547 cm-0.0547 cm-1-1 = 1/d = 1/dii

-18.3 cm = -18.3 cm = ddii

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex distance of 8.30 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the Determine the image size. image size.

((ddi i = -18.3 cm)= -18.3 cm)

hhii/h/hoo = - d = - dii/d/doo

hhi i /(4.00 cm) = - (-18.3 cm)/(8.30 cm)/(4.00 cm) = - (-18.3 cm)/(8.30 cm)

hhi i = - (4.00 cm) • (-18.3 cm)/(8.30 cm)= - (4.00 cm) • (-18.3 cm)/(8.30 cm)

hhii = = 8.81 cm8.81 cm

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex distance of 8.30 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the image distance.Determine the image distance.

1/f = 1/do + 1/d1/f = 1/do + 1/dii

1/(15.2 cm) = 1/(8.30 cm) + 1/d1/(15.2 cm) = 1/(8.30 cm) + 1/dii

0.0658 cm0.0658 cm-1-1 = 0.120 cm = 0.120 cm-1 -1 + 1/d+ 1/dii

-0.0547 cm-0.0547 cm-1-1 = 1/d = 1/dii

-18.3 cm-18.3 cm= d= dii

A 4.00-cm tall light bulb is placed a A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex distance of 8.30 cm from a double convex

lens having a focal length of 15.2 cm. lens having a focal length of 15.2 cm. Determine the Determine the image size.image size.

(-18.3 cm(-18.3 cm= d= dii))

hhii/h/hoo = - d = - dii/d/doo

hhi i /(4.00 cm) = - (-18.3 cm)/(8.30 cm)/(4.00 cm) = - (-18.3 cm)/(8.30 cm)

hhi i = - (4.00 cm) • (-18.3 cm)/(8.30 cm)= - (4.00 cm) • (-18.3 cm)/(8.30 cm)

hhi i = = 8.81 cm8.81 cm