Optics Review #1 LCHS Dr.E

In a vacuum, all electromagnetic waves have the same (A) speed (B) phase (C) frequency (D) wavelength (A) speed

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 (D) speed

The diagram represents a light ray reflecting from a plane mirror 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° (A) 25°

(A) The frequency decreases and the speed increases. 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. (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 mirror f is - if the mirror is a convex mirror di 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 4. 0-cm tall light bulb is placed a distance of 8 A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance. 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.3 cm) + 1/di 0.0658 cm-1 = 0.120 cm-1 + 1/di -0.0547 cm-1 = 1/di 18.3 = di

A 4. 0-cm tall light bulb is placed a distance of 8 A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the image size. (18.3 = di) hi/ho = - di/do hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm) hi = - (4.0 cm) • (-18.2 cm)/(8.3 cm) hi = 8.8 cm

Determine the image distance for a 5. 00-cm tall object placed 45 Determine the image distance for a 5.00-cm tall object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm. di = 22.5 cm

Determine the image height for a 5. 00-cm tall object placed 45 Determine the image height for a 5.00-cm tall object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm. (di = 22.5 cm) hi = -2.5 cm

Determine the image distance for a 5. 00-cm tall object placed 30 Determine the image distance for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm. di = 30.0 cm

Determine the image height for a 5. 00-cm tall object placed 30 Determine the image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm. (di = 30.0 cm) hi = -5.0 cm

Determine the image distance for a 5. 00-cm tall object placed 20 Determine the image distance for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm. di = 60.0 cm

Determine the image height for a 5. 00-cm tall object placed 20 Determine the image height for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm. (di = 60.0 cm ) hi = -15.0 cm

Snell’s Law

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

1.33 • sin (60 degrees) = 2.42 • sin 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

air - flint glass: 18 degrees flint glass - water: 22 degrees The angle of incidence at first boundary is 30 degrees. Use the given n values and Snell's Law to calculate the θr values at each boundary. The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e.g., the θr at the air-flint glass boundary is the θi at the flint glass-water boundary. air - flint glass: 18 degrees flint glass - water: 22 degrees water - diamond: 12 degrees diamond - zirconium: 13 degrees cubic zirconium - air: 30 degrees

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

Calculate the critical angle for an 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 (1.58) – air boundary. Critical Angle = sin-1 (nout/nin)) CA = sin-1 (1.0 / 1.58) CA = 39.3 degrees

Lens

A 4. 00-cm tall light bulb is placed a distance of 45 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance. 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(45.7 cm) + 1/di 0.0658 cm-1 = 0.0219 cm-1 + 1/di 0.0439 cm-1 = 1/di 22.8 cm = di

A 4. 00-cm tall light bulb is placed a distance of 45 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image size. (22.8 cm = di) hi/ho = - di/do hi /(4.00 cm) = - (22.8 cm)/(45.7 cm) hi = - (4.00 cm) • (22.8 cm)/(45.7 cm) hi = -1.99 cm

A 4. 00-cm tall light bulb is placed a distance of 8 A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance. 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.30 cm) + 1/di 0.0658 cm-1 = 0.120 cm-1 + 1/di -0.0547 cm-1 = 1/di -18.3 cm = di

A 4. 00-cm tall light bulb is placed a distance of 8 A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image size. (di = -18.3 cm) hi/ho = - di/do hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm) hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm) hi = 8.81 cm

A 4. 00-cm tall light bulb is placed a distance of 8 A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance. 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(8.30 cm) + 1/di 0.0658 cm-1 = 0.120 cm-1 + 1/di -0.0547 cm-1 = 1/di -18.3 cm= di

A 4. 00-cm tall light bulb is placed a distance of 8 A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. Determine the image size. (-18.3 cm= di) hi/ho = - di/do hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm) hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm) hi = 8.81 cm