Physics: Optics: Lenses, Mirrors, and Ray Diagrams
Using ray diagrams and equations to predict image properties
Physics: Optics: Lenses, Mirrors, and Ray Diagrams
Using ray diagrams and equations to predict image properties
Physics - Grade 9-12
- 1
A 3 cm tall object is placed beyond 2F in front of a converging lens. Describe the image location, orientation, size, and type.
The image forms between F and 2F on the opposite side of the lens. It is inverted, smaller than the object, and real. - 2
A candle is placed between the focal point F and the center of curvature C of a concave mirror. Describe the image location, orientation, size, and type.
For a concave mirror, an object between F and C produces an enlarged real image.
The image forms beyond C in front of the mirror. It is inverted, larger than the object, and real. - 3
A converging lens has a focal length of 10 cm. An object is placed 30 cm from the lens. Calculate the image distance and magnification.
First solve for 1/di, then take the reciprocal.
Using 1/f = 1/do + 1/di, 1/10 = 1/30 + 1/di, so di = 15 cm. The magnification is m = -di/do = -15/30 = -0.5, so the image is real, inverted, and half the object size. - 4
A diverging lens has a focal length of -12 cm. An object is placed 24 cm from the lens. Calculate the image distance and describe the image.
A diverging lens has a negative focal length.
Using 1/f = 1/do + 1/di, 1/-12 = 1/24 + 1/di, so di = -8 cm. The image is virtual because the image distance is negative, upright because the magnification is positive, and reduced because m = 1/3. - 5
Draw or describe the ray diagram for an object in front of a convex mirror. What type of image does a convex mirror always form?
A convex mirror always forms a virtual, upright, reduced image behind the mirror. The image appears between the mirror and the focal point. - 6
A concave mirror has a focal length of 15 cm. An object is placed 10 cm in front of the mirror. Find the image distance and state whether the image is real or virtual.
If the object is inside the focal length of a concave mirror, the image distance will be negative.
Using 1/f = 1/do + 1/di, 1/15 = 1/10 + 1/di, so di = -30 cm. The negative image distance means the image is virtual and forms behind the mirror. - 7
List the three principal rays commonly used to draw a ray diagram for a converging lens.
A ray parallel to the principal axis refracts through the far focal point. A ray through the near focal point refracts parallel to the principal axis. A ray through the center of the lens continues nearly straight. - 8
A lens forms an image that can be projected onto a screen. Is the image real or virtual? Explain how you know.
The image is real because only real images can be projected onto a screen. Real images are formed where light rays actually meet. - 9
An object is placed 2 m in front of a plane mirror. Where is the image located, and what are its size and orientation compared with the object?
The image is located 2 m behind the mirror. It is the same size as the object, upright, virtual, and laterally reversed. - 10
A light ray enters glass from air at an angle of 30 degrees to the normal. If the index of refraction of air is 1.00 and the index of refraction of glass is 1.50, find the angle of refraction.
Use sin 30 degrees = 0.5.
Using Snell's law, n1 sin theta1 = n2 sin theta2. Since 1.00 sin 30 degrees = 1.50 sin theta2, sin theta2 = 0.333, so theta2 is about 19.5 degrees. The ray bends toward the normal. - 11
A small insect is 5 cm from a converging lens with a focal length of 10 cm. Find the image distance and magnification. Describe why this lens can act as a magnifying glass.
A converging lens acts as a magnifying glass when the object is inside the focal length.
Using 1/f = 1/do + 1/di, 1/10 = 1/5 + 1/di, so di = -10 cm. The magnification is m = -di/do = 2, so the image is virtual, upright, and twice as large as the object. This is how a magnifying glass makes nearby objects appear larger. - 12
A student calculates an image distance of -18 cm for a convex mirror. Interpret the negative sign and describe the image type.
The negative image distance means the image forms behind the mirror and is virtual. For a convex mirror, the image is also upright and reduced. - 13
A converging lens projects a sharp image on a screen 30 cm from the lens. The object is 60 cm from the lens on the other side. Find the focal length and magnification.
Both object distance and real image distance are positive for a converging lens in this setup.
Using 1/f = 1/do + 1/di, 1/f = 1/60 + 1/30 = 3/60 = 1/20, so f = 20 cm. The magnification is m = -di/do = -30/60 = -0.5, so the image is inverted and half the object size. - 14
In a ray diagram for a diverging lens, a student draws a ray parallel to the principal axis and then refracts it through the far focal point. Identify the error and correct it.
For a diverging lens, refracted rays spread apart, and dashed back-traced rays show where they appear to come from.
The error is that a diverging lens does not send a parallel ray through the far focal point. The refracted ray should spread away from the principal axis as if it came from the near focal point on the object side of the lens. - 15
A 4 cm tall object is placed 25 cm in front of a plane mirror. State the image distance, image height, orientation, and image type.
A plane mirror produces an image the same distance behind the mirror as the object is in front of it.
The image is 25 cm behind the mirror and is 4 cm tall. It is upright, virtual, the same size as the object, and laterally reversed.