Physics: Optics: Lenses, Mirrors, and Ray Diagrams Worksheet

Question 1

A 3 cm tall object is placed beyond 2F in front of a converging lens. Describe the image location, orientation, size, and type.

Accepted Answer

The image forms between F and 2F on the opposite side of the lens. It is inverted, smaller than the object, and real.

Question 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.

Accepted Answer

The image forms beyond C in front of the mirror. It is inverted, larger than the object, and real.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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?

Accepted Answer

A convex mirror always forms a virtual, upright, reduced image behind the mirror. The image appears between the mirror and the focal point.

Question 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.

Accepted Answer

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.

Question 7

List the three principal rays commonly used to draw a ray diagram for a converging lens.

Accepted Answer

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.

Question 8

A lens forms an image that can be projected onto a screen. Is the image real or virtual? Explain how you know.

Accepted Answer

The image is real because only real images can be projected onto a screen. Real images are formed where light rays actually meet.

Question 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?

Accepted Answer

The image is located 2 m behind the mirror. It is the same size as the object, upright, virtual, and laterally reversed.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 12

A student calculates an image distance of -18 cm for a convex mirror. Interpret the negative sign and describe the image type.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Physics: Optics: Lenses, Mirrors, and Ray Diagrams

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