Convex mirror – problems and solutions

1. The focal length of a convex mirror is 10 cm and the object distance is 20 cm. Determine (a) the image distance (b) the magnification of image

Known :

The focal length (f) = -10 cm

The minus sign indicates that the focal point of convex mirror is virtual

The object distance (d_o) = 20 cm

Solution :

Formation of image by concave mirror :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/10 – 1/20 = -2/20 – 1/20 = -3/20

d_i = -20/3 = -6.7 cm

The minus sign indicates that the image is virtual.

The magnification of image :

m = – d_i / d_o = -(-6.7)/20 = 6.7/20 = 0.3

m = 0,3 time smaller than the object.

The plus sign indicates that the image is upward.

2. A 10-cm high object is placed in front of a convex mirror with focal length 20 cm. Determine the image height if the object distance is (a) 10 cm (b) 30 cm (c) 40 cm (d) 50 cm

Known :

The focal length of convex mirror (f) = -20 cm

The minus sign indicates that the focal point is virtual

The radius of curvature (r) = 2 f = 2(20) = 40 cm

The object height (h) = 10 cm

Solution :

a) the focal length (f) = -20 cm and the object distance (d_o) = 10 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/10 = -1/20 – 2/20 = -3/20

d_i = -20/3 = -6.7

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image (m) :

m = –d_i / d_o = -(-6.7)/10 = 6.7/10 = 0.67

The plus sign indicates that the image is upright.

The image is 0.67 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.67) = 6.7 cm

b) the focal length (f) = -20 cm and the object distance (d_o) = 30 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/30 = -3/60 – 2/60 = -5/60

d_i = -60/5 = -12

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image (m) :

m = –d_i / d_o = -(-12)/30 = 12/30 = 0.4

The plus sign indicates that the image is upright.

The image is 0,4 times smaller the object.

The height of image (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.4) = 4 cm

c) The focal length (f) = -20 cm and the object distance (d_o) = 40 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/40 = -2/40 – 1/40 = -3/40

d_i = -40/3 = -13.3

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

The magnification of image (m) :

m = – d_i / d_o = -(-13.3)/40 = 13.3/40 = 0.3

The plus sign indicates that the image is upright.

The image is 0.3 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.3) = 3 cm

d) The focal length (f) = -20 cm and the object distance (d_o) = 50 cm

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = -1/20 – 1/50 = -5/100 – 2/100 = -7/100

d_i = -100/7 = -14.3

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

The magnification of image (m) :

m = – d_i / d_o = -(-14.3)/50 = 14.3/50 = 0.3

The plus sign indicates that the image is upright.

The image is 0.3 smaller than the object.

The image height (h_i) :

m = h_i / h_o

h_i = h_o m = (10 cm)(0.3) = 3 cm

3. On object is 20 cm in front of convex mirror. If the image height is 1/5 times the object height, determine (a) the image length b) the focal length c) the properties of image

Known :

The object distance (d_o) = 20 cm

The image height (h_i) = 1/5 h = 0.2 h

The object height (h) = h

Solution :

a) the image distance (d_i)

Formula of the magnification of image :

m = h_i / h_o = 0.2h / h = 0.2

The plus sign indicates that the image is upright.

The image is 0.2 smaller than the object.

The image distance (d_i) :

–d_i = m d_o

d_i = – m d_o = -(0.2)(20 cm) = -4 cm

The minus sign indicates that the image is virtual or the image is behind the convex mirror.

b) The focal length (f)

The focal length (f) :

1/f = 1/d_o + 1/ d_i = 1/20 – 1/4 = 1/20 – 5/20 = -4/20

f = -20/4 = -5 cm

The minus sign indicates that the focal point is virtual.

c) The properties of image :

– Upright

– Smaller

– virtual

4. Light strikes a convex mirror parallel to the axis will be reflected ….

A. towards the focal point of the mirror

B. as from the focal point of the mirror

C. through the center of the curvature of the mirror

D. perpendicular to the mirror plane

Solution

The problem drew in the figure below.

The correct answer is B.

5. A biker sees the image of a motorcycle behind it 1/6 times its original size when the distance between the biker and motorcycle is 30 meters. Determine the radius of curvature of the rear view mirror…

A. 7.14 m

B. 8.57 m

C. 12.00 m

D. 24.00 m

Known :

Magnification of image (M) = 1/6 times

Object distance (d) = 30 meter

Wanted: The radius of curvature of the rear view mirror (R)

Solution :

Calculate the distance of the image (d’)

Since the magnification of image (M) and the object distance (s) has been known, the image distance can be known using the formula of image magnification :

A negative sign means the image is virtual. The image is 5 meters behind the convex mirror.

Calculate the focal length (f)

Since the object distance (d) and the image distance (d’), then the focal length can be calculated using the formula of the mirror :

The radius of curvature (R)

The radius of curvature of a convex mirror is twice the focal length of a convex mirror.
R = 2 f = 2 (6 meters) = 12 meters
The radius of curvature of the convex mirror is 12 meters.
The correct answer is C.

6. The convex mirror is chosen as the rearview mirror of a motorcycle because the properties of the image produced by the mirror are…

A. real, upright, minimized

B. real, upright, enlarged

C. virtual, upright, minimized

D. virtual, upright, enlarged

Solution :

Based on the two figure above can concluded that the properties of the image are virtual, upright, minimized.

The correct answer is C.

7. An object is 12 cm in front of a convex mirror with a radius of 6 cm. The properties of the image are…

A. real, inverted at a distance of 12 cm

B. real, upright at a distance of 4 cm

C. virtual, upright at a distance of 2.4 cm

D. virtual, inverted at a distance of 6 cm

Known :

Object’s distance (d) = 12 cm

Radius of the convex mirror (r) = 6 cm.

The focal length of the convex mirror (f) = 6 cm / 2 = -3 cm

The focal length of a convex mirror is signed negative because it is virtual. Virtual because it is not passed by light.

Wanted : Properties of image

Solution :

Distance of image (d’) :

1/d’ = 1/f – 1/d = -1/3 – 1/12 = -4/12 – 1/12 = -5/12

d’ = -12/5 = -2.4 cm

The image distance signed negative means image is virtual.

Magnification of image (m):

m = -d’ / d = -(-2.4) / 12 = 2.4 / 12 = 0.2 times

The magnification of image signed positive means that the image is upright and the magnification of image is 0.2 means the image size is smaller than the object size (reduced).

The correct answer is C.

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