Physics Mirror Worksheet Please Help me With Question 2

Question

Please Help With Question 2, and could you also see if i did 1 a-e correctly, i can’t afford to fail this assignment

1. Find the distance to the image in the following curved mirror problems and state whether the image is virtual or real AND whether it is upright or inverted. (4 points each)

a. An object is placed 1.0 cm from a convex mirror with a focal length of 4.0 cm.

 

1/f = 1/so + 1/si

F= focal length of the convex mirror (4.0 cm)

so= object distance (1.0 cm)

si= image distance (unknown)

1/4.0=1/1.0+1/si   1/4.0 – 1/1.0 = 1/si    1/si = 1/4.0 – 1/1.0  si= 4.0 – 1.0 = 3.0 cm  

The distance to the image is 3.0 cm.

Since the image distance is positive, the image formed by a convex mirror is virtual. The image is also upright. 

 

 

b. An object is placed 12 cm from a convex mirror with a focal length of 5.0 cm.

 

1/f = 1/so + 1/si   

F= focal length of the convex mirror (5.0 cm)

so= object distance (12 cm)

si= image distance (unknown)

1/5.0 = 1/12 + 1/si   1/5.0 – 1/12 = 1/si   1/si = 1/5.0 – 1/12 si= 60/13 cm 

The distance to the image is approximately 4.62 cm.

Since the image distance is positive, the image formed by a convex mirror is virtual. The image is also upright. 

 

 

c. An object is placed 110 mm from a concave mirror with a focal length of 5.0 cm.

 

1/f = 1/so+1/si  f= focal length of the concave mirror (5.0 cm)  so= object distance (110mm = 11.0 cm)  si= image distance (unknown)

1/5.0=1/11.0+1/si  1/si=1/5.0-1/11.0   si= 55/3 cm  

The distance to the image is approximately 18.33 cm.

Since the image distance is positive, the image formed by a concave mirror is real. The image is also inverted. 

d. An object is placed 9.0 cm from a concave mirror with a focal length of 9.0 cm.

 

1/f=1/so+1/si   f= focal length of the concave mirror (9.0 cm)  so= object distance (9.0cm)   si= image distance (unknown)

1/9.0=1/9.0+1/si   1/9.0 – 1/9.0 = 1/si  0= 1/si   si= infinity 

The distance to the image is infinite.

Since the image distance is infinite, the image formed by a concave mirror is virtual. The image is also upright. 

 

 

e. An object is placed 1.5 cm from a concave mirror with a focal length of 3.0 cm.

 

1/f=1/so+1/si   f=focal length of the concave mirror (3.0 cm)  so= object distance (1.5 cm)  si= image distance (unknown)   

1/3.0=1/1.5+1/si   -1/3.0=1/si   si= -3.0cm

The distance to the image is -3.0cm.

Since the image distance is negative, the image formed by a concave mirror is virtual. The image is also upright. 

2. Choose three of the above mirror problems and draw a scale diagram for each one that shows geometrically how the distance to the image is found. (5 points each)

 

Be sure to: 

• Label so, si, and f clearly.

• Use a ruler or straight edge for neatness

• Draw to scale.

 

 

 

 

Answer 1

It is important to use the correct sign conventions for the mirror type.

For a spherical mirror, the focus is given a negative sign. The image sign calculated will be negative.

1. Find the distance to the image in the following curved mirror problems and state whether the image is virtual or real AND whether it is upright or inverted.

1/f = 1/so + 1/si In this case the numbers should be - 1/ 4.0 cm = 1 / 1.0 cm + 1/si

1/s_i = - 1/4 - 1/1

1/si = - 1/4 - 1 = -1 125 invert this to get

s_i = - 0.8 cm. The image is virtual (negative sign).

Magnification is defined as - s_i / s_o = - -0.8 / 1.0 = + 0.8. So, the image is upright (pos. sign) and reduced in size (|M| < 1). Note that s_i ≠ 4.0 -1.0.

For a concave mirror, the focus is positive, and the image sign can be either positive or negative, depending on the object distance If s_o < f, then the image is upright and virtual. If s_o > f and < R, the image is real, inverted and enlarged.

If s_o is > R, image is real, inverted, and reduced in size compared to the object.

c. An object is placed 110 mm from a concave mirror with a focal length of 5.0 cm. s_o > R where R = 2f.

I / 5.0 = 1 / 11.0 cm + 1 / s_i

1/ s_i = 1/5.0 - 1/11.0 = 0.109, image is real. (positive sign). Now invert 0.109.

s_i = 9.2 cm. Magnification = - 9.2 / 11.0 cm = - 0.83, and the image is inverted and reduced.

I have done two examples. Hopefully you can figure out how to do the rest.