A pen placed 13.0 cm from a concave spherical mirror produces a real image 13.5 cm from the mirror.
a) What is the focal length of the mirror?
Answer in units of cm.
b) Calculate the magnification of the image.
Assume the pen is placed 28.9 cm from the mirror.
c) What is the position of the new image?
(Answer with −1000 if the image does not exist.)
Answer in units of cm.
Assume the pen is placed 28.9 cm from the
mirror.
c) What is the position of the new image?
(Answer with −1000 if the image does not exist.)
Answer in units of cm.
e) Describe the new image.
1. virtual, inverted, larger
2. real, upright, smaller
3. real, inverted, larger
4. real, upright, larger
5. None of these
6. virtual, upright, smaller
7. real, inverted, smaller
8. virtual, inverted, smaller
9. virtual, upright, larger
a convex mirror has a focal length of
-0.90m. An object with a height of 2.5m from the mirror. Could u help me please!! :D
a) To find the focal length of the mirror, we can use the mirror equation:
1/f = 1/d_o + 1/d_i
Where f is the focal length, d_o is the object distance, and d_i is the image distance.
Given that the object distance (d_o) is 13.0 cm and the image distance (d_i) is 13.5 cm, we can plug in these values into the equation and solve for the focal length:
1/f = 1/13.0 + 1/13.5
Simplifying the equation, we get:
1/f = 0.0769 + 0.0741
1/f = 0.151
Dividing both sides of the equation by 1, we get:
f = 1 / 0.151
f ≈ 6.6238 cm
Therefore, the focal length of the mirror is approximately 6.6238 cm.
b) The magnification (M) of the image can be found using the magnification formula:
M = -d_i / d_o
Given that the image distance (d_i) is 13.5 cm and the object distance (d_o) is 13.0 cm, we can plug in these values and calculate the magnification:
M = -13.5 / 13.0
M ≈ -1.0385
Therefore, the magnification of the image is approximately -1.0385.
c) To find the position of the new image, we can use the mirror equation again, with the new object distance (d_o) of 28.9 cm:
1/f = 1/d_o + 1/d_i
Substituting the known values, we get:
1/6.6238 = 1/28.9 + 1/d_i
Simplifying the equation, we get:
0.151 = 0.0346 + 1/d_i
Subtracting 0.0346 from both sides of the equation, we get:
0.151 - 0.0346 = 1/d_i
0.1164 = 1/d_i
Taking the reciprocal of both sides, we get:
1/0.1164 = d_i
d_i ≈ 8.5944 cm
Therefore, the position of the new image is approximately 8.5944 cm.
e) To describe the new image, we need to consider its characteristics. From the given options:
1. virtual, inverted, larger
2. real, upright, smaller
3. real, inverted, larger
4. real, upright, larger
5. None of these
6. virtual, upright, smaller
7. real, inverted, smaller
8. virtual, inverted, smaller
9. virtual, upright, larger
From the mirror equation, we can determine the properties of the image. The given equation is for a concave mirror, which produces a real and inverted image when the object distance is greater than the focal length.
Since the object is placed beyond the focal point (28.9 cm > 6.6238 cm), we can conclude that the image is real and inverted.
Additionally, the magnification calculated in part b) is negative (-1.0385), indicating that the image is also smaller in size compared to the object.
Therefore, the description that matches the characteristics of the new image is option 7: real, inverted, smaller.