1/p + 1/q = 1/f
Insert p for each and solve for q
magnification is -q/p and orientation is upright or invert based on mag
(a)
p = 9 cm
image distance cm
image orientation
magnification
(b)
p = 18 cm
image distance cm
image orientation
magnification
(c)
p = 31 cm
image distance cm
image orientation
magnification
Insert p for each and solve for q
magnification is -q/p and orientation is upright or invert based on mag
The mirror equation is given by:
1/f = 1/p + 1/q
where f is the focal length of the mirror, p is the object distance, and q is the image distance.
For a concave mirror, the focal length (f) is half the radius of curvature (R):
f = R/2
Given that the radius of curvature (R) is 18 cm, the focal length is:
f = 18 cm / 2 = 9 cm
Now let's solve each part of the question:
(a) For p = 9 cm:
Using the mirror equation:
1/9 = 1/9 + 1/q
Simplifying the equation:
1/q = 1/9 - 1/9
1/q = 0
Since 1/q = 0, the image distance (q) approaches infinity.
Therefore, the image is formed at infinity.
Since the image is formed at infinity, it is a virtual image.
Since the image is virtual, it will be upright.
The magnification (m) can be determined using the formula:
m = -q / p
Substituting the values:
m = -∞ / 9
The magnification is -∞ (negative infinity).
(b) For p = 18 cm:
Using the mirror equation:
1/9 = 1/18 + 1/q
Simplifying the equation:
1/q = 1/9 - 1/18
1/q = 1/18
Multiplying both sides by 18:
18/q = 1
q = 18 cm
The image distance (q) is 18 cm.
Since the image distance is positive, the image is real.
Using the magnification formula:
m = -q / p
Substituting the values:
m = -18 / 18
m = -1
The magnification is -1.
(c) For p = 31 cm:
Using the mirror equation:
1/9 = 1/31 + 1/q
Simplifying the equation:
1/q = 1/31 - 1/9
1/q = 3/279
Multiplying both sides by 279:
279/q = 3
q = 93 cm
The image distance (q) is 93 cm.
Since the image distance is positive, the image is real.
Using the magnification formula:
m = -q / p
Substituting the values:
m = -93 / 31
m = -3
The magnification is -3.
Summary:
(a) Image distance: Infinity
Image orientation: Virtual
Magnification: Negative infinity
(b) Image distance: 18 cm
Image orientation: Real
Magnification: -1
(c) Image distance: 93 cm
Image orientation: Real
Magnification: -3
1/p + 1/q = 1/f
where p is the object distance, q is the image distance, and f is the focal length.
Given that the radius of curvature (R) is 18 cm, the focal length (f) can be calculated using the formula:
f = R/2
Substituting the value of R = 18 cm, we get:
f = 18/2 = 9 cm
Using the mirror equation, we can find the image distance (q) for each object distance (p) given:
(a) p = 9 cm:
Substituting p = 9 cm and f = 9 cm in the mirror equation:
1/9 + 1/q = 1/9
Simplifying, we get:
1/q = 1/9 - 1/9 = 0
Since 1/q = 0, q = infinity.
The image distance is infinity, indicating that the image is formed at infinity.
The image orientation for a concave mirror is inverted (upside down).
To find the magnification (m), we can use the formula:
m = -q/p
Substituting q = infinity and p = 9 cm, we get:
m = -infinity
The magnification is negative infinity, indicating an infinitely magnified inverted image.
(b) p = 18 cm:
Using the same steps as above, substituting p = 18 cm and f = 9 cm in the mirror equation, we get:
1/18 + 1/q = 1/9
Simplifying, we have:
1/q = 1/9 - 1/18 = 1/18
q = 18 cm
The image distance is 18 cm.
The image orientation for a concave mirror is inverted (upside down).
To find the magnification (m), we can use the formula:
m = -q/p
Substituting q = 18 cm and p = 18 cm, we get:
m = -18/18 = -1
The magnification is -1, indicating a real inverted image with the same size as the object.
(c) p = 31 cm:
Using the same steps as above, substituting p = 31 cm and f = 9 cm in the mirror equation, we get:
1/31 + 1/q = 1/9
Simplifying, we have:
1/q = 1/9 - 1/31
Finding the least common denominator (LCD) of 9 and 31, we have:
1/q = (31 - 9)/(279)
Simplifying, we get:
1/q = 22/279
q = 279/22 ≈ 12.68 cm
The image distance is approximately 12.68 cm.
The image orientation for a concave mirror is inverted (upside down).
To find the magnification (m), we can use the formula:
m = -q/p
Substituting q = 12.68 cm and p = 31 cm, we get:
m = -12.68/31 ≈ -0.41
The magnification is approximately -0.41, indicating a reduced real inverted image.