Ed(height of 2.3 m) is 30 m away from the passenger side of a car. The radius of curvature of the mirror is -4m.
How far from the mirror is the image?
Is the image virtual or real?
How tall is the image
To find the distance from the mirror to the image, we can use the mirror equation:
1/f = 1/dₒ + 1/dᵢ
Where:
- f is the focal length of the mirror
- dₒ is the object distance
- dᵢ is the image distance
Given that the radius of curvature of the mirror is -4m, we can find the focal length (f) by dividing the radius of curvature by 2:
f = R/2
f = -4m/2
f = -2m
Next, we need to determine the object distance (dₒ). In this case, the object distance is the distance between Ed and the mirror, which is 30m.
dₒ = 30m
Now, we can solve for the image distance (dᵢ) using the mirror equation:
1/-2m = 1/30m + 1/dᵢ
Simplifying the equation:
-1/2m = 1/30m + 1/dᵢ
To make the calculation easier, let's find a common denominator:
-1/2m = (1 + 30m) / 30m + 1/dᵢ
Now, we can combine the fractions:
-1/2m = (30m + 1) / 30m + 1/dᵢ
To simplify further, let's get rid of the fractions by cross-multiplying:
-15m + 1/dᵢ = 30m + 1
Now, let's isolate the term with the image distance:
1/dᵢ = 30m -15m + 1
Simplifying:
1/dᵢ = 15m + 1
Now, let's invert both sides of the equation to solve for the image distance:
dᵢ = 1 / (15m + 1)
Plugging in the values:
dᵢ = 1 / (15 * 30 + 1)
dᵢ = 1 / (450 + 1)
dᵢ = 1 / 451
dᵢ ≈ 0.0022m
Therefore, the image is approximately 0.0022 meters away from the mirror.
Now, let's determine if the image is virtual or real. In this case, since the image distance (dᵢ) is positive, the image is real.
Lastly, to find the height of the image, we can use the magnification equation:
magnification (m) = -dᵢ / dₒ
Plugging in the values:
magnification (m) = -0.0022m / 30m
magnification (m) ≈ -0.000073
Since the magnification is negative, it means the image is inverted. To calculate the height of the image, we can use the magnification equation:
m = hᵢ / hₒ
Rearranging the equation to solve for the height of the image (hᵢ):
hᵢ = m * hₒ
Given that Ed's height is 2.3m, we can plug in the values:
hᵢ = -0.000073 * 2.3m
hᵢ ≈ -0.000168m
Therefore, the height of the image is approximately -0.000168 meters. Note that the negative sign indicates that the image is inverted.