Answer :
To solve this problem, we will utilize the mirror formula and the magnification formula.
First, let's calculate the magnification (m), which is the ratio of the height of the image (h_i) to the height of the object (h_o), and it is also equal to the negative ratio of the image distance (v) to the object distance (u).
Given:
Object distance (u) = -40 cm (negative according to the sign convention, which states that distances measured against the direction of incident light are negative)
Image distance (v) = 80 cm (positive, indicating that the image is formed on the same side as the object, which is the case for a real image)
Height of the object (h_o) = 3 cm
Now let's use the magnification formula:
[tex]\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \][/tex]
Plugging in the known values:
[tex]\[ m = -\frac{80}{-40} = \frac{80}{40} = 2 \][/tex]
The magnification, m, is positive 2, which means the image is twice the size of the object.
Now let's find the height of the image (h_i) using the magnification and the object's height:
[tex]\[ h_i = m \cdot h_o \][/tex]
[tex]\[ h_i = 2 \cdot 3 \text{ cm} = 6 \text{ cm} \][/tex]
However, since magnification was determined using the mirror formula where the image distance was taken as positive for a real image, it implicitly means that the magnification is negative (since in the mirror formula we'd normally have a negative image distance for a real image). Therefore, the image height is actually -6 cm. The negative sign indicates that the image is inverted.
The nature of the image can be described as:
- Real (because it is formed on the same side of the mirror as the object)
- Inverted (because the magnification is negative, indicating the image is upside-down compared to the object)
- Enlarged (because the absolute value of magnification |m| is greater than 1)
The type of mirror that produces a real, inverted, and enlarged image when the object is placed beyond the focal point is a concave mirror. A convex mirror would not create a real image; it would only create virtual, diminished images irrespective of the object's position.
In summary, the image is 6 cm high and inverted. The type of mirror used is a concave mirror.
First, let's calculate the magnification (m), which is the ratio of the height of the image (h_i) to the height of the object (h_o), and it is also equal to the negative ratio of the image distance (v) to the object distance (u).
Given:
Object distance (u) = -40 cm (negative according to the sign convention, which states that distances measured against the direction of incident light are negative)
Image distance (v) = 80 cm (positive, indicating that the image is formed on the same side as the object, which is the case for a real image)
Height of the object (h_o) = 3 cm
Now let's use the magnification formula:
[tex]\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \][/tex]
Plugging in the known values:
[tex]\[ m = -\frac{80}{-40} = \frac{80}{40} = 2 \][/tex]
The magnification, m, is positive 2, which means the image is twice the size of the object.
Now let's find the height of the image (h_i) using the magnification and the object's height:
[tex]\[ h_i = m \cdot h_o \][/tex]
[tex]\[ h_i = 2 \cdot 3 \text{ cm} = 6 \text{ cm} \][/tex]
However, since magnification was determined using the mirror formula where the image distance was taken as positive for a real image, it implicitly means that the magnification is negative (since in the mirror formula we'd normally have a negative image distance for a real image). Therefore, the image height is actually -6 cm. The negative sign indicates that the image is inverted.
The nature of the image can be described as:
- Real (because it is formed on the same side of the mirror as the object)
- Inverted (because the magnification is negative, indicating the image is upside-down compared to the object)
- Enlarged (because the absolute value of magnification |m| is greater than 1)
The type of mirror that produces a real, inverted, and enlarged image when the object is placed beyond the focal point is a concave mirror. A convex mirror would not create a real image; it would only create virtual, diminished images irrespective of the object's position.
In summary, the image is 6 cm high and inverted. The type of mirror used is a concave mirror.
