A concave mirror produce six times magnified image virtual of an object placed at 5cm in front in it. Where is the image formed ?
To determine where the image is formed in a concave mirror, we need to use the magnification formula:
Magnification (M) = - (Image Distance) / (Object Distance)
Given that the magnification (M) is 6 times magnified, we can substitute this value into the formula:
6 = - (Image Distance) / 5cm
To find the image distance, we can solve for it by rearranging the formula:
(Image Distance) = -6 * 5cm
(Image Distance) = -30cm
The negative sign indicates that the image formed by the concave mirror is virtual. Now we know that the image distance is -30cm.
To determine the position of the image, we need to consider the sign conventions used for mirror equation calculations. In this case, since the object is placed at 5cm in front of the mirror, the object distance is positive (+5cm).
Using the mirror equation:
(1 / Object Distance) + (1 / Image Distance) = (1 / Focal Length)
Substituting in the known values:
(1 / 5cm) + (1 / -30cm) = (1 / Focal Length)
Simplifying the equation:
(1 / 5cm) - (1 / 30cm) = (1 / Focal Length)
Multiplying both sides of the equation by 30cm * 5cm:
6 - 1 = (30cm * 5cm) / (Focal Length)
Simplifying further:
5 = 150cm / (Focal Length)
Rearranging the equation:
(Focal Length) = 150cm / 5
(Focal Length) = 30cm
Therefore, based on the calculations, the image is formed at a position of -30cm (behind the mirror), indicating that it is a virtual image.
A concave mirror produce six times magnified image virtual of an object placed at 5cm in front in it. Where is the image formed ?
If ‘ho’ is the height of the object and ‘hi’ is the height of the image, then the magnification M produced by a spherical mirror is given by
M=hi/ho.
The magnification M is also related to the object distance ‘do’ and the image distance ‘di’.
M= - di/do.
ho is positive, as the object is usually places above the principle axis;
hi > 0 - for virtual images,hi<0 - for real images
When the image is virtual, it is erect so the height of the image hi >0 which results M >0
M=di/do => di = M•do=6•5=30 cm