A passenger in a bus looks at the rear view convex mirror with a focal length of 15 cm. If a toy is hanging at a distance of 30 cm from the mirror, where will the image of the toy appear and what will be the nature of the image?+10 cm; virtual, erect and diminished+15 cm; virtual, erect and enlarged-10 cm; real, erect and diminished-15 cm; virtual, erect and enlarged
Question
A passenger in a bus looks at the rear view convex mirror with a focal length of 15 cm. If a toy is hanging at a distance of 30 cm from the mirror, where will the image of the toy appear and what will be the nature of the image?+10 cm; virtual, erect and diminished+15 cm; virtual, erect and enlarged-10 cm; real, erect and diminished-15 cm; virtual, erect and enlarged
Solution
To solve this problem, we can use the mirror formula which is 1/f = 1/v + 1/u. Here, f is the focal length of the mirror, v is the image distance, and u is the object distance.
Given: f = -15 cm (The focal length of a convex mirror is always taken as negative) u = -30 cm (The object distance is also negative as the object is on the same side as the light source)
Substituting these values into the mirror formula:
1/v = 1/f - 1/u 1/v = 1/(-15) - 1/(-30) 1/v = -2/30 v = -15 cm
The negative sign indicates that the image is formed on the same side as the object. This means the image is virtual and erect.
As for the size of the image, we can use the magnification formula which is m = -v/u.
m = -(-15)/-30 m = -0.5
The negative sign indicates that the image is erect. The value 0.5 (which is less than 1) indicates that the image is diminished.
So, the image of the toy will appear at -15 cm, and the nature of the image will be virtual, erect, and diminished. Therefore, the correct answer is "-15 cm; virtual, erect and diminished".
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