Solve the following examples
An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?

According to the question,
Object distance (u) = -25 cm;
Height of object (Ho) = + 7 cm;
Focal length (f) = -15 cm;
Image distance = v cm;
By Mirror Formula:
![]()
1/v + 1/u = 1/f.
Putting values we get,
![]()
⇒ ![]()
⇒ ![]()
Taking the lcm, we get,
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ v = -37.5 cm
∴ the image distance = 37.5 cm (The negative sign indicates that the image is formed on the left side of the mirror)
Now, image distance (v) = -37.5 cm
Magnification = ![]()
Hi is the image height.
HO is the object height.
“v” is the image distance
“u” is the object distance
⇒ ![]()
⇒ ![]()
⇒ Hi = -10.5 cm.
Size of image is 10.5 cm which is negative hence image is real .
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