A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal lengh 20 cm . the distance of the object from the lens is 15 cm . Find the nature, position and the size of the image ? And also tell why would you take focal length positive?
Dear student,
Height of object ho = 4 cm
Focal length of convex lens = 20 cm
Object distance u = -15 cm
Image distance v = ?
Image height hi =?
Lens formula: 1/v - 1/u = 1/f
1/v = 1/f + 1/u
= 1/20 + 1/(-15)
= -1/60
⇒ v = - 60 cm
The image distance is 60 cm on the same side of the lens as the object.
Magnification m = v/u = (-60)/(-15) = 4 = hi/ho = hi/4
⇒ hi = 4 x 4 = 16 cm
The image is enlarged and erect.
So a virtual, erect, enlarged image is formed at 60 cm from the optic center on the same side of the lens as the object is positioned.
The focal length of the convex lens lies to the right of the lens that is why it is taken as positive.
Regards
Height of object ho = 4 cm
Focal length of convex lens = 20 cm
Object distance u = -15 cm
Image distance v = ?
Image height hi =?
Lens formula: 1/v - 1/u = 1/f
1/v = 1/f + 1/u
= 1/20 + 1/(-15)
= -1/60
⇒ v = - 60 cm
The image distance is 60 cm on the same side of the lens as the object.
Magnification m = v/u = (-60)/(-15) = 4 = hi/ho = hi/4
⇒ hi = 4 x 4 = 16 cm
The image is enlarged and erect.
So a virtual, erect, enlarged image is formed at 60 cm from the optic center on the same side of the lens as the object is positioned.
The focal length of the convex lens lies to the right of the lens that is why it is taken as positive.
Regards
