an object 3cm high is placed prependicularto the principal axis of concave lens of focal length 15cm.the image is form at a distance of 10cm from the lens.find the distance at which object is placed and size and nature of image.?

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Solution

Focal length (f) = -20 cm
Here,
Object Distance (u) = ?
h_o = 3 cm
Image Distance, v = -10 cm
now according to lens formula,
1/f =1/v - 1/u
1/u = 1/v - 1/f
This gives,
u = -30 cm
Magnification,
m = h_i/h_o= v/u
m = 10/30 = h_i/h_o
h_i=1/3 x 3=1 cm
Image is virtual , erect and smaller.

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an object 3cm high is placed prependicularto the principal axis of concave lens of focal length 15cm.the image is form at a distance of 10cm from the lens.find the distance at which object is placed and size and nature of image.?

Focal length (f) = -20 cmHere,Object Distance (u) = ?ho = 3 cmImage Distance, v = -10 cmnow according to lens formula,1/f =1/v - 1/u1/u = 1/v - 1/fThis gives, u = -30 cmMagnification,m = hi/ho= v/u m = 10/30 = hi/hohi =1/3 x 3=1 cmImage is virtual , erect and smaller.

Focal length (f) = -20 cm
Here,
Object Distance (u) = ?
h_o = 3 cm
Image Distance, v = -10 cm
now according to lens formula,
1/f =1/v - 1/u
1/u = 1/v - 1/f
This gives,
u = -30 cm
Magnification,
m = h_i/h_o= v/u
m = 10/30 = h_i/h_o
h_i=1/3 x 3=1 cm
Image is virtual , erect and smaller.