Example 10.4A 2.0 cm tall object is placed perpendicular to the princi

1.	Example 10.4A 2.0 cm tall object is placed perpendicular to the principal axts of aconvex lens of focal length 10 cm. The distance of the object from thelens is 15 em. Find the nature, position and size of the image. Alsofind its magniflcation.
Answer» Object's size (h1) = 2 cmFocal length of convex lens (f) = 10 cmObject distance from the lens (u) = -15 cm Image distance ( v) = ?Image size (h2) = ? We know, 1/v - 1/u= 1/f 1/v = 1/f + 1/u 1/v = 1/10 - 1/15 1/v = 1/30 Thus, v = 30 cm Now, v = 30 cm, 'v' is positive, that means the image is formed on the right side of the lens. That's why it is REAL and INVERTED. Now, magnification =? Linear magnification (m) = Size of image / Size of object = h2 / h1 = v/u m = h2/2 = 30/-15 m= h2 × -15 = 30×2 m = -15 h2 = 60 h2 = 60 / -15 h2 = -4 cm Size of the image = -4 cmIt's in negative , that means that the image is INVERTED. Like my answer if you find it useful!

Example 10.4A 2.0 cm tall object is placed perpendicular to the principal axts of aconvex lens of focal length 10 cm. The distance of the object from thelens is 15 em. Find the nature, position and size of the image. Alsofind its magniflcation.

Answer»

Object's size (h1) = 2 cmFocal length of convex lens (f) = 10 cmObject distance from the lens (u) = -15 cm Image distance ( v) = ?Image size (h2) = ?

We know,

1/v - 1/u= 1/f

1/v = 1/f + 1/u

1/v = 1/10 - 1/15

1/v = 1/30

Thus, v = 30 cm

Now, v = 30 cm, 'v' is positive, that means the image is formed on the right side of the lens. That's why it is REAL and INVERTED.

Now, magnification =?

Linear magnification (m) = Size of image / Size of object = h2 / h1 = v/u

m = h2/2 = 30/-15

m= h2 × -15 = 30×2

m = -15 h2 = 60

h2 = 60 / -15

h2 = -4 cm

Size of the image = -4 cmIt's in negative , that means that the image is INVERTED.

Like my answer if you find it useful!

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