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An object AB of size 2cm is placed at a distance of 20cm from a concave lens of focal length 20cm. The object is placed perpendicular to the principal axis with the end A on the principal axis.Taking 1cm of drawing to represent 2cm of actual length draw a neat diagram showing the paths of two rays from B one parallel to the principal axis and the other going through the optical centre of the lens. ​

Answer 1

Explanation:

An object AB of size 2cm is placed at a distance of 20cm from a concave lens of focal length 20cm. The object is placed perpendicular to the principal axis with the end A on the principal axis.Taking 1cm of drawing to represent 2cm of actual length draw a neat diagram showing the paths of two rays from B one parallel to the princip

Answer 2

Object's size (h1) = 2 cm

Focal length of convex lens (f) = 10 cm Object distance from the lens (u) = -15 cm Image distance (v) = ? Image size (h2) = ?

• We know,

• 1/v1/u= 1/f

• 1/v = 1/f + 1/u

• 1/v1/10 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 x 15 = 30x2

• m = -15 h2 = 60

• h2= 60 / -15.