In this problem a lens of focal length 15 cm is being used to magnify an object

Question

In this problem a lens of focal length 15 cm is being used to magnify an object 1 cm high. The object is 29.3 cm away from the viewer's eye and x1 = 7.9 cm from the lens. (So, x2 = 21.4 cm.) See the figure above.

1) Locate the image of the object formed by the lens. Give both the distance of the image from the lens and from the eye. Also give the height of the image. Distance from lens = ___ cm

2) Distance from eye = ___ cm

3) Height of Image = ___ cm

4) Find the angular magnification caused by the lens. Do this by finding two angles, the angle subtended by the object at the eye and the angle subtended at the eye by the image. Angle subtended by the object = ___ o

5) Angle subtended by the image = ___o

6) Angular magnification is defined as the ratio of the angle subtended by the image and the angle subtended by the object. Find the angular magnification

Angular magnification =

Explanation / Answer

given

f = 15 cm

object height, h = 1 cm

object distance, u = 7.9 m

let v is the image distance.

1) apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/15 - 1/7.9

v = -16.7 cm (in the same side as object)

distance from lense, |v| = 16.7 cm

2) distance from eye = 16.7 + 21.4 = 38.1 cm

3) magnification, m = -v/u

= -(-16.7)/7.9

= 2.11

Height of the image = m*object height

= 2.11*1

= 2.11 cm

4) angle sutended by the object = tan^-1( h/(x1+x2))

= tan^-1(1/(7.9+21.4)

= 1.95 degrees

5) angle subtended by the object = tan^-1( 2.11/(21.4 + 16.7))

= 3.17 degrees

6) angular magnification = 3.17/1.95

= 1.62

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