Question

A diverging lens has a focal length of ?12.0 cm.

Part 1 out of 2 (a) Find the image distance for objects placed these distances from the lens: 3.00 cm, 12.0 cm, 14.0 cm, 24.0 cm, and 33.0 cm. In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size (select E for enlarged or D for diminished) 12.0 14.0 24.0 33.0 p (cm) q (cm) m =-q/p Real or virtual | | (select) (select) # (select) # (select) # (select) # Orientation (select) (select) (select) (select) (select) Relative size (select) # (select) # (select) # (select) # (select) #

Real or virtual choices = real or virtual

Orientation= upright or inverted

Relative Size= D(diminished) or E(enlarged)

Find the image distance for objects placed these distances from the lens: 2.00 cm, 8.00 cm, 13.0 cm, 16.0 cm, and 17.0 cm. In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size (select E for enlarged or D for diminished).

Answer 1

For object placed at distance x = 2.0 cm
p = 2.0 cm
f = - 12.0 cm

We know,
1/f = 1/p + 1/q
-1/12.0 = 1/2.0 + 1/q
q = - 1.7 cm

M = -q/p
M = 1.7/3.0
M = 0.57

From above values it is clear,
Image is Virtual !!
Orientation is Upright !!
Image is Diminished !!

P.S in case of Diverging lens the last 3 properties will allways remain same !!