Answer:
a) q = -9.23 cm, b) h’= 0.577 mm
, c) image is right and virtual
Explanation:
This is an optical exercise, where the constructor equation should be used
1 / f = 1 / p + 1 / q
Where f is the focal length, p the distance to the object and q the distance to the image
A) The cocal distance is framed with the relationship
1 / f = (n₂-1) (1 /R₁ -1 /R₂)
In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave
1 / f = (1.60 -1) (1 /∞ - 1 / (-4))
1 / f = 0.6 / 4 = 0.15
f = 6.67 cm
We have the distance to the object p = 24.0 cm, let's calculate
1 / q = 1 / f - 1 / p
1 / q = 1 / 6.67 - 1/24
1 / q = 0.15 - 0.04167 = 0.10833
q = -9.23 cm
distance to the negative image is before the lens
B) the magnification of the lenses is given by
M = h ’/ h = - q / p
h’= - q / p h
h’= - (-9.23) / 24.0 0.150
h’= 0.05759 cm
h’= 0.577 mm
C) the object is after the focal length, therefore, the image is right and virtual
