Question

1. an object 5cm high forms a virtual image of 1.25cm high, when placed in front of aconvex mirror at a distance of 24 cm.calculate:a.the position of the image b.the focal length of the mirror
2. an object forms a virtual image 1/8 the of the size of the object. if the object is placed at a distance of 40cm from the convex mirror, calculate:
a.the position of the image
b.the focal length of the convex mirror

Answer 1

1.

Height Of Object(h) = 5cm.

Height Of Image(h') = 1.25cm.(Positive because image is virtual and therefore erect)

Object Distance(u) = -24cm.

M=h'/h = -v/u

1.25/5 = -v/-24

0.25 = v/24

v = 6 cm.

By Mirror Formula,

1/f = 1/v + 1/u

1/f = 1/6 + 1/-24

1/f = 1/6 - 1/24

1/f = 4-1/24

1/f = 3/24

f = 24/3 = 8cm.

a. Position Of Image= 6cm from mirror (Between Pole and Principal Focus)

b. Focal Length = 8cm.

2.

h' = h/8.

Therefore, h = 8h'.

u = -40cm.

M = h'/h = -v/u

h'/8h' = -v/-40

1/8 = v/40

v = 40/8 = 5cm.

By Mirror Formula,

1/f = 1/v + 1/f

1/f = 1/5 + 1/-40

1/f = 1/5 - 1/40

1/f = 8-1/40

1/f = 7/40

f = 40/7 = 5.71cm.

a. Position Of Image: 5cm from mirror (Between Pole and Principal Focus)

b. Focal Length = 5.71cm.