Question

A 1-cm high object is placed 10 cm from a concave mirror with the focal length, f = 15 cm. Determine :

A. The image distance?

B. The image height?

C. The properties of image formed by the concave mirror?



​

Answer 1

f=-15 cm

Focal length of concave mirror is taken negative.

u=-10cm

Object distance is always negative

v=? (image distance)

1/f=1/u+1/v (Mirror Formula)

Putting values

1/-15=1/-10+1/v

Taking 1/-10 left side and it will become negative.

1/-15-(1/-10)=1/v

➖✖️➖=➕
1/-15+1/10=1/v

Taking LCM of 15 and 10 i.e. 60

-1*4/60 +1*6/60

-4/60+6/60

-4+6/60

2/60=1/30=0.03 cm

Image distance is 0.03cm

Magnification=image height/object height=-image distance/object distance

h'/h= -v/u

h'/1= -0.03/-10

1 gets multiplied when taken to right side and - - cancelled

h'=0.003

The image formed by the is real and inverted and diminished

Answer 2

Answer:

a) 1/v = 1/f - 1/u

= -1/15+1/10

= -2+3/30

v/1 (ie.V) = 30/1

therefore V = 30.

B) h(img) = hi

hi= -v/u * ho

= -30/-10* -1

= -3cm

u is less than f , so image placed between p and f .

c) image placed between p and f= image formed beyond mirror.

enlarged

virtual and erect.