Question

A student has focused the image of a candle flame on a white screen
using a concave mirror. The situation is given below
Length of the flame = 1.5 cm
Focal length of the mirror= 12 cm, distance of the flame from
the mirror = 18 cm. If the flame is perpendicular to principle axis of
the mirror then calculate image distance and focal length if the
distance between the mirror and flame reduced to 10 cm then what
would be observed on the screen. Draw a ray diagram to justify your
answer for this situation.​

Answer 1

Answer:Given height of the flame  h1=1.5cm

f=-12cm         u=18cm

Using the mirror formula

1/f=1/u+1/v

1/v=1/f-1/v

 1/v=1/-12 +1/18

1/v   =-18+12/18*12

1/v=-1/36

v=-36cm (image distance)

m=-v/u

m=-(-36)/18

m=-2

so the image formed is real inverted and enlarged

height of the image  m=h2/h1 (h1-- height of the object    h2-- height of the image)

   -2=h2/1.5

h2=3cm

 so the image height is 3cm and the negative sign indicates that the image is formed below the principal axis.

                           OR

Given,

ho=1.5cm

f=-12cm

u= -18cm

1/f=1/v+1/u

1/-12=1/v-1/18

1/18-1/12=1/v

2-3/36=1/v (LCM)

-1/36=1/v

v=-36cm.

m= -v/u

m=hi/ho

hi/ho=-v/u

hi/1.5=-(-36)/-18

hi/1.5=-2

hi=-2*1.5

hi=-3cm.

In the other answer they written+3cm which is wrong.

Answer 2

Answer:

please see the attachment