Question

A student has focused the image of a candle flame on a white screen using a concave mirror. The situation is as given below.
Length of the flame = 1.5 cm
Focal length of the mirror = 12 cm
Distance of flame from the mirror = 18 cm

If the flame is perpendicular to the principal axis of the mirror, then calculate the following.

a) distance of the image from the mirror
b) length of the image

if thr distance between the mirror and the flame is reduced to 10 cm, then what would be observed on the screen. Draw a ray diagram to justify your answer for this situation.

(Please give answer to this queation)

Answer 1

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.

Answer 2

Answer:

Explanation:

Given :-

h = + 1.5 cm

f = - 12 cm

u = - 18 cm

To Find :-

v = ? and h' = ?

$\sf By,\bf\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\sf\implies\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\sf\implies\dfrac{1}{v}=\dfrac{1}{(-12)}-\dfrac{1}{(-18)}$

$\sf\implies\dfrac{1}{v}=\dfrac{-3+2}{(36)}$

$\sf\implies\dfrac{1}{v}=\dfrac{-1}{36}$

$\bf\implies v=-36 \: cm$

(b) $\sf By, \bf h'=-\dfrac{v}{u}\times h$

$\sf\implies h'=\dfrac{-36 \: cm}{-18 \: cm}\times1.5$

$\bf\implies h'=-3 \: cm$

(ii) No,distinct image would be formed on the screen. In this case the image formed will be virtual.

(iii) Refer the Diagram provided in Attachment,