Question

an object 5cm in size is placed at 20cm infront of a concave mirror of focal length 15cm at what distance from mirror would a screen be placed in order to obtain a sharp image?find the nature and size of image ​

Answer 1

Answer:

Height of image (h2) =

Height of object (h1) = 5cm

Focal length of mirror(f)=-15cm

Object distance (u)=--20cm

Image distance(v)=

Now we will apply mirror formulae

1/f=1/v+1/u

-1/15=1/v-1/20

shifting -1/20 to r.h.s

1/20-1/15=1/v

1/v=-1/60

v=-60

∴ image will be formed at 60 cm in front of mirror

Now, we will calculate magnification.

Since we have two formulas for magnification w will use both the formulas to calculate height of the image.

magnification(m)=-v/u=-(-60/-20)=-3

magnification(m)=h2/h1

-3=h2/5

-15=h2

h2=-15

∴ size of image will be 12cm in size/height and since size is in negative image will be formed below principal axis.

Explanation:

Answer 2

Explanation:

Height of image (h2) =?

Height of object (h1) = 5cm

Focal length of mirror(f)=-15cm

Object distance (u)=--20cm

Image distance(v)=?

Now we will apply mirror formulae

1/f=1/v+1/u

-1/15=1/v-1/20

shifting -1/20 to r.h.s

1/20-1/15=1/v

1/v=-1/60

v=-60  

Therefore image will be formed at 60 cm in front of mirror

Now, we will calculate magnification.  

Since we have two formulas for magnification w will use both the formulas to calculate height of the image.  

magnification(m)=-v/u=-(-60/-20)=-3

magnification(m)=h2/h1

                   -3=h2/5

                    -15=h2

                     h2=-15  

Therefore size of image will be 12cm in size/height and since size is in negative image will be formed below principal axis.