Question

03. A 5 cm tall object is placed perpendicular to the axis of convex lens of Focal length 10 cm. the dist of the object from lens is 15cm find the Nature, position and size of image and Also find its magnification. ​

Answer 1

Answer:

Given:

'u' is object distance = -15

'v' is image distance  = ?

height of the object  =  5 cm

focal length = 10 cm

distance of object  =  - 15 cm      

Using lens formula

1 / f  = 1 / v  - 1 / u

transposing as  1 / f + 1 / u = 1 / v

1 / 10 + 1 / u = 1 / v

1 / 10  -  1 / 15  =    1 / 30

1 / 30 =   1 / v

v =  30 cm

magnification = h′  /  h  =  v /  u

h′= height of image  = ?

h = height of object = 5 cm -  given

v = 30 cm

u = -15 cm

magnification =   h′ /  5   =   30 / - 15

                     h′- size of image = - 10 cm

magnification  is  -2

ANS :

the Nature is virtual and erect

Position of image is front of lens

size of image is -10 cm

magnification is -2

HOPE IT HELPS U !!

Answer 2

Answer:

Given:

'u' is object distance = -15

'v' is image distance  = ?

height of the object  =  5 cm

focal length = 10 cm

distance of object  =  - 15 cm      

Using lens formula

1 / f  = 1 / v  - 1 / u

transposing as  1 / f + 1 / u = 1 / v

1 / 10 + 1 / u = 1 / v

1 / 10  -  1 / 15  =    1 / 30

1 / 30 =   1 / v

v =  30 cm

magnification = h′  /  h  =  v /  u

h′= height of image  = ?

h = height of object = 5 cm -  given

v = 30 cm

u = -15 cm

magnification =   h′ /  5   =   30 / - 15

                    h′- size of image = - 10 cm

magnification  is  -2

ANS :

the Nature is virtual and erect

Position of image is front of lens

size of image is -10 cm

magnification is -2

Explanation: