Question

An object placed 50 centimetre from a lens produce a virtual image at a distance of 10 centimetre in front of the lens draw a diagram to show the formation of image calculate focal length of the lens and magnification produced Need diagram only.....

Answer 1

Answer:

→ focal length of lens = -12.5 cm

→ magnification produced by lens = 0.2

Explanation:

position of object , u = -50 cm

lens is forming virtual and erect image

position of image , v = -10 cm

Let, focal length of image = f

Using lens formula

→ 1 /f = 1/v - 1/u

→ 1/f = 1/(-10) - 1/(-50)

→ 1 / f = 1 / (-10) + ( 1 / 50 )

→ 1 / f = (-5 + 1 ) / 50

→ 1 / f = -4 / 50

→ f = -12.5 cm

Therefore,

Focal length of lens will be - 12.5 cm .

since focal length is negative therefore , Lens will be concave .

Using formula for magnification (m) of lens

→ m = v / u

→ m = -10 / -50

→ m = 0.2

Therefore,

Magnification produced by lens is 0.2

( positive magnification shows that image formed is virtual and erect )

▶Refer to the attachment for ray diagram