A convex lens has focal length equal to 8 cm. An object is placed at a

1.	A convex lens has focal length equal to 8 cm. An object is placed at a distance 12 cm from the lens. Calculate the image distance and image height if the height of the object is 24 cm . State the nature of the image
Answer» Answer : position of IMAGE is 24 cm at POSITIVE principal axis. height of image is 48 cm below principal axis. nature of image is real and inverted. Given : focal length of convex lens , f = 8 cm position of object , u = -12 cm height of the object , h₁ = 24 cm To find : position of image, v = ? height of image, h₂ = ? Nature of the image Formulae required : Lens formula 1/f = 1/v - 1/u Formula for MAGNIFICATION of lens m = v/u = h₂/h₁ [ where f is focal length of lens , v is position of image , u is position of object ,m is magnification, h₂ is height of image and h₁ is height of object ] Solution :- Using lens formula to calculate position of image → 1/f = 1/v - 1/u → 1/8 = 1/v - 1/(-12) → 1/v = 1/8 - 1/12 → 1/v = (3 - 2) / 24 → 1/v = 1 / 24 → v = 24 cm Therefore, image will be formed at 24 cm behind the lens. (i.e, at positive principal axis) Using formula for magnification of lens to find the height of image → m = v/u = h₂ / h₁ → v / u = h₂ / h₁ → 24 / (-12) = h₂ / 24 → -2 = h₂ / 24 → h₂ = - 48 cm Therefore, height of image will be 48 cm below the principal axis. And, Nature of image will be real and inverted.

A convex lens has focal length equal to 8 cm. An object is placed at a distance 12 cm from the lens. Calculate the image distance and image height if the height of the object is 24 cm . State the nature of the image

Answer»

Answer :

position of IMAGE is 24 cm at POSITIVE principal axis.
height of image is 48 cm below principal axis.
nature of image is real and inverted.

Given :

focal length of convex lens , f = 8 cm
position of object , u = -12 cm
height of the object , h₁ = 24 cm

To find :

position of image, v = ?
height of image, h₂ = ?
Nature of the image

Formulae required :

Lens formula

1/f = 1/v - 1/u

Formula for MAGNIFICATION of lens

m = v/u = h₂/h₁

[ where f is focal length of lens , v is position of image , u is position of object ,m is magnification, h₂ is height of image and h₁ is height of object ]

Solution :-

Using lens formula to calculate position of image

→ 1/f = 1/v - 1/u

→ 1/8 = 1/v - 1/(-12)

→ 1/v = 1/8 - 1/12

→ 1/v = (3 - 2) / 24

→ 1/v = 1 / 24

→ v = 24 cm

Therefore,

image will be formed at 24 cm behind the lens. (i.e, at positive principal axis)

Using formula for magnification of lens to find the height of image

→ m = v/u = h₂ / h₁

→ v / u = h₂ / h₁

→ 24 / (-12) = h₂ / 24

→ -2 = h₂ / 24

→ h₂ = - 48 cm

Therefore,

height of image will be 48 cm below the principal axis.

And,

Nature of image will be real and inverted.