Question

an object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine position, size and nature of the image if the distance of the object from the lens is 10cm.

Answer 1

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LENS       :  A  transparent material.

LENS FORMULA         =   1/f = 1/v ₋ 1/u

MAGNIFICATION, m   = + v/u

TYPES OF SPHERICAL LENS : Convex lens and Concave lens

CONCAVE LENS :

• A lens bounded by two spherical surfaces, curved inwards is known as double concave lens ( or simply concave lens .

• It is also known as diverging lens because it diverges the light.

• Thinner at the middle and thicker at the edges.

• Used in Myopic eye defect correction.

SIGN CONVENTION FOR CONCAVE LENS :

1. OBJECT DISTANCE " u "      =    ₋ve

2. IMAGE DISTANCE   " v "       =    ₋ve

3. FOCAL LENGTH     " f "         =   ₋ve

4. HEIGHT OF THE OBJECT     =   + ve

5. HEIGHT OF THE IMAGE        =   +ve

→ Given in the question  :

Height of the object     =     6 cm

Focal length                  =  ₋ 5 cm

Object distance ( u )     =   ₋ 10cm

Now using Lens Formula :

       1/f           =     1/v ₋ 1/u

∴    1/₋5          =     1/v ₋ 1/₋10

or    1/v           =      1/₋5 ₋1/10 = ₋2₋1/10

      1/v           =      ₋3/10

       v            =       ₋ 10/3 cm

• Thus, the image is formed on the same side of the object at a distance of ₋10/3 cm from the optical centre of the lens.

• The negative sign indicates that the image is Virtual.

USING THE FORMULA,  m = height of the image / height of the object  = v/u

⇒                        Height of the image  = v/u × height of the object

                           =   ₋ 10 / 3 × ( ₋ 10 ) × 6

                           =     + 2cm

• THE POSITIVE SIGN INDICATES THAT THE IMAGE IS ERECT.