Question

A convex lens has focal length of 2.0cm .find its magnifying power if images is formed at DDV ​

Answer 1

Focal length of the objective lens (f1) = 2.0 cm

Focal length of the eyepiece (f2) = 6.25 cm

Distance between the objective lens and the eyepiece (d)= 15 cm

(a) Least distance of distinct vision, (d1) = 25 cm

Therefore, Image distance for the eyepiece (v2) = – 25 cm

Object distance for the eyepiece =  u2

According to the lens formula,

1/ v2 - 1/u2 = 1/ f2

1/u2 = 1/ v2 -1/ f2  = 1/-25 - 1/6.25 = -1-4/ 25 = -5/25

 u2 = -5 cm

Image distance for the objective lens, (v1) =- d + u2 = 15-5 = 10 cm

Object distance for the objective lens = u2

According to the lens formula,

1/ v1 - 1/u1 = 1/f1

1/u1 = 1/ v1 - 1/f1

        = 1/10 - 1/2 = 1-5/10 = -4/10

u1 = -2.5 cm

Magnitude of the object distance (u1) = 2.5 cm

The magnifying power of a compound microscope

m = v1/ u1 (1 + d1/f2)

    = 10/2.5 (1 + 25/6.25)

    = 4 (1 +4) = 20

Hence, the magnifying power of the microscope is 20.

Answer 2

here is your answer

P=1/f
so
P=1/2
P=0.5 D
therefore power of lens is 0.5D

hope it helps