Question

Power of a lens is -4D. A virtual, erect image is obtained at a distance of 5 cm from it. Find the position of the object. If you can answer then only answer other wise I willl report​

Answer 1

Since,the power of the lens is -4D it must be a concave lens,therefore;

as we know:

=> 1/D=f

=> f= -(1/4)

=> f= -0.25m or 25cm

Now;

1/f= (1/v) - (1/u)

where u is object distance and v is image distance;

Therefore:

(1/25)= (1/5) -(1/u)

(1/5)-(1/25)= (1/u)

(5-1/25)= 1/u

4/25 = 1/u

u= (25/4)

u= 6.25

The object distance is 6.25cm or ~ 6cm. hope this helps.....