Question

an object is kept at a distance of 100cm in front of convex lens of focal lenght 50cm calucate image distance and magnification​

Answer 1

Explanation:

Using lens formula,

1/f=1/v-1/u

⇒1/50=1/v-1/-100

⇒1/50=1/v+1/100

⇒1/50-1/100=1/v

⇒1/100=1/v

⇒v=100cm

since v is +ve, the image is real.

Magnification M= v/u

⇒M=100/-100

⇒M= -1

For -4D lens,

we know that power=1/focal length (in m)

-4D=1/f

f= -1/4D

f= -0.25m= -25cm (it's a concave lens)

u= -100cm (using sign conventions)

Using lens formula,

1/f=1/v-1/u

⇒ 1/-25=1/v-1/-100

⇒ -1/25=1/v+1/100

⇒ -1/25-1/100=1/v

⇒ -4-1/100=1/v

⇒ -5/100=1/v

⇒1/v= -1/20

⇒v= -20cm 

since v is -ve, image is virtual.

Magnification M= v/u

M=-20/-100

M=0.2