Question

If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image? 

Answer 1

 Object distance (u) =-3 

`1/f=1/v-1/u` 

⇒`1/8=1/v-1/-3` 

⇒`1/8=1/v+1/3` 

⇒`1/8-1/3=1/v` 

⇒`(3-8)/24=1/v` 

⇒`-5/24=1/

⇒`v=-24/5` 

⇒`v=-4.8`cm 

The image will be at a distance of 4.8 cm in front of the lens. 

Magnification `(m) v/u` 

 ⇒  `m=(-4.8)/-3` 

     ⇒m=1.6 

Positive value of magnifiction shows that the image is virtual and erect. 

`m=h_i/h_o` 

⇒`1.6=h_i/3` 

⇒`h_i=3xx1.6=4.8` cm 

positive sign of the image shows that the image will be formed above the principal axis.