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 

${\displaystyle \frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$ 

⇒${\displaystyle \frac{1}{8}=\frac{1}{v}-\frac{1}{-3}}$ 

⇒${\displaystyle \frac{1}{8}=\frac{1}{v}+\frac{1}{3}}$ 

⇒${\displaystyle \frac{1}{8}-\frac{1}{3}=\frac{1}{v}}$ 

⇒${\displaystyle \frac{3-8}{24}=\frac{1}{v}}$ 

⇒`-5/24=1/

⇒${\displaystyle v=-\frac{24}{5}}$ 

⇒${\displaystyle v=-4.8}$cm 

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

Magnification ${\displaystyle \left(m\right)\frac{v}{u}}$ 

 ⇒  ${\displaystyle m=\frac{-4.8}{-3}}$ 

     ⇒m=1.6 

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

${\displaystyle m=\frac{h_i}{h_o}}$ 

⇒${\displaystyle 1.6=\frac{h_i}{3}}$ 

⇒${\displaystyle h_i=3\times 1.6=4.8}$ cm 

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