Question

A double convex lens made of glass of refractive index $1.5$ has both radii of curvature of $20cm$. An object $2cm$ high is placed at $10cm$ from the lens. Find the position, nature and size of the image

Answer 1

• Lens maker formula is as follows $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
• 
  

Put $R_1=20cm$ $R_2=-20cm$, and $n=1.5$ we will get $f=20cm$

$\text{Thin Lens Formula}$ is as follows $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$v$ = image distance , $u$ = object distance = $-10cm$ and $f=20cm$ we get $v=-20cm$

$\text{Nature of the image}$

The image is on the same side as the object because u and v both are negative. As the lens is convex so for being on the same side the image will be virtual and thus erect.

Size =magnification$\times$object size= $|\frac{-v}{u}|\times 2cm$ = $4cm$